Chapter-09 Differential Equations

He who seeks for methods without having a definite problem in mind seeks for the most part in vain. - D. HILBERT

9.1 Introduction

In Class XI and in Chapter 5 of the present book, we discussed how to differentiate a given function $f$ with respect to an independent variable, i.e., how to find $f^{\prime}(x)$ for a given function $f$ at each $x$ in its domain of definition. Further, in the chapter on Integral Calculus, we discussed how to find a function $f$ whose derivative is the function $g$, which may also be formulated as follows:

For a given function $g$, find a function $f$ such that

$$ \frac{d y}{d x}=g(x) \text { where } y=f(x) $$

Henri Poincare $(1854-1912)$

An equation of the form (1) is known as a differential equation. A formal definition will be given later.

These equations arise in a variety of applications, may it be in Physics, Chemistry, Biology, Anthropology, Geology, Economics etc. Hence, an indepth study of differential equations has assumed prime importance in all modern scientific investigations.

In this chapter, we will study some basic concepts related to differential equation, general and particular solutions of a differential equation, formation of differential equations, some methods to solve a first order - first degree differential equation and some applications of differential equations in different areas.

9.2 Basic Concepts

We are already familiar with the equations of the type:

$$ \begin{align*} x^{2}-3 x+3=0 \tag{1} \\ \sin x+\cos x=0 \tag{2} \\ x+y=7 \tag{3} \end{align*} $$

Let us consider the equation:

$$ \begin{equation*} x \frac{d y}{d x}+y=0 \tag{4} \end{equation*} $$

We see that equations (1), (2) and (3) involve independent and/or dependent variable (variables) only but equation (4) involves variables as well as derivative of the dependent variable $y$ with respect to the independent variable $x$. Such an equation is called a differential equation.

In general, an equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation.

A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation, e.g.,

$ 2 \frac{d^{2} y}{d x^{2}}+(\frac{d y}{d x})^{3}=0 \text{ is an ordinary differential equation } $

Of course, there are differential equations involving derivatives with respect to more than one independent variables, called partial differential equations but at this stage we shall confine ourselves to the study of ordinary differential equations only. Now onward, we will use the term ‘differential equation’ for ‘ordinary differential equation’.

Note

1. We shall prefer to use the following notations for derivatives:

$$ \frac{d y}{d x}=y^{\prime}, \frac{d^{2} y}{d x^{2}}=y^{\prime \prime}, \frac{d^{3} y}{d x^{3}}=y^{\prime \prime \prime} $$

2. For derivatives of higher order, it will be inconvenient to use so many dashes as supersuffix therefore, we use the notation $y_n$ for $n$th order derivative $\frac{d^{n} y}{d x^{n}}$.

9.2.1 Order of a differential equation

Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.

Consider the following differential equations:

$$ \begin{align*} & \frac{d y}{d x}=e^{x} \tag{6}\\ & \frac{d^{2} y}{d x^{2}}+y=0 \tag{7}\\ & \frac{d^{3} y}{d x^{3}}+x^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=0 \tag{8} \end{align*} $$

The equations (6), (7) and (8) involve the highest derivative of first, second and third order respectively. Therefore, the order of these equations are 1,2 and 3 respectively.

9.2.2 Degree of a differential equation

To study the degree of a differential equation, the key point is that the differential equation must be a polynomial equation in derivatives, i.e., $y^{\prime}, y^{\prime \prime}, y^{\prime \prime \prime}$ etc. Consider the following differential equations:

$ \begin{aligned} \frac{d^{3} y}{d x^{3}}+2(\frac{d^{2} y}{d x^{2}})^{2}-\frac{d y}{d x}+y & =0 \\ (\frac{d y}{d x})^{2}+(\frac{d y}{d x})-\sin ^{2} y & =0 \\ \frac{d y}{d x}+\sin (\frac{d y}{d x}) & =0 \end{aligned} $

We observe that equation (9) is a polynomial equation in $y^{\prime \prime \prime}, y^{\prime \prime}$ and $y^{\prime}$, equation (10) is a polynomial equation in $y^{\prime}$ (not a polynomial in $y$ though). Degree of such differential equations can be defined. But equation (11) is not a polynomial equation in $y^{\prime}$ and degree of such a differential equation can not be defined.

By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation.

In view of the above definition, one may observe that differential equations (6), (7), (8) and (9) each are of degree one, equation (10) is of degree two while the degree of differential equation (11) is not defined.

Note Order and degree (if defined) of a differential equation are always positive integers.

Example 1 Find the order and degree, if defined, of each of the following differential equations:

(i) $\frac{d y}{d x}-\cos x=0$

(ii) $x y \frac{d^{2} y}{d x^{2}}+x(\frac{d y}{d x})^{2}-y \frac{d y}{d x}=0$

(iii) $y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0$

Solution

(i) The highest order derivative present in the differential equation is $\frac{d y}{d x}$, so its order is one. It is a polynomial equation in $y^{\prime}$ and the highest power raised to $\frac{d y}{d x}$ is one, so its degree is one.

(ii) The highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$, so its order is two. It is a polynomial equation in $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ and the highest power raised to $\frac{d^{2} y}{d x^{2}}$ is one, so its degree is one.

(iii) The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$, so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

EXERCISE 9.1

Determine order and degree (if defined) of differential equations given in Exercises 1 to 10 .

1. $\frac{d^{4} y}{d x^{4}}+\sin (y^{\prime \prime \prime})=0$

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Solution

$\Rightarrow y^{\prime \prime \prime \prime}+\sin (y^{\prime \prime \prime})=0$

The highest order derivative present in the differential equation is $y^{\prime \prime \prime \prime}$. Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

2. $y^{\prime}+5 y=0$

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Solution

The given differential equation is:

$y^{\prime}+5 y=0$

The highest order derivative present in the differential equation is $y^{\prime}$. Therefore, its order is one.

It is a polynomial equation in $y^{\prime}$. The highest power raised to $y^{\prime}$ is 1 . Hence, its degree is one.

3. $(\frac{d s}{d t})^{4}+3 s \frac{d^{2} s}{d t^{2}}=0$

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Solution

$(\frac{d s}{d t})^{4}+3 \frac{d^{2} s}{d t^{2}}=0$

The highest order derivative present in the given differential equation is $\frac{d^{2} s}{d t^{2}}$. Therefore, its order is two.

It is a polynomial equation in $\frac{d^{2} s}{d t^{2}}$ and $\frac{d s}{d t}$. The power raised to $\frac{d^{2} s}{d t^{2}}$ is 1 . Hence, its degree is one.

4. $(\frac{d^{2} y}{d x^{2}})^{2}+\cos (\frac{d y}{d x})=0$

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Solution

$(\frac{d^{2} y}{d x^{2}})^{2}+\cos (\frac{d y}{d x})=0$

The highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$. Therefore, its order is 2.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

5. $\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x$

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Solution

$\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x$

$\Rightarrow \frac{d^{2} y}{d x^{2}}-\cos 3 x-\sin 3 x=0$

The highest order derivative present in the differential equation is $\frac{d^{2} y}{d x^{2}}$. Therefore, its order is two.

It is a polynomial equation in $\frac{d^{2} y}{d x^{2}}$ and the power raised to $\frac{d^{2} y}{d x^{2}}$ is 1 .

Hence, its degree is one.

6. $(y^{\prime \prime \prime})^{2}+(y^{\prime \prime})^{3}+(y^{\prime})^{4}+y^{5}=0$

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Solution

$(y^{\prime \prime \prime})^{2}+(y^{\prime \prime})^{3}+(y^{\prime})+y^{5}=0$

The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$. Therefore, its order is three.

The given differential equation is a polynomial equation in $y^{\prime \prime \prime}, y^{\prime \prime}$, and $y^{\prime}$.

The highest power raised to $y^{\prime \prime \prime}$ is 2 . Hence, its degree is 2 .

7. $y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$

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Solution

$y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$

The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$. Therefore, its order is three.

It is a polynomial equation in $y^{\prime \prime \prime}, y^{\prime \prime}$ and $y^{\prime}$. The highest power raised to $y^{\prime \prime \prime}$ is 1 . Hence, its degree is 1 .

8. $y^{\prime}+y=e^{x}$

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Solution

$ \begin{aligned} & y^{\prime}+y=e^{x} \\ & \Rightarrow y^{\prime}+y-e^{x}=0 \end{aligned} $

The highest order derivative present in the differential equation is $y^{\prime}$. Therefore, its order is one.

The given differential equation is a polynomial equation in $y^{\prime}$ and the highest power raised to $y^{\prime}$ is one. Hence, its degree is one.

9. $y^{\prime \prime}+(y^{\prime})^{2}+2 y=0$

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Solution

$y^{\prime \prime}+(y^{\prime})^{2}+2 y=0$

The highest order derivative present in the differential equation is $y^{\prime \prime}$. Therefore, its order is two.

The given differential equation is a polynomial equation in $y^{\prime \prime}$ and $y^{\prime}$ and the highest power raised to $y^{\prime \prime}$ is one.

Hence, its degree is one.

10. $y^{\prime \prime}+2 y^{\prime}+\sin y=0$

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Solution

$y^{\prime \prime}+2 y^{\prime}+\sin y=0$

The highest order derivative present in the differential equation is $y^{\prime \prime}$. Therefore, its order is two.

This is a polynomial equation in $y^{\prime \prime}$ and $y^{\prime}$ and the highest power raised to $y^{\prime \prime}$ is one. Hence, its degree is one.

11. The degree of the differential equation

$\quad\quad$ $ (\frac{d^{2} y}{d x^{2}})^{3}+(\frac{d y}{d x})^{2}+\sin (\frac{d y}{d x})+1=0 \text{ is } $

$\quad\quad$(A) 3

$\quad\quad$(B) 2

$\quad\quad$(C) 1

$\quad\quad$(D) not defined

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Solution

$(\frac{d^{2} y}{d x^{2}})^{3}+(\frac{d y}{d x})^{2}+\sin (\frac{d y}{d x})+1=0$

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

12. The order of the differential equation

$\quad\quad$ $ 2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0 \text{ is } $

$\quad\quad$(A) 2

$\quad\quad$(B) 1

$\quad\quad$(C) 0

$\quad\quad$(D) not defined

Show Answer

Solution

$2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$

The highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$. Therefore, its order is two.

Hence, the correct answer is A.

9.3 General and Particular Solutions of a Differential Equation

In earlier Classes, we have solved the equations of the type:

$$ \begin{align*} x^{2}+1=0 \tag{1} \\ \sin ^{2} x-\cos x=0 \tag{2} \end{align*} $$

Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the given equation i.e., when that number is substituted for the unknown $x$ in the given equation, L.H.S. becomes equal to the R.H.S..

Now consider the differential equation

$\frac{d^{2} y}{d x^{2}}+y=0$

In contrast to the first two equations, the solution of this differential equation is a function $\phi$ that will satisfy it i.e., when the function $\phi$ is substituted for the unknown $y$ (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S..

The curve $y=\phi(x)$ is called the solution curve (integral curve) of the given differential equation. Consider the function given by

$$ \begin{equation*} y=\phi(x)=a \sin (x+b) \tag{4} \end{equation*} $$

where $a, b \in \mathbf{R}$. When this function and its derivative are substituted in equation (3), L.H.S. = R.H.S.. So it is a solution of the differential equation (3).

Let $a$ and $b$ be given some particular values say $a=2$ and $b=\frac{\pi}{4}$, then we get a function

$$ \begin{equation*} y=\phi _{1}(x)=2 \sin \left(x+\frac{\pi}{4}\right) \tag{5} \end{equation*} $$

When this function and its derivative are substituted in equation (3) again L.H.S. = R.H.S.. Therefore $\phi_1$ is also a solution of equation (3).

Function $\phi$ consists of two arbitrary constants (parameters) $a, b$ and it is called general solution of the given differential equation. Whereas function $\phi_1$ contains no arbitrary constants but only the particular values of the parameters $a$ and $b$ and hence is called a particular solution of the given differential equation. The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation.

The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Example 2 Verify that the function $y=e^{-3 x}$ is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-6 y=0$

Solution Given function is $y=e^{-3 x}$. Differentiating both sides of equation with respect to $x$, we get

$$ \begin{equation*} \frac{d y}{d x}=3 e^{-3 x} \tag{1} \end{equation*} $$

Now, differentiating (1) with respect to $x$, we have

$$ \frac{d^{2} y}{d x^{2}}=9 e^{-3 x} $$

Substituting the values of $\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ and $y$ in the given differential equation, we get

L.H.S. $=9 e^{-3 x}+(-3 e^{-3 x})-6 . e^{-3 x}=9 e^{-3 x}-9 e^{-3 x}=0=$ R.H.S..

Therefore, the given function is a solution of the given differential equation.

Example 3 Verify that the function $y=a \cos x+b \sin x$, where, $a, b \in \mathbf{R}$ is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}+y=0$

Solution The given function is

$$ \begin{equation*} y=a \cos x+b \sin x \tag{1} \end{equation*} $$

Differentiating both sides of equation (1) with respect to $x$, successively, we get

$$ \begin{aligned} \frac{d y}{d x} & =-a \sin x+b \cos x \\ \frac{d^{2} y}{d x^{2}} & =-a \cos x-b \sin x \end{aligned} $$

Substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $y$ in the given differential equation, we get

L.H.S. $=(-a \cos x-b \sin x)+(a \cos x+b \sin x)=0=$ R.H.S.

Therefore, the given function is a solution of the given differential equation.

EXERCISE 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

1. $y=e^{x}+1 \quad \quad \quad \quad \quad \quad\quad\quad: y^{\prime \prime}-y^{\prime}=0$

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Solution

$y=e^{x}+1$

Differentiating both sides of this equation with respect to $x$, we get:

$\frac{d y}{d x}=\frac{d}{d x}(e^{x}+1)$

$\Rightarrow y^{\prime}=e^{x}$

Now, differentiating equation (1) with respect to $x$, we get:

$\frac{d}{d x}(y^{\prime})=\frac{d}{d x}(e^{x})$

$\Rightarrow y^{\prime \prime}=e^{x}$

Substituting the values of $y^{\prime}$ and $y^{\prime \prime}$ in the given differential equation, we get the L.H.S. as:

$y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=0=$ R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

2. $y=x^{2}+2 x+C \quad \quad \quad \quad \quad : y^{\prime}-2 x-2=0$

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Solution

$y=x^{2}+2 x+C$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(x^{2}+2 x+C)$

$\Rightarrow y^{\prime}=2 x+2$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{\prime}-2 x-2=2 x+2-2 x-2=0=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

3. $y=\cos x+C \quad \quad \quad \quad \quad \quad: y^{\prime}+\sin x=0$

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Solution

$y=\cos x+C$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(\cos x+C)$

$\Rightarrow y^{\prime}=-\sin x$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{\prime}+\sin x=-\sin x+\sin x=0=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

4. $y=\sqrt{1+x^{2}} \quad \quad \quad \quad \quad \quad\quad: y^{\prime}=\frac{x y}{1+x^{2}}$

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Solution

$y=\sqrt{1+x^{2}}$

Differentiating both sides of the equation with respect to $x$, we get: $y^{\prime}=\frac{d}{d x}(\sqrt{1+x^{2}})$

$y^{\prime}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot \frac{d}{d x}(1+x^{2})$

$y^{\prime}=\frac{2 x}{2 \sqrt{1+x^{2}}}$

$y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \times \sqrt{1+x^{2}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \cdot y$

$\Rightarrow y^{\prime}=\frac{x y}{1+x^{2}}$

$\therefore$ L.H.S. $=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

5. $y=A x \quad \quad \quad \quad \quad \quad \quad \quad \quad: x y^{\prime}=y(x \neq 0)$

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Solution

$y=A x$

Differentiating both sides with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(A x)$

$\Rightarrow y^{\prime}=A$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=x y^{\prime}=x \cdot A=A x=y=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

6. $y=x \sin x \quad \quad \quad \quad \quad \quad \quad \quad: x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0 \text{ and } x>y \text{ or } x<-y)$

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Solution

$y=x \sin x$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(x \sin x)$

$\Rightarrow y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow y^{\prime}=\sin x+x \cos x$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

$ \begin{aligned} \text{ L.H.S. }=x y^{\prime} & =x(\sin x+x \cos x) \\ & =x \sin x+x^{2} \cos x \\ & =y+x^{2} \cdot \sqrt{1-\sin ^{2}} x \\ & =y+x^{2} \sqrt{1-(\frac{y}{x})^{2}} \\ & =y+x \sqrt{y^{2}-x^{2}} \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

7. $x y=\log y+C \quad \quad \quad \quad \quad \quad : y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)$

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Solution

$x y=\log y+C$

Differentiating both sides of this equation with respect to $x$, we get: $\frac{d}{d x}(x y)=\frac{d}{d x}(\log y)$

$\Rightarrow y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}$

$\Rightarrow y+x y^{\prime}=\frac{1}{y} y^{\prime}$

$\Rightarrow y^{2}+x y y^{\prime}=y^{\prime}$

$\Rightarrow(x y-1) y^{\prime}=-y^{2}$

$\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y}$

$\therefore$ L.H.S. $=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

8. $y-\cos y=x \quad \quad \quad \quad \quad \quad \quad : y \sin y+\cos y+x y^{\prime}=y$

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Solution

$y-\cos y=x$

Differentiating both sides of the equation with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}-\frac{d}{d x}(\cos y)=\frac{d}{d x}(x) \\ & \Rightarrow y^{\prime}+\sin y \cdot y^{\prime}=1 \\ & \Rightarrow y^{\prime}(1+\sin y)=1 \\ & \Rightarrow y^{\prime}=\frac{1}{1+\sin y} \end{aligned} $

Substituting the value of $y^{\prime}$ in equation (1), we get:

L.H.S. $=(y \sin y+\cos y+x) y^{\prime}$

$ \begin{aligned} & =(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y} \\ & =y(1+\sin y) \cdot \frac{1}{1+\sin y} \\ & =y \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

9. $x+y=\tan ^{-1} y \quad \quad \quad \quad \quad \quad : y^{2} y^{\prime}+y^{2}+1=0$

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Solution

$x+y=\tan ^{-1} y$

Differentiating both sides of this equation with respect to $x$, we get:

$ \frac{d}{d x}(x+y)=\frac{d}{d x}(\tan ^{-1} y) $

$ \Rightarrow 1+y^{\prime}=[\frac{1}{1+y^{2}}] y^{\prime} $

$\Rightarrow y^{\prime}[\frac{1}{1+y^{2}}-1]=1$

$\Rightarrow y^{\prime}[\frac{1-(1+y^{2})}{1+y^{2}}]=1$

$\Rightarrow y^{\prime}[\frac{-y^{2}}{1+y^{2}}]=1$

$\Rightarrow y^{\prime}=\frac{-(1+y^{2})}{y^{2}}$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{2} y^{\prime}+y^{2}+1=y^{2}[\frac{-(1+y^{2})}{y^{2}}]+y^{2}+1$

$ \begin{aligned} & =-1-y^{2}+y^{2}+1 \\ & =0 \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

10. $y=\sqrt{a^{2}-x^{2}} x \in(-a, a) \quad\quad: x+y \frac{d y}{d x}=0(y \neq 0)$

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Solution

$y=\sqrt{a^{2}-x^{2}}$

Differentiating both sides of this equation with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(\sqrt{a^{2}-x^{2}}) \\ & \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}(a^{2}-x^{2}) \\ & =\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x) \\ & =\frac{-x}{\sqrt{a^{2}-x^{2}}} \end{aligned} $

Substituting the value of $\frac{d y}{d x}$ in the given differential equation, we get:

L.H.S. $=x+y \frac{d y}{d x}=x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}}$

$ \begin{aligned} & =x-x \\ & =0 \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0

(B) 2

(C) 3

(D) 4

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Solution

We know that the number of constants in the general solution of a differential equation of order $n$ is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is D.

12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3

(B) 2

(C) 1

(D) 0

Show Answer

Solution

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is D.

9.4 Methods of Solving First Order, First Degree Differential Equations

In this section we shall discuss three methods of solving first order first degree differential equations.

9.4.1 Differential equations with variables separable

A first order-first degree differential equation is of the form

$$ \begin{equation*} \frac{d y}{d x}=\mathrm{F}(x, y) \tag{1} \end{equation*} $$

If $F(x, y)$ can be expressed as a product $g(x) h(y)$, where, $g(x)$ is a function of $x$ and $h(y)$ is a function of $y$, then the differential equation (1) is said to be of variable separable type. The differential equation (1) then has the form

$$ \begin{equation*} \frac{d y}{d x}=h(y) \cdot g(x) \tag{2} \end{equation*} $$

If $h(y) \neq 0$, separating the variables, (2) can be rewritten as

$$ \begin{equation*} \frac{1}{h(y)} d y=g(x) d x \tag{3} \end{equation*} $$

Integrating both sides of (3), we get

$$ \begin{equation*} \int \frac{1}{h(y)} d y=\int g(x) d x \tag{4} \end{equation*} $$

Thus, (4) provides the solutions of given differential equation in the form

$$ \begin{equation*} \mathrm{H}(y)=\mathrm{G}(x)+\mathrm{C} \tag{5} \end{equation*} $$

Here, $H(y)$ and $G(x)$ are the anti derivatives of $\frac{1}{h(y)}$ and $g(x)$ respectively and $C$ is the arbitrary constant.

Example 4 Find the general solution of the differential equation $\frac{d y}{d x}=\frac{x+1}{2-y},(y \neq 2)$

Solution We have

$$ \begin{equation*} \frac{d y}{d x}=\frac{x+1}{2-y}(y \neq 2) \tag{1} \end{equation*} $$

Separating the variables in equation (1), we get

$$ \begin{equation*} (2-y) d y=(x+1) d x \tag{2} \end{equation*} $$

Integrating both sides of equation (2), we get

$$ \int(2-y) d y=\int(x+1) d x $$

$$ \text{ or } \qquad 2 y-\frac{y^{2}}{2}=\frac{x^{2}}{2}+x+\mathrm{C} _{1} $$

$$ \text{ or } \qquad x^{2}+y^{2}+2 x-4 y+2 \mathrm{C} _{1}=0 $$

$\text{ or } \qquad x^{2}+y^{2}+2 x-4 y+\mathrm{C}=0 \text { where } \mathrm{C}=2 \mathrm{C} _{1}$

which is the general solution of equation (1).

Example 5 Find the general solution of the differential equation $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$.

Solution Since $1+y^{2} \neq 0$, therefore separating the variables, the given differential equation can be written as

$$ \begin{equation*} \frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}} \tag{1} \end{equation*} $$

Integrating both sides of equation (1), we get

$$ \int \frac{d y}{1+y^{2}}=\int \frac{d x}{1+x^{2}} $$

$$\text{ or }\qquad \tan ^{-1} y=\tan ^{-1} x+\mathrm{C} $$

which is the general solution of equation (1).

Example 6 Find the particular solution of the differential equation $\frac{d y}{d x}=-4 x y^{2}$ given that $y=1$, when $x=0$.

Solution If $y \neq 0$, the given differential equation can be written as

$$ \begin{equation*} \frac{d y}{y^{2}}=-4 x d x \tag{1} \end{equation*} $$

Integrating both sides of equation (1), we get

$ \begin{aligned} \int \frac{d y}{y^{2}} & =-4 \int x d x \\ \frac{1}{y} & =-2 x^{2}+C \\ \text{ or } \quad y & =\frac{1}{2 x^{2}-C} \end{aligned} $

Substituting $y=1$ and $x=0$ in equation (2), we get, $C=-1$.

Now substituting the value of $C$ in equation (2), we get the particular solution of the given differential equation as $y=\frac{1}{2 x^{2}+1}$.

Example 7 Find the equation of the curve passing through the point $(1,1)$ whose differential equation is $x d y=(2 x^{2}+1) d x(x \neq 0)$.

Solution The given differential equation can be expressed as

$\text{ or } \qquad dy $ $ =(\frac{2x^2+1}{x}) dx \\ dy =(2 x+\frac{1}{x}) d x $

Integrating both sides of equation (1), we get

$$ \int d y=\int\left(2 x+\frac{1}{x}\right) d x $$

$ \begin{equation*} \text{ or }\qquad y=x^{2}+\log |x|+\mathrm{C} \tag{2} \end{equation*} $

Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point $(1,1)$. Therefore substituting $x=1, y=1$ in equation (2), we get $C=0$.

Now substituting the value of $C$ in equation (2) we get the equation of the required curve as $y=x^{2}+\log |x|$.

Example 8 Find the equation of a curve passing through the point $(-2,3)$, given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2 x}{y^{2}}$.

Solution We know that the slope of the tangent to a curve is given by $\frac{d y}{d x}$.

$$ \begin{equation*} \frac{d y}{d x}=\frac{2 x}{y^{2}} \tag{1} \end{equation*} $$

Separating the variables, equation (1) can be written as

$$ \begin{equation*} y^{2} d y=2 x d x \tag{2} \end{equation*} $$

Integrating both sides of equation (2), we get

$$ \int y^{2} d y=\int 2 x d x $$

$$ \begin{equation*} \text{ or } \qquad \frac{y^{3}}{3}=x^{2}+\mathrm{C} \tag{3} \end{equation*} $$

Substituting $x=-2, y=3$ in equation (3), we get $C=5$.

Substituting the value of $C$ in equation (3), we get the equation of the required curve as

$$ \frac{y^{3}}{3}=x^{2}+5 \quad \text{ or } \quad y=(3 x^{2}+15)^{\frac{1}{3}} $$

Example 9 In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?

Solution Let $P$ be the principal at any time $t$. According to the given problem,

$$ \begin{align*} & \frac{d \mathrm{P}}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P} \\ & \frac{d \mathrm{P}}{d t}=\frac{\mathrm{P}}{20} \tag{1} \end{align*} $$

separating the variables in equation (1), we get

$$ \begin{equation*} \frac{d \mathrm{P}}{\mathrm{P}}=\frac{d t}{20} \tag{2} \end{equation*} $$

Integrating both sides of equation (2), we get

$$ \begin{aligned} \text{ or } \qquad \log P & =\frac{t}{20}+C_1 \\ P & =e^{\frac{t}{20}} \cdot e^{C_1} \end{aligned} $$

$$ \begin{equation*} \text{ or } \qquad \mathrm{P}=\mathrm{C} e^{\frac{t}{20}} \quad\left(\text { where } e^{\mathrm{C} _{1}}=\mathrm{C}\right) \tag{3} \end{equation*} $$

Now $\qquad \mathrm{P}=1000, \quad \text { when } t=0$

Substituting the values of $P$ and $t$ in (3), we get $C=1000$.

Therefore, equation (3), gives

$$ P=1000 e^{\frac{t}{20}} $$

Let $t$ years be the time required to double the principal. Then

$$ 2000=1000 e^{\frac{t}{20}} \Rightarrow t=20 \log _{e} 2 $$

EXERCISE 9.3

For each of the differential equations in Exercises 1 to 10, find the general solution:

1. $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

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Solution

The given differential equation is:

$\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2}$

$\Rightarrow \frac{d y}{d x}=(\sec ^{2} \frac{x}{2}-1)$

Separating the variables,we get:

$d y=(\sec ^{2} \frac{x}{2}-1) d x$

Now, integrating both sides of this equation, we get:

$ \begin{aligned} & \int d y=\int(\sec ^{2} \frac{x}{2}-1) d x=\int \sec ^{2} \frac{x}{2} d x-\int d x \\ & \Rightarrow y=2 \tan \frac{x}{2}-x+C \end{aligned} $

This is the required general solution of the given differential equation.

2. $\frac{d y}{d x}=\sqrt{4-y^{2}}(-2<y<2)$

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Solution

The given differential equation is: $\frac{d y}{d x}=\sqrt{4-y^{2}}$

Separating the variables, we get:

$\Rightarrow \frac{d y}{\sqrt{4-y^{2}}}=d x$

Now, integrating both sides of this equation, we get:

$ \begin{aligned} & \int \frac{d y}{\sqrt{4-y^{2}}}=\int d x \\ & \Rightarrow \sin ^{-1} \frac{y}{2}=x+C \\ & \Rightarrow \frac{y}{2}=\sin (x+C) \\ & \Rightarrow y=2 \sin (x+C) \end{aligned} $

This is the required general solution of the given differential equation.

3. $\frac{d y}{d x}+y=1(y \neq 1)$

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Solution

The given differential equation is:

$ \begin{aligned} & \frac{d y}{d x}+y=1 \\ & \Rightarrow d y+y d x=d x \\ & \Rightarrow d y=(1-y) d x \end{aligned} $

Separating the variables, we get:

$\Rightarrow \frac{d y}{1-y}=d x$

Now, integrating both sides, we get: $\int \frac{d y}{1-y}=\int d x$

$\Rightarrow -\log (1-y)=x+\log (C)$

$\Rightarrow-\log C-\log (1-y)=x$

$\Rightarrow \log C(1-y)=-x$

$\Rightarrow C(1-y)=e^{-x}$

$\Rightarrow 1-y=\frac{1}{C} e^{-x}$

$\Rightarrow y=1-\frac{1}{C} e^{-x}$

$\Rightarrow y=1+A e^{-x}$ (where $A=-\frac{1}{C}$ )

This is the required general solution of the given differential equation.

4. $\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

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Solution

The given differential equation is:

$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

$\Rightarrow \frac{\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y}{\tan x \tan y}=0$

$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0$

$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$

Integrating both sides of this equation, we get:

$$ \begin{equation*} \int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y \tag{1} \end{equation*} $$

Let $\tan x=t$.

$\therefore \frac{d}{d x}(\tan x)=\frac{d t}{d x}$

$\Rightarrow \sec ^{2} x=\frac{d t}{d x}$

$\Rightarrow \sec ^{2} x d x=d t$

Now, $\int \frac{\sec ^{2} x}{\tan x} d x=\int_t^1 d t$.

$ \begin{aligned} & =\log t \\ & =\log (\tan x) \end{aligned} $

Similarly, $\int \frac{\sec ^{2} x}{\tan x} d y=\log (\tan y)$.

Substituting these values in equation (1), we get:

$ \begin{aligned} & \log (\tan x)=-\log (\tan y)+\log C \\ & \Rightarrow \log (\tan x)=\log (\frac{C}{\tan y}) \\ & \Rightarrow \tan x=\frac{C}{\tan y} \\ & \Rightarrow \tan x \tan y=C \end{aligned} $

This is the required general solution of the given differential equation.

5. $(e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0$

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Solution

The given differential equation is:

$ \begin{aligned} & (e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0 \\ & \Rightarrow(e^{x}+e^{-x}) d y=(e^{x}-e^{-x}) d x \\ & \Rightarrow d y=[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}] d x \end{aligned} $

Integrating both sides of this equation, we get: $\int d y=\int[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}] d x+C$

$\Rightarrow y=\int[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}] d x+C$

Let $(e^{x}+e^{-x})=t$.

Differentiating both sides with respect to $x$, we get:

$\frac{d}{d x}(e^{x}+e^{-x})=\frac{d t}{d x}$

$\Rightarrow e^{x}-e^{-x}=\frac{d t}{d t}$

$\Rightarrow(e^{x}-e^{-x}) d x=d t$

Substituting this value in equation (1), we get:

$ \begin{aligned} & y=\int \frac{1}{t} d t+C \\ & \Rightarrow y=\log (t)+C \\ & \Rightarrow y=\log (e^{x}+e^{-x})+C \end{aligned} $

This is the required general solution of the given differential equation.

6. $\frac{d y}{d x}=(1+x^{2})(1+y^{2})$

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Solution

The given differential equation is:

$ \begin{aligned} & \frac{d y}{d x}=(1+x^{2})(1+y^{2}) \\ & \Rightarrow \frac{d y}{1+y^{2}}=(1+x^{2}) d x \end{aligned} $

Integrating both sides of this equation, we get: $\int \frac{d y}{1+y^{2}}=\int(1+x^{2}) d x$

$\Rightarrow \tan ^{-1} y=\int d x+\int x^{2} d x$

$\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+C$

This is the required general solution of the given differential equation.

7. $y \log y d x-x d y=0$

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Solution

The given differential equation is:

$y \log y d x-x d y=0$

$\Rightarrow y \log y d x=x d y$

$\Rightarrow \frac{d y}{y \log y}=\frac{d x}{x}$

Integrating both sides, we get:

$\int \frac{d y}{y \log y}=\int \frac{d x}{x}$

Let $\log y=t$.

$\therefore \frac{d}{d y}(\log y)=\frac{d t}{d y}$

$\Rightarrow \frac{1}{y}=\frac{d t}{d y}$

$\Rightarrow \frac{1}{y} d y=d t$

Substituting this value in equation (1), we get:

$ \begin{aligned} & \int \frac{d t}{t}=\int \frac{d x}{x} \\ & \Rightarrow \log t=\log x+\log C \\ & \Rightarrow \log (\log y)=\log C x \\ & \Rightarrow \log y=C x \\ & \Rightarrow y=e^{C x} \end{aligned} $

This is the required general solution of the given differential equation.

8. $x^{5} \frac{d y}{d x}=-y^{5}$

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Solution

The given differential equation is:

$x^{5} \frac{d y}{d x}=-y^{5}$

$\Rightarrow \frac{d y}{y^{5}}=-\frac{d x}{x^{5}}$

$\Rightarrow \frac{d x}{x^{5}}+\frac{d y}{y^{5}}=0$

Integrating both sides, we get:

$ \begin{aligned} & \int \frac{d x}{x^{5}}+\int \frac{d y}{y^{5}}=k \quad \text{ (where } k \text{ is any constant) } \\ & \Rightarrow \int x^{-5} d x+\int y^{-5} d y=k \\ & \Rightarrow \frac{x^{-4}}{-4}+\frac{y^{-4}}{-4}=k \\ & \Rightarrow x^{-4}+y^{-4}=-4 k \\ & \Rightarrow x^{-4}+y^{-4}=C \quad(C=-4 k) \end{aligned} $

This is the required general solution of the given differential equation.

9. $\frac{d y}{d x}=\sin ^{-1} x$

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Solution

The given differential equation is:

$\frac{d y}{d x}=\sin ^{-1} x$

$\Rightarrow d y=\sin ^{-1} x d x$

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\int \sin ^{-1} x d x \\ & \Rightarrow y=\int(\sin ^{-1} x \cdot 1) d x \\ & \Rightarrow y=\sin ^{-1} x \cdot \int(1) d x-\int[(\frac{d}{d x}(\sin ^{-1} x) \cdot \int(1) d x)] d x \\ & \Rightarrow y=\sin ^{-1} x \cdot x-\int(\frac{1}{\sqrt{1-x^{2}}} \cdot x) d x \\ & \Rightarrow y=x \sin ^{-1} x+\int \frac{-x}{\sqrt{1-x^{2}}} d x \tag{1} \end{align*} $$

Let $1-x^{2}=t$.

$\Rightarrow \frac{d}{d x}(1-x^{2})=\frac{d t}{d x}$

$\Rightarrow-2 x=\frac{d t}{d x}$

$\Rightarrow x d x=-\frac{1}{2} d t$

Substituting this value in equation (1), we get:

$ \begin{aligned} & y=x \sin ^{-1} x+\int \frac{1}{2 \sqrt{t}} d t \\ & \Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \int(t)^{-\frac{1}{2}} d t \\ & \Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\ & \Rightarrow y=x \sin ^{-1} x+\sqrt{t}+C \\ & \Rightarrow y=x \sin ^{-1} x+\sqrt{1-x^{2}}+C \end{aligned} $

This is the required general solution of the given differential equation.

10. $e^{x} \tan y d x+(1-e^{x}) \sec ^{2} y d y=0$

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Solution

The given differential equation is:

$e^{x} \tan y d x+(1-e^{x}) \sec ^{2} y d y=0$

$(1-e^{x}) \sec ^{2} y d y=-e^{x} \tan y d x$

Separating the variables, we get:

$ \frac{\sec ^{2} y}{\tan y} d y=\frac{-e^{x}}{1-e^{x}} d x $

Integrating both sides, we get:

$$ \begin{equation*} \int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{-e^{x}}{1-e^{x}} d x \tag{1} \end{equation*} $$

Let $\tan y=u$.

$ \begin{aligned} & \Rightarrow \frac{d}{d y}(\tan y)=\frac{d u}{d y} \\ & \Rightarrow \sec ^{2} y=\frac{d u}{d y} \\ & \Rightarrow \sec ^{2} y d y=d u \\ & \therefore \int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{d u}{u}=\log u=\log (\tan y) \end{aligned} $

Now, let $1-e^{x}=t$.

$\therefore \frac{d}{d x}(1-e^{x})=\frac{d t}{d x}$

$\Rightarrow-e^{x}=\frac{d t}{d x}$

$\Rightarrow-e^{x} d x=d t$

$\Rightarrow \int \frac{-e^{x}}{1-e^{x}} d x=\int \frac{d t}{t}=\log t=\log (1-e^{x})$

Substituting the values of $\int \frac{\sec ^{2} y}{\tan y} d y$ and $\int \frac{-e^{x}}{1-e^{x}} d x$ in equation (1), we get:

$\Rightarrow \log (\tan y)=\log (1-e^{x})+\log C$

$\Rightarrow \log (\tan y)=\log [C(1-e^{x})]$

$\Rightarrow \tan y=C(1-e^{x})$

This is the required general solution of the given differential equation.

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

11. $(x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x ; y=1$ when $x=0$

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Solution

The given differential equation is:

$ \begin{aligned} & (x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x \\ & \Rightarrow \frac{d y}{d x}=\frac{2 x^{2}+x}{(x^{3}+x^{2}+x+1)} \\ & \Rightarrow d y=\frac{2 x^{2}+x}{(x+1)(x^{2}+1)} d x \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\int \frac{2 x^{2}+x}{(x+1)(x^{2}+1)} d x \tag{1}\\ & \text{ Let } \frac{2 x^{2}+x}{(x+1)(x^{2}+1)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1} . \tag{2}\\ & \Rightarrow \frac{2 x^{2}+x}{(x+1)(x^{2}+1)}=\frac{A x^{2}+A+(B x+C)(x+1)}{(x+1)(x^{2}+1)} \\ & \Rightarrow 2 x^{2}+x=A x^{2}+A+B x^{2}+B x+C x+C \\ & \Rightarrow 2 x^{2}+x=(A+B) x^{2}+(B+C) x+(A+C) \end{align*} $$

Comparing the coefficients of $x^{2}$ and $x$, we get:

$A+B=2$

$B+C=1$

$A+C=0$

Solving these equations, we get:

$ A=\frac{1}{2}, B=\frac{3}{2} \text{ and } C=\frac{-1}{2} $

Substituting the values of $A, B$, and $C$ in equation (2), we get:

$ \frac{2 x^{2}+x}{(x+1)(x^{2}+1)}=\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \frac{(3 x-1)}{(x^{2}+1)} $

Therefore, equation (1) becomes:

$$ \begin{align*} & \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^{2}+1} d x \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{1}{x^{2}+1} d x \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \cdot \int \frac{2 x}{x^{2}+1} d x-\frac{1}{2} \tan ^{-1} x+C \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log (x^{2}+1)-\frac{1}{2} \tan ^{-1} x+C \\ & \Rightarrow y=\frac{1}{4}[2 \log (x+1)+3 \log (x^{2}+1)]-\frac{1}{2} \tan ^{-1} x+C \\ & \Rightarrow y=\frac{1}{4}\log (x+1)^{2}(x^{2}+1)^{3}]-\frac{1}{2} \tan ^{-1} x+C \tag{3} \end{align*} $$

Now, $y=1$ when $x=0$.

$\Rightarrow 1=\frac{1}{4} \log (1)-\frac{1}{2} \tan ^{-1} 0+C$

$\Rightarrow 1=\frac{1}{4} \times 0-\frac{1}{2} \times 0+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (3), we get:

$y=\frac{1}{4}[\log (x+1)^{2}(x^{2}+1)^{3}]-\frac{1}{2} \tan ^{-1} x+1$

12. $x(x^{2}-1) \frac{d y}{d x}=1 ; y=0$ when $x=2$

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Solution

$ \begin{aligned} & x(x^{2}-1) \frac{d y}{d x}=1 \\ & \Rightarrow d y=\frac{d x}{x(x^{2}-1)} \\ & \Rightarrow d y=\frac{1}{x(x-1)(x+1)} d x \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\int \frac{1}{x(x-1)(x+1)} d x \tag{1}\\ & \text{ Let } \frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} . \tag{2}\\ & \Rightarrow \frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)} \\ & =\frac{(A+B+C) x^{2}+(B-C) x-A}{x(x-1)(x+1)} \end{align*} $$

Comparing the coefficients of $x^{2}, x$, and constant, we get:

$A=-1$

$B-C=0$

$A+B+C=0$

Solving these equations, we get $B=\frac{1}{2}$ and $C=\frac{1}{2}$.

Substituting the values of $A, B$, and $C$ in equation (2), we get:

$ \frac{1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)} $

Therefore, equation (1) becomes:

$$ \begin{align*} & \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x+1} d x \\ & \Rightarrow y=-\log x+\frac{1}{2} \log (x-1)+\frac{1}{2} \log (x+1)+\log k \\ & \Rightarrow y=\frac{1}{2} \log [\frac{k^{2}(x-1)(x+1)}{x^{2}}] \tag{3} \end{align*} $$

Now, $y=0$ when $x=2$.

$\Rightarrow 0=\frac{1}{2} \log [\frac{k^{2}(2-1)(2+1)}{4}]$

$\Rightarrow \log (\frac{3 k^{2}}{4})=0$

$\Rightarrow \frac{3 k^{2}}{4}=1$

$\Rightarrow 3 k^{2}=4$

$\Rightarrow k^{2}=\frac{4}{3}$

Substituting the value of $k^{2}$ in equation (3), we get:

$y=\frac{1}{2} \log [\frac{4(x-1)(x+1)}{3 x^{2}}]$

$y=\frac{1}{2} \log [\frac{4(x^{2}-1)}{3 x^{2}}]$

13. $\cos (\frac{d y}{d x})=a(a \in \mathbf{R}) ; y=1$ when $x=0$

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Solution

$\cos (\frac{d y}{d x})=a$

$\Rightarrow \frac{d y}{d x}=\cos ^{-1} a$

$\Rightarrow d y=\cos ^{-1} a d x$

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\cos ^{-1} a \int d x \\ & \Rightarrow y=\cos ^{-1} a \cdot x+C \\ & \Rightarrow y=x \cos ^{-1} a+C \tag{1} \end{align*} $$

Now, $y=1$ when $x=0$.

$\Rightarrow 1=0 \cdot \cos ^{-1} a+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (1), we get:

$y=x \cos ^{-1} a+1$ $\Rightarrow \frac{y-1}{x}=\cos ^{-1} a$ $\Rightarrow \cos (\frac{y-1}{x})=a$

14. $\frac{d y}{d x}=y \tan x ; y=1$ when $x=0$

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Solution

$\frac{d y}{d x}=y \tan x$

$\Rightarrow \frac{d y}{y}=\tan x d x$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{d y}{y}=\int \tan x d x \\ & \Rightarrow \log y=\log (\sec x)+\log C \\ & \Rightarrow \log y=\log (C \sec x) \\ & \Rightarrow y=C \sec x \tag{1} \end{align*} $$

Now, $y=1$ when $x=0$.

$\Rightarrow 1=C \times \sec 0$

$\Rightarrow 1=C \times 1$

$\Rightarrow C=1$

Substituting $C=1$ in equation (1), we get:

$y=\sec x$

15. Find the equation of a curve passing through the point $(0,0)$ and whose differential equation is $y^{\prime}=e^{x} \sin x$.

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Solution

The differential equation of the curve is:

$$ \begin{aligned} & y^{\prime}=e^{x} \sin x \\ & \Rightarrow \frac{d y}{d x}=e^{x} \sin x \\ & \Rightarrow d y=e^{x} \sin x \end{aligned} $$

Integrating both sides, we get:

$$ \begin{equation*} \int d y=\int e^{x} \sin x d x \tag{1} \end{equation*} $$

Let $I=\int e^{x} \sin x d x$.

$\Rightarrow I=\sin x \int e^{x} d x-\int(\frac{d}{d x}(\sin x) \cdot \int e^{x} d x) d x$ $\Rightarrow I=\sin x \cdot e^{x}-\int \cos x \cdot e^{x} d x$

$\Rightarrow I=\sin x \cdot e^{x}-[\cos x \cdot \int e^{x} d x-\int(\frac{d}{d x}(\cos x) \cdot \int e^{x} d x) d x]$

$\Rightarrow I=\sin x \cdot e^{x}-[\cos x \cdot e^{x}-\int(-\sin x) \cdot e^{x} d x]$

$\Rightarrow I=e^{x} \sin x-e^{x} \cos x-I$

$\Rightarrow 2 I=e^{x}(\sin x-\cos x)$

$\Rightarrow I=\frac{e^{x}(\sin x-\cos x)}{2}$

Substituting this value in equation (1), we get:

$y=\frac{e^{x}(\sin x-\cos x)}{2}+C$

Now, the curve passes through point $(0,0)$.

$\therefore 0=\frac{e^{0}(\sin 0-\cos 0)}{2}+C$

$\Rightarrow 0=\frac{1(0-1)}{2}+C$

$\Rightarrow C=\frac{1}{2}$

Substituting $C=\frac{1}{2}$ in equation (2), we get:

$y=\frac{e^{x}(\sin x-\cos x)}{2}+\frac{1}{2}$

$\Rightarrow 2 y=e^{x}(\sin x-\cos x)+1$

$\Rightarrow 2 y-1=e^{x}(\sin x-\cos x)$

Hence, the required equation of the curve is $2 y-1=e^{x}(\sin x-\cos x)$.

16. For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$, find the solution curve passing through the point $(1,-1)$.

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Solution

The differential equation of the given curve is:

$x y \frac{d y}{d x}=(x+2)(y+2)$

$\Rightarrow(\frac{y}{y+2}) d y=(\frac{x+2}{x}) d x$

$\Rightarrow(1-\frac{2}{y+2}) d y=(1+\frac{2}{x}) d x$

Integrating both sides, we get:

$$ \begin{align*} & \int(1-\frac{2}{y+2}) d y=\int(1+\frac{2}{x}) d x \\ & \Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x \\ & \Rightarrow y-2 \log (y+2)=x+2 \log x+C \\ & \Rightarrow y-x-C=\log x^{2}+\log (y+2)^{2} \\ & \Rightarrow y-x-C=\log [x^{2}(y+2)^{2}] \tag{1} \end{align*} $$

Now, the curve passes through point $(1,-1)$.

$ \begin{aligned} & \Rightarrow-1-1-C=\log [(1)^{2}(-1+2)^{2}] \\ & \Rightarrow-2-C=\log 1=0 \\ & \Rightarrow C=-2 \end{aligned} $

Substituting $C=-2$ in equation (1), we get:

$ y-x+2=\log [x^{2}(y+2)^{2}] $

This is the required solution of the given curve.

17. Find the equation of a curve passing through the point $(0,-2)$ given that at any point $(x, y)$ on the curve, the product of the slope of its tangent and $y$ coordinate of the point is equal to the $x$ coordinate of the point.

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Solution

Let $x$ and $y$ be the $x$-coordinate and $y$-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the $\frac{d y}{d x}$

According to the given information, we get:

$y \cdot \frac{d y}{d x}=x$

$\Rightarrow y d y=x d x$

Integrating both sides, we get:

$\int y d y=\int x d x$

$\Rightarrow \frac{y^{2}}{2}=\frac{x^{2}}{2}+C$

$\Rightarrow y^{2}-x^{2}=2 C$

Now, the curve passes through point $(0,-2)$.

$\therefore(-2)^{2}-0^{2}=2 C$

$\Rightarrow 2 C=4$

Substituting $2 C=4$ in equation (1), we get:

$y^{2}-x^{2}=4$

This is the required equation of the curve.

18. At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$. Find the equation of the curve given that it passes through $(-2,1)$.

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Solution

It is given that $(x, y)$ is the point of contact of the curve and its tangent.

The slope $(m_1)$ of the line segment joining $(x, y)$ and $(-4,-3)$ is $\frac{y+3}{x+4}$.

We know that the slope of the tangent to the curve is given by the relation, $\frac{d y}{d x}$

$\therefore$ Slope $(m_2)$ of the tangent $=\frac{d y}{d x}$

According to the given information:

$m_2=2 m_1$

$\Rightarrow \frac{d y}{d x}=\frac{2(y+3)}{x+4}$

$\Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{d y}{y+3}=2 \int \frac{d x}{x+4} \\ & \Rightarrow \log (y+3)=2 \log (x+4)+\log C \\ & \Rightarrow \log (y+3) \log C(x+4)^{2} \\ & \Rightarrow y+3=C(x+4)^{2} \tag{1} \end{align*} $$

This is the general equation of the curve.

It is given that it passes through point $(-2,1)$.

$ \begin{aligned} & \Rightarrow 1+3=C(-2+4)^{2} \\ & \Rightarrow 4=4 C \\ & \Rightarrow C=1 \end{aligned} $

Substituting $C=1$ in equation (1), we get:

$y+3=(x+4)^{2}$

This is the required equation of the curve.

19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after $t$ seconds.

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Solution

Let the rate of change of the volume of the balloon be $k$ (where $k$ is a constant).

$\Rightarrow \frac{d v}{d t}=k$

$\Rightarrow \frac{d}{d t}(\frac{4}{3} \pi r^{3})=k$

$[.$ Volume of sphere $.=\frac{4}{3} \pi r^{3}]$

$\Rightarrow \frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=k$

$\Rightarrow 4 \pi r^{2} d r=k d t$

Integrating both sides, we get:

$4 \pi \int r^{2} d r=k \int d t$

$\Rightarrow 4 \pi \cdot \frac{r^{3}}{3}=k t+C$

$\Rightarrow 4 \pi r^{3}=3(k t+C)$

Now, at $t=0, r=3$ :

$\Rightarrow 4 \pi \times 3^{3}=3(k \times 0+C)$

$\Rightarrow 108 \pi=3 C$

$\Rightarrow C=36 \Pi$

At $t=3, r=6:$

$\Rightarrow 4 \pi \times 6^{3}=3(k \times 3+C)$

$\Rightarrow 864 \pi=3(3 k+36 \pi)$

$\Rightarrow 3 k=-288 \pi-36 \pi=252 \pi$

$\Rightarrow k=84 \pi$

Substituting the values of $k$ and $C$ in equation (1), we get:

$ \begin{aligned} & 4 \pi r^{3}=3[84 \pi t+36 \pi] \\ & \Rightarrow 4 \pi r^{3}=4 \pi(63 t+27) \\ & \Rightarrow r^{3}=63 t+27 \\ & \Rightarrow r=(63 t+27)^{\frac{1}{3}} \end{aligned} $

Thus, the radius of the balloon after $t$ seconds is $(63 t+27)^{\frac{1}{3}}$.

20. In a bank, principal increases continuously at the rate of $r %$ per year. Find the value of $r$ if Rs 100 double itself in 10 years $(\log _{e} 2=0.6931)$.

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Solution

Let $p, t$, and $r$ represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of $r %$ per year.

$\Rightarrow \frac{d p}{d t}=(\frac{r}{100}) p$

$\Rightarrow \frac{d p}{p}=(\frac{r}{100}) d t$

Integrating both sides, we get:

$\int \frac{d p}{p}=\frac{r}{100} \int d t$

$\Rightarrow \log p=\frac{r t}{100}+k$

$\Rightarrow p=e^{\frac{r t}{100}+k}$

It is given that when $t=0, p=100$.

$\Rightarrow 100=e^{k}$

Now, if $t=10$, then $p=2 \times 100=200$.

Therefore, equation (1) becomes:

$ \begin{aligned} & 200=e^{\frac{r}{10}+k} \\ & \Rightarrow 200=e^{\frac{r}{10}} \cdot e^{k} \\ & \Rightarrow 200=e^{\frac{r}{10}} \cdot 100 \\ & \Rightarrow e^{\frac{r}{10}}=2 \\ & \Rightarrow \frac{r}{10}=\log _{e} 2 \\ & \Rightarrow \frac{r}{10}=0.6931 \\ & \Rightarrow r=6.931 \end{aligned} $

Hence, the value of $r$ is $6.93 %$.

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years $(e^{0.5}=1.648)$.

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Solution

Let $p$ and $t$ be the principal and time respectively.

It is given that the principal increases continuously at the rate of $5 %$ per year.

$\Rightarrow \frac{d p}{d t}=(\frac{5}{100}) p$

$\Rightarrow \frac{d p}{d t}=\frac{p}{20}$

$\Rightarrow \frac{d p}{p}=\frac{d t}{20}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{d p}{p}=\frac{1}{20} \int d t \\ & \Rightarrow \log p=\frac{t}{20}+C \\ & \Rightarrow p=e^{\frac{1}{20}+C} \tag{1} \end{align*} $$

Now, when $t=0, p=1000$.

$\Rightarrow 1000=e^{C}$

At $t=10$, equation (1) becomes:

$ \begin{aligned} & p=e^{\frac{1}{2}+C} \\ & \Rightarrow p=e^{0.5} \times e^{C} \\ & \Rightarrow p=1.648 \times 1000 \\ & \Rightarrow p=1648 \end{aligned} $

Hence, after 10 years the amount will worth Rs 1648.

22. In a culture, the bacteria count is $1,00,000$. The number is increased by $10 %$ in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

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Solution

Let $y$ be the number of bacteria at any instant $t$.

It is given that the rate of growth of the bacteria is proportional to the number present.

$\therefore \frac{d y}{d t} \propto y$

$\Rightarrow \frac{d y}{d t}=k y$ (where $k$ is a constant)

$\Rightarrow \frac{d y}{y}=k d t$

Integrating both sides, we get:

$\int \frac{d y}{y}=k \int d t$

$\Rightarrow \log y=k t+C$

Let $y_0$ be the number of bacteria at $t=0$.

$\Rightarrow \log y_0=C$

Substituting the value of $C$ in equation (1), we get:

$$ \begin{align*} & \log y=k t+\log y_0 \\ & \Rightarrow \log y-\log y_0=k t \\ & \Rightarrow \log (\frac{y}{y_0})=k t \\ & \Rightarrow k t=\log (\frac{y}{y_0}) \tag{2} \end{align*} $$

Also, it is given that the number of bacteria increases by $10 %$ in 2 hours. $\Rightarrow y=\frac{110}{100} y_0$

$\Rightarrow \frac{y}{y_0}=\frac{11}{10}$

Substituting this value in equation (2), we get:

$$ \begin{aligned} & k \cdot 2=\log (\frac{11}{10}) \\ & \Rightarrow k=\frac{1}{2} \log (\frac{11}{10}) \end{aligned} $$

Therefore, equation (2) becomes:

$$ \begin{align*} & \frac{1}{2} \log (\frac{11}{10}) \cdot t=\log (\frac{y}{y_0}) \\ & \Rightarrow t=\frac{2 \log (\frac{y}{y_0})}{\log (\frac{11}{10})} \tag{4} \end{align*} $$

Now, let the time when the number of bacteria increases from 100000 to 200000 be $t_1$.

$\Rightarrow y=2 y_0$ at $t=t_1$

From equation (4), we get:

$t_1=\frac{2 \log (\frac{y}{y_0})}{\log (\frac{11}{10})}=\frac{2 \log 2}{\log (\frac{11}{10})}$

Hence, in $\frac{2 \log 2}{\log (\frac{11}{10})}$ hours the number of bacteria increases from 100000 to 200000.

23. The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is

$\quad\quad$ (A) $e^{x}+e^{-y}=C$

$\quad\quad$ (B) $e^{x}+e^{y}=C$

$\quad\quad$ (C) $e^{-x}+e^{y}=C$

$\quad\quad$ (D) $e^{-x}+e^{-y}=C$

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Solution

$ \begin{aligned} & \frac{d y}{d x}=e^{x+y}=e^{x} \cdot e^{y} \\ & \Rightarrow \frac{d y}{e^{y}}=e^{x} d x \\ & \Rightarrow e^{-y} d y=e^{x} d x \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & \int e^{-y} d y=\int e^{x} d x \\ & \Rightarrow-e^{-y}=e^{x}+k \\ & \Rightarrow e^{x}+e^{-y}=-k \\ & \Rightarrow e^{x}+e^{-y}=c \end{aligned} $

Hence, the correct answer is A.

9.4.2 Homogeneous differential equations

Consider the following functions in $x$ and $y$

$$ \begin{matrix} F_1(x, y)=y^{2}+2 x y, & F_2(x, y)=2 x-3 y, \\ F_3(x, y)=\cos (\frac{y}{x}), & F_4(x, y)=\sin x+\cos y \end{matrix} $$

If we replace $x$ and $y$ by $\lambda x$ and $\lambda y$ respectively in the above functions, for any nonzero constant $\lambda$, we get

$ \begin{aligned} & F_1(\lambda x, \lambda y)=\lambda^{2}(y^{2}+2 x y)=\lambda^{2} F_1(x, y) \\ & F_2(\lambda x, \lambda y)=\lambda(2 x-3 y)=\lambda F_2(x, y) \\ & F_3(\lambda x, \lambda y)=\cos (\frac{\lambda y}{\lambda x})=\cos (\frac{y}{x})=\lambda^{0} \quad F_3(x, y) \\ & F_4(\lambda x, \lambda y)=\sin \lambda x+\cos \lambda y \neq \lambda^{n} F_4(x, y), \text{ for any } n \in \mathbf{N} \end{aligned} $

Here, we observe that the functions $F_1, F_2, F_3$ can be written in the form $F(\lambda x, \lambda y)=\lambda^{n} F(x, y)$ but $F_4$ can not be written in this form. This leads to the following definition:

A function $F(x, y)$ is said to be homogeneous function of degree $n$ if $F(\lambda x, \lambda y)=\lambda^{n} F(x, y)$ for any nonzero constant $\lambda$.

We note that in the above examples, $F_1, F_2, F_3$ are homogeneous functions of degree 2, 1, 0 respectively but $F_4$ is not a homogeneous function.

We also observe that

$ \begin{aligned} & \qquad F_1(x, y)=x^{2}(\frac{y^{2}}{x^{2}}+\frac{2 y}{x})=x^{2} h_1(\frac{y}{x}) \\ & \qquad F_1(x, y)=y^{2}(1+\frac{2 x}{y})=y^{2} h_2(\frac{x}{y}) \\ & \text{or}\qquad F_2(x, y)=x^{1}(2-\frac{3 y}{x})=x^{1} h_3(\frac{y}{x}) \\ & \qquad F_2(x, y)=y^{1}(2 \frac{x}{y}-3)=y^{1} h_4(\frac{x}{y}) \\ & \qquad F_3(x, y)=x^{0} \cos (\frac{y}{x})=x^{0} h_5(\frac{y}{x}) \\ & \qquad F_4(x, y) \neq x^{n} h_6(\frac{y}{x}), \text{ for any } n \in \mathbf{N} \\ & \qquad F_4(x, y) \neq y^{n} h_7(\frac{x}{y}), \text{ for any } n \in \mathbf{N} \end{aligned} $

$$ \begin{aligned} & \text{or}\qquad\mathrm{F} _{4}(x, y) \neq x^{n} h _{6}\left(\frac{y}{x}\right), n \in \mathbf{N} \text { के किसी भी मान के लिए } \\ & \qquad \mathrm{F} _{4}(x, y) \neq y^{n} h _{7}\left(\frac{x}{y}\right), n \in \mathbf{N} \end{aligned} $$

Therefore, a function $F(x, y)$ is a homogeneous function of degree $n$ if A differential equation of the form $\frac{d y}{d x}=F(x, y)$ is said to be homogenous if $F(x, y)$ is a homogenous function of degree zero.

To solve a homogeneous differential equation of the type

$$ \mathrm{F}(x, y)=x^{n} g\left(\frac{y}{x}\right) \quad \text { or } \quad y^{n} h\left(\frac{x}{y}\right) $$

$$ \begin{equation*} \frac{d y}{d x}=\mathrm{F}(x, y)=g\left(\frac{y}{x}\right) \tag{1} \end{equation*} $$

We make the substitution $\qquad y=v \cdot x \tag{2}$

Differentiating equation (2) with respect to $x$, we get

$$ \frac{d y}{d x}=v+x \frac{d v}{d x} \tag{3} $$

Substituting the value of $\frac{d y}{d x}$ from equation (3) in equation (1), we get or

$$ v+x \frac{d v}{d x}=g(v) $$

$$ \begin{equation*} x \frac{d v}{d x}=g(v)-v \tag{4} \end{equation*} $$

Separating the variables in equation (4), we get

$$ \frac{d v}{g(v)-v}=\frac{d x}{x} \tag{5} $$

Integrating both sides of equation (5), we get

$$ \int \frac{d v}{g(v)-v}=\int \frac{1}{x} d x+C \tag{6} $$

Equation (6) gives general solution (primitive) of the differential equation (1) when we replace $v$ by $\frac{y}{x}$.

Note If the homogeneous differential equation is in the form $\frac{d x}{d y}=F(x, y)$ where, $F(x, y)$ is homogenous function of degree zero, then we make substitution $\frac{x}{y}=v$ i.e., $x=v y$ and we proceed further to find the general solution as discussed above by writing $\frac{d x}{d y}=F(x, y)=h(\frac{x}{y})$.

Example 10 Show that the differential equation $(x-y) \frac{d y}{d x}=x+2 y$ is homogeneous and solve it.

Solution The given differential equation can be expressed as

$$ \begin{equation*} \frac{d y}{d x}=\frac{x+2 y}{x-y} \tag{1} \end{equation*} $$

Let $$ \mathrm{F}(x, y)=\frac{x+2 y}{x-y} $$

Now $$ F(\lambda x, \lambda y)=\frac{\lambda(x+2 y)}{\lambda(x-y)}=\lambda^{0} \cdot f(x, y) $$

Therefore, $F(x, y)$ is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.

Alternatively,

$$ \frac{d y}{d x}=(\frac{1+\frac{2 y}{x}}{1-\frac{y}{x}})=g(\frac{y}{x}) \tag{2} $$

R.H.S. of differential equation (2) is of the form $g(\frac{y}{x})$ and so it is a homogeneous function of degree zero. Therefore, equation (1) is a homogeneous differential equation. To solve it we make the substitution

$$ y=v x \tag{3} $$

Differentiating equation (3) with respect to, $x$ we get

$$ \frac{d y}{d x}=v+x \frac{d v}{d x} \tag{4} $$

Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1) we get

$$ v+x \frac{d v}{d x}=\frac{1+2 v}{1-v} $$

$\text{ or }\qquad x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v $

$\text{ or }\qquad x \frac{d v}{d x}=\frac{v^{2}+v+1}{1-v} $

$\text{ or }\qquad \frac{v-1}{v^{2}+v+1} d v=\frac{-d x}{x} $

Integrating both sides of equation (5), we get

$$ \int \frac{v-1}{v^{2}+v+1} d v=-\int \frac{d x}{x} $$

$$\text{ or }\qquad \frac{1}{2} \int \frac{2 v+1-3}{v^{2}+v+1} d v=-\log |x|+C_1 $$

$$\text{ or }\qquad \frac{1}{2} \int \frac{2 v+1}{v^{2}+v+1} d v-\frac{3}{2} \int \frac{1}{v^{2}+v+1} d v=-\log |x|+C_1 $$

$$ \frac{1}{2} \log |v^{2}+v+1|-\frac{3}{2} \int \frac{1}{(v+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} d v=-\log |x|+C_1 $$

or $ \begin{aligned} & \frac{1}{2} \log |v^{2}+v+1|-\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}(\frac{2 v+1}{\sqrt{3}})=-\log |x|+C_1 \\ & \frac{1}{2} \log |v^{2}+v+1|+\frac{1}{2} \log x^{2}=\sqrt{3} \tan ^{-1}(\frac{2 v+1}{\sqrt{3}})+C_1 \end{aligned} $

Replacing $v$ by $\frac{y}{x}$, we get

or $$ \frac{1}{2} \log |\frac{y^{2}}{x^{2}}+\frac{y}{x}+1|+\frac{1}{2} \log x^{2}=\sqrt{3} \tan ^{-1}(\frac{2 y+x}{\sqrt{3} x})+C_1 $$

$$ \frac{1}{2} \log |(\frac{y^{2}}{x^{2}}+\frac{y}{x}+1) x^{2}|=\sqrt{3} \tan ^{-1}(\frac{2 y+x}{\sqrt{3} x})+C_1 $$

or $$ \log |(y^{2}+x y+x^{2})|=2 \sqrt{3} \tan ^{-1}(\frac{2 y+x}{\sqrt{3} x})+2 C_1 $$

$$ \log |(x^{2}+x y+y^{2})|=2 \sqrt{3} \tan ^{-1}(\frac{x+2 y}{\sqrt{3} x})+C $$

which is the general solution of the differential equation (1)

Example 11 Show that the differential equation $x \cos (\frac{y}{x}) \frac{d y}{d x}=y \cos (\frac{y}{x})+x$ is homogeneous and solve it.

Solution The given differential equation can be written as

$$ \frac{d y}{d x}=\frac{y \cos (\frac{y}{x})+x}{x \cos (\frac{y}{x})} \tag{1} $$

It is a differential equation of the form $\frac{d y}{d x}=F(x, y)$.

Here $ F(x, y)=\frac{y \cos (\frac{y}{x})+x}{x \cos (\frac{y}{x})} $

Replacing $x$ by $\lambda x$ and $y$ by $\lambda y$, we get

$ F(\lambda x, \lambda y)=\frac{\lambda[y \cos (\frac{y}{x})+x]}{\lambda(x \cos \frac{y}{x})}=\lambda^{0}[F(x, y)] $

Thus, $F(x, y)$ is a homogeneous function of degree zero.Therefore, the given differential equation is a homogeneous differential equation.To solve it we make the substitution

$$ y=v x \tag{2} $$

Differentiating equation (2) with respect to $x$, we get

$$ \frac{d y}{d x}=v+x \frac{d v}{d x} \tag{3} $$

Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1), we get

$$ v+x \frac{d v}{d x}=\frac{v \cos v+1}{\cos v} $$

$$\text{ or }\qquad x \frac{d v}{d x}=\frac{v \cos v+1}{\cos v}-v $$

$$\text{ or }\qquad x \frac{d v}{d x}=\frac{1}{\cos v} $$

$$\text{ or }\qquad \cos v d v=\frac{d x}{x} $$

$$\text{ Therefore }\qquad \int \cos v d v=\int \frac{1}{x} d x $$

$\text{ or }\qquad \sin v=\log |x|+\log |\mathrm{C}| $ $ \sin v=\log |\mathrm{C} x| $

Replacing $v$ by $\frac{y}{x}$, we get

$$ \sin (\frac{y}{x})=\log |C x| $$

which is the general solution of the differential equation (1).

Example 12 Show that the differential equation $2 y e^{\frac{x}{y}} d x+(y-2 x e^{\frac{x}{y}}) d y=0$ is homogeneous and find its particular solution, given that, $x=0$ when $y=1$.

Solution The given differential equation can be written as

$$\text{ Let }\qquad \begin{equation*} \frac{d x}{d y}=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}} \tag{1} \end{equation*} $$

$$ \mathrm{F}(x, y)=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}} \text { तब } \mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left(2 x e^{\frac{x}{y}}-y\right)}{\lambda\left(2 y e^{\frac{x}{y}}\right)}=\lambda^{\circ}[\mathrm{F}(x, y)] $$

Thus, $F(x, y)$ is a homogeneous function of degree zero.

Therefore, the given differential equation is a homogeneous differential equation.

To solve it, we make the substitution $ x=v y $

Differentiating equation (2) with respect to $y$, we get

$$ \frac{d x}{d y}=v+y \frac{d v}{d y} \tag{2} $$

Substituting the value of $x$ and $\frac{d x}{d y}$ in equation (1), we get

or $ \begin{aligned} v+y \frac{d v}{d y} & =\frac{2 v e^{v}-1}{2 e^{v}} \\ y \frac{d v}{d y} & =\frac{2 v e^{v}-1}{2 e^{v}}-v \\ y \frac{d v}{d y} & =-\frac{1}{2 e^{v}} \\ 2 e^{v} d v & =\frac{-d y}{y} \\ \int 2 e^{v} \cdot d v & =-\int \frac{d y}{y} \\ 2 e^{v} & =-\log |y|+C \end{aligned} $

or and replacing $v$ by $\frac{x}{y}$, we get

$$ 2 e^{\frac{x}{y}}+\log |y|=C \tag{3} $$

Substituting $x=0$ and $y=1$ in equation (3), we get

$$ 2 e^{0}+\log |1|=C \Rightarrow C=2 $$

Substituting the value of $C$ in equation (3), we get

$$ 2 e^{\frac{x}{y}}+\log |y|=2 $$

which is the particular solution of the given differential equation.

Example 13 Show that the family of curves for which the slope of the tangent at any point $(x, y)$ on it is $\frac{x^{2}+y^{2}}{2 x y}$, is given by $x^{2}-y^{2}=c x$.

Solution We know that the slope of the tangent at any point on a curve is $\frac{d y}{d x}$.

$$ \begin{equation*} \text{ Therefore, } \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y} \text { or } \frac{d y}{d x}=\frac{1+\frac{y^{2}}{x^{2}}}{\frac{2 y}{x}} \tag{1} \end{equation*} $$

Clearly, (1) is a homogenous differential equation. To solve it we make substitution $ y=v x $ Differentiating $y=v x$ with respect to $x$,

we get $$ \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \text { or } v+x \frac{d v}{d x}=\frac{1+v^{2}}{2 v} \\ & x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \text { or } \frac{2 v}{1-v^{2}} d v=\frac{d x}{x} \text { or } \frac{2 v}{v^{2}-1} d v=-\frac{d x}{x} \end{aligned} $$

$$ \text{ Therefore } \qquad \int \frac{2 v}{v^{2}-1} d v=-\int \frac{1}{x} d x $$

$$ \text{ or } \qquad \log |v^{2}-1|=-\log |x|+\log |C_1| $$

$$ \text{ or } \qquad \log |(v^{2}-1)(x)|=\log |C_1| $$

$$ \text{ or } \qquad (v^{2}-1) x= \pm C_1 $$

Replacing $v$ by $\frac{y}{x}$, we get

$$ (\frac{y^{2}}{x^{2}}-1) x= \pm C_1 $$

$$ \text{ or } \qquad (y^{2}-x^{2})= \pm C_1 x \text{ or } x^{2}-y^{2}=C x $$

EXERCISE 9.4

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

1. $(x^{2}+x y) d y=(x^{2}+y^{2}) d x$

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Solution

The given differential equation i.e., $(x^{2}+x y) d y=(x^{2}+y^{2}) d x$ can be written as:

$\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}$

Let $F(x, y)=\frac{x^{2}+y^{2}}{x^{2}+x y}$.

Now, $F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}=\frac{x^{2}+y^{2}}{x^{2}+x y}=\lambda^{0} \cdot F(x, y)$

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{x^{2}+(v x)^{2}}{x^{2}+x(v x)} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{1+v^{2}}{1+v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}-v=\frac{(1+v^{2})-v(1+v)}{1+v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1-v}{1+v} \\ & \Rightarrow(\frac{1+v}{1-v})=d v=\frac{d x}{x} \\ & \Rightarrow(\frac{2-1+v}{1-v}) d v=\frac{d x}{x} \\ & \Rightarrow(\frac{2}{1-v}-1) d v=\frac{d x}{x} \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & -2 \log (1-v)-v=\log x-\log k \\ & \Rightarrow v=-2 \log (1-v)-\log x+\log k \\ & \Rightarrow v=\log [\frac{k}{x(1-v)^{2}}] \\ & \Rightarrow \frac{y}{x}=\log [\frac{k}{x(1-\frac{y}{x})^{2}}] \\ & \Rightarrow \frac{y}{x}=\log [\frac{k x}{(x-y)^{2}}] \\ & \Rightarrow \frac{k x}{(x-y)^{2}}=e^{\frac{y}{x}} \\ & \Rightarrow(x-y)^{2}=k x e^{-\frac{y}{x}} \end{aligned} $

This is the required solution of the given differential equation.

2. $y^{\prime}=\frac{x+y}{x}$

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Solution

The given differential equation is:

$y^{\prime}=\frac{x+y}{x}$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x}$

Let $F(x, y)=\frac{x+y}{x}$.

Now, $F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}=\lambda^{0} F(x, y)$

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x}$

$\Rightarrow v+x \frac{d v}{d x}=1+v$

$x \frac{d v}{d x}=1$

$\Rightarrow d v=\frac{d x}{x}$

Integrating both sides, we get:

$v=\log x+C$

$\Rightarrow \frac{y}{x}=\log x+C$

$\Rightarrow y=x \log x+C x$

This is the required solution of the given differential equation.

3. $(x-y) d y-(x+y) d x=0$

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Solution

The given differential equation is:

$(x-y) d y-(x+y) d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x-y}$

Let $F(x, y)=\frac{x+y}{x-y}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}=\lambda^{0} \cdot F(x, y)$

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v}$

$x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1-v}$

$\Rightarrow \frac{1-v}{(1+v^{2})} d v=\frac{d x}{x}$

$\Rightarrow(\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}}) d v=\frac{d x}{x}$

Integrating both sides, we get:

$ \begin{aligned} & \tan ^{-1} v-\frac{1}{2} \log (1+v^{2})=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2} \log [1+(\frac{y}{x})^{2}]=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2} \log (\frac{x^{2}+y^{2}}{x^{2}})=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2}[\log (x^{2}+y^{2})-\log x^{2}]=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})=\frac{1}{2} \log (x^{2}+y^{2})+C \end{aligned} $

This is the required solution of the given differential equation.

4. $(x^{2}-y^{2}) d x+2 x y d y=0$

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Solution

The given differential equation is:

$(x^{2}-y^{2}) d x+2 x y d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-(x^{2}-y^{2})}{2 x y}$

Let $F(x, y)=\frac{-(x^{2}-y^{2})}{2 x y}$.

$\therefore F(\lambda x, \lambda y)=[\frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)}]=\frac{-(x^{2}-y^{2})}{2 x y}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$ $\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=-[\frac{x^{2}-(v x)^{2}}{2 x \cdot(v x)}]$

$v+x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v=\frac{v^{2}-1-2 v^{2}}{2 v}$

$\Rightarrow x \frac{d v}{d x}=-\frac{(1+v^{2})}{2 v}$

$\Rightarrow \frac{2 v}{1+v^{2}} d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log (1+v^{2})=-\log x+\log C=\log \frac{C}{x}$

$\Rightarrow 1+v^{2}=\frac{C}{x}$

$\Rightarrow[1+\frac{y^{2}}{x^{2}}]=\frac{C}{x}$

$\Rightarrow x^{2}+y^{2}=C x$

This is the required solution of the given differential equation.

5. $x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$

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Solution

The given differential equation is:

$x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$ $\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$

Let $F(x, y)=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda x)(\lambda y)}{(\lambda x)^{2}}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x^{2}-2(v x)^{2}+x \cdot(v x)}{x^{2}}$

$\Rightarrow v+x \frac{d v}{d x}=1-2 v^{2}+v$

$\Rightarrow x \frac{d v}{d x}=1-2 v^{2}$

$\Rightarrow \frac{d v}{1-2 v^{2}}=\frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot \frac{d v}{\frac{1}{2}-v^{2}}=\frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot[\frac{d v}{(\frac{1}{\sqrt{2}})^{2}-v^{2}}]=\frac{d x}{x}$

Integrating both sides, we get:

$ \begin{aligned} & \frac{1}{2} \cdot \frac{1}{2 \times \frac{1}{\sqrt{2}}} \log |\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}|=\log |x|+C \\ & \Rightarrow \frac{1}{2 \sqrt{2}} \log |\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}|=\log |x|+C \\ & \Rightarrow \frac{1}{2 \sqrt{2}} \log |\frac{x+\sqrt{2} y}{x-\sqrt{2} y}|=\log |x|+C \end{aligned} $

This is the required solution for the given differential equation.

6. $x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

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Solution

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

$\Rightarrow x d y=[y+\sqrt{x^{2}+y^{2}}] d x$

$\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$

Let $F(x, y)=\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+(v x)^{2}}}{x}$

$\Rightarrow v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}}$

$\Rightarrow \frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}$

Integrating both sides, we get:

$\log |v+\sqrt{1+v^{2}}|=\log |x|+\log C$

$\Rightarrow \log |\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}|=\log |C x|$

$\Rightarrow \log |\frac{y+\sqrt{x^{2}+y^{2}}}{x}|=\log |C x|$

$\Rightarrow y+\sqrt{x^{2}+y^{2}}=C x^{2}$

This is the required solution of the given differential equation.

7. ${x \cos (\frac{y}{x})+y \sin (\frac{y}{x})} y d x={y \sin (\frac{y}{x})-x \cos (\frac{y}{x})} x d y$

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Solution

The given differential equation is:

$$x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right) y d x=y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right) x d y$$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d y}{d x}=v+x=\frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{(x \cos v+v x \sin v) \cdot v x}{(v x \sin v-x \cos v) \cdot x} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v}-v \\ & \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v-v^{2} \sin v+v \cos v}{v \sin v-\cos v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\ & \Rightarrow[\frac{v \sin v-\cos v}{v \cos v}] d v=\frac{2 d x}{x} \\ & \Rightarrow(\tan v-\frac{1}{v}) d v=\frac{2 d x}{x} \end{aligned} $

Integrating both sides, we get:

$\log (\sec v)-\log v=2 \log x+\log C$

$\Rightarrow \log (\frac{\sec v}{v})=\log (C x^{2})$

$\Rightarrow(\frac{\sec v}{v})=C x^{2}$

$\Rightarrow \sec v=C x^{2} v$

$\Rightarrow \sec (\frac{y}{x})=C \cdot x^{2} \cdot \frac{y}{x}$

$\Rightarrow \sec (\frac{y}{x})=C x y$

$\Rightarrow \cos (\frac{y}{x})=\frac{1}{C x y}=\frac{1}{C} \cdot \frac{1}{x y}$

$\Rightarrow x y \cos (\frac{y}{x})=k \quad(k=\frac{1}{C})$

This is the required solution of the given differential equation.

8. $x \frac{d y}{d x}-y+x \sin (\frac{y}{x})=0$

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Solution

$x \frac{d y}{d x}-y+x \sin (\frac{y}{x})=0$

$\Rightarrow x \frac{d y}{d x}=y-x \sin (\frac{y}{x})$

$\Rightarrow \frac{d y}{d x}=\frac{y-x \sin (\frac{y}{x})}{x}$

Let $F(x, y)=\frac{y-x \sin (\frac{y}{x})}{x}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y-\lambda x \sin (\frac{\lambda y}{\lambda x})}{\lambda x}=\frac{y-x \sin (\frac{y}{x})}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x-x \sin v}{x}$

$\Rightarrow v+x \frac{d v}{d x}=v-\sin v$

$\Rightarrow-\frac{d v}{\sin v}=\frac{d x}{x}$

$\Rightarrow cosec v d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log |cosec v-\cot v|=-\log x+\log C=\log \frac{C}{x}$

$\Rightarrow cosec(\frac{y}{x})-\cot (\frac{y}{x})=\frac{C}{x}$

$\Rightarrow \frac{1}{\sin (\frac{y}{x})}-\frac{\cos (\frac{y}{x})}{\sin (\frac{y}{x})}=\frac{C}{x}$

$\Rightarrow x[1-\cos (\frac{y}{x})]=C \sin (\frac{y}{x})$

This is the required solution of the given differential equation.

9. $y d x+x \log (\frac{y}{x}) d y-2 x d y=0$

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Solution

$y d x+x \log (\frac{y}{x}) d y-2 x d y=0$

$\Rightarrow y d x=[2 x-x \log (\frac{y}{x})] d y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{2 x-x \log (\frac{y}{x})}$

Let $F(x, y)=\frac{y}{2 x-x \log (\frac{y}{x})}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{2(\lambda x)-(\lambda x) \log (\frac{\lambda y}{\lambda x})}=\frac{y}{2 x-\log (\frac{y}{x})}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$ $\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x}{2 x-x \log v}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v$

$\Rightarrow x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v}$

$\Rightarrow \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x}$

$\Rightarrow[\frac{1+(1-\log v)}{v(\log v-1)}] d v=\frac{d x}{x}$

$\Rightarrow[\frac{1}{v(\log v-1)}-\frac{1}{v}] d v=\frac{d x}{x}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\int \frac{1}{x} d x \\ & \Rightarrow \int \frac{d v}{v(\log v-1)}-\log v=\log x+\log C \tag{2} \end{align*} $$

$\Rightarrow$ Let $\log v-1=t$

$\Rightarrow \frac{d}{d v}(\log v-1)=\frac{d t}{d v}$

$\Rightarrow \frac{1}{v}=\frac{d t}{d v}$

$\Rightarrow \frac{d v}{v}=d t$

Therefore, equation (1) becomes:

$\Rightarrow \int \frac{d t}{t}-\log v=\log x+\log C$

$\Rightarrow \log t-\log (\frac{y}{x})=\log (C x)$

$\Rightarrow \log [\log (\frac{y}{x})-1]-\log (\frac{y}{x})=\log (C x)$

$\Rightarrow \log [\frac{\log (\frac{y}{x})-1}{\frac{y}{x}}]=\log (C x)$

$\Rightarrow \frac{x}{y}[\log (\frac{y}{x})-1]=C x$

$\Rightarrow \log (\frac{y}{x})-1=C y$

This is the required solution of the given differential equation.

10. $(1+e^{\frac{x}{y}}) d x+e^{\frac{x}{y}}(1-\frac{x}{y}) d y=0$

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Solution

$(1+e^{\frac{x}{y}}) d x+e^{\frac{x}{y}}(1-\frac{x}{y}) d y=0$

$\Rightarrow(1+e^{\frac{x}{y}}) d x=-e^{\frac{x}{y}}(1-\frac{x}{y}) d y$ $\Rightarrow \frac{d x}{d y}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$

Let $F(x, y)=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$.

$\therefore F(\lambda x, \lambda y)=\frac{-e^{\frac{\lambda x}{\lambda y}}(1-\frac{\lambda x}{\lambda y})}{1+e^{\frac{\lambda x}{\lambda y}}}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$x=v y$

$\Rightarrow \frac{d}{d y}(x)=\frac{d}{d y}(v y)$

$\Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}$

Substituting the values of $x$ and $\frac{d x}{d y}$ in equation (1), we get:

$v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}}{1+e^{v}}-v$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}-v-v e^{v}}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=-[\frac{v+e^{v}}{1+e^{v}}]$

$\Rightarrow[\frac{1+e^{v}}{v+e^{v}}] d v=-\frac{d y}{y}$

Integrating both sides, we get: $\Rightarrow \log (v+e^{v})=-\log y+\log C=\log (\frac{C}{y})$

$\Rightarrow[\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}$

$\Rightarrow x+y e^{\frac{x}{y}}=C$

This is the required solution of the given differential equation.

For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

11. $(x+y) d y+(x-y) d x=0 ; y=1$ when $x=1$

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Solution

$(x+y) d y+(x-y) d x=0$

$\Rightarrow(x+y) d y=-(x-y) d x$

$\Rightarrow \frac{d y}{d x}=\frac{-(x-y)}{x+y}$

Let $F(x, y)=\frac{-(x-y)}{x+y}$.

$\therefore F(\lambda x, \lambda y)=\frac{-(\lambda x-\lambda y)}{\lambda x-\lambda y}=\frac{-(x-y)}{x+y}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get: $v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v-1}{v+1}$

$\Rightarrow x \frac{d v}{d x}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1}$

$\Rightarrow x \frac{d v}{d x}=\frac{v-1-v^{2}-v}{v+1}=\frac{-(1+v^{2})}{v+1}$

$\Rightarrow \frac{(v+1)}{1+v^{2}} d v=-\frac{d x}{x}$

$\Rightarrow[\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}] d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\frac{1}{2} \log (1+v^{2})+\tan ^{-1} v=-\log x+k$

$\Rightarrow \log (1+v^{2})+2 \tan ^{-1} v=-2 \log x+2 k$

$\Rightarrow \log [(1+v^{2}) \cdot x^{2}]+2 \tan ^{-1} v=2 k$

$\Rightarrow \log [(1+\frac{y^{2}}{x^{2}}) \cdot x^{2}]+2 \tan ^{-1} \frac{y}{x}=2 k$

$\Rightarrow \log (x^{2}+y^{2})+2 \tan ^{-1} \frac{y}{x}=2 k$

Now, $y=1$ at $x=1$.

$\Rightarrow \log 2+2 \tan ^{-1} 1=2 k$

$\Rightarrow \log 2+2 \times \frac{\pi}{4}=2 k$

$\Rightarrow \frac{\pi}{2}+\log 2=2 k$

Substituting the value of $2 k$ in equation (2), we get:

$\log (x^{2}+y^{2})+2 \tan ^{-1}(\frac{y}{x})=\frac{\pi}{2}+\log 2$

This is the required solution of the given differential equation.

12. $x^{2} d y+(x y+y^{2}) d x=0 ; y=1$ when $x=1$

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Solution

$x^{2} d y+(x y+y^{2}) d x=0$

$\Rightarrow x^{2} d y=-(x y+y^{2}) d x$

$\Rightarrow \frac{d y}{d x}=\frac{-(x y+y^{2})}{x^{2}}$

Let $F(x, y)=\frac{-(x y+y^{2})}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{[\lambda x \cdot \lambda y+(\lambda y)^{2}]}{(\lambda x)^{2}}=\frac{-(x y+y^{2})}{x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{-[x \cdot v x+(v x)^{2}]}{x^{2}}=-v-v^{2} \\ & \Rightarrow x \frac{d v}{d x}=-v^{2}-2 v=-v(v+2) \\ & \Rightarrow \frac{d v}{v(v+2)}=-\frac{d x}{x} \\ & \Rightarrow \frac{1}{2}[\frac{(v+2)-v}{v(v+2)}] d v=-\frac{d x}{x} \\ & \Rightarrow \frac{1}{2}[\frac{1}{v}-\frac{1}{v+2}] d v=-\frac{d x}{x} \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \frac{1}{2}[\log v-\log (v+2)]=-\log x+\log C \\ & \Rightarrow \frac{1}{2} \log (\frac{v}{v+2})=\log \frac{C}{x} \\ & \Rightarrow \frac{v}{v+2}=(\frac{C}{x})^{2} \\ & \Rightarrow \frac{\frac{y}{y}}{\frac{y}{x}+2}=(\frac{C}{x})^{2} \\ & \Rightarrow \frac{y}{y+2 x}=\frac{C^{2}}{x^{2}} \\ & \Rightarrow \frac{x^{2} y}{y+2 x}=C^{2} \tag{2} \end{align*} $$

Now, $y=1$ at $x=1$.

$\Rightarrow \frac{1}{1+2}=C^{2}$

$\Rightarrow C^{2}=\frac{1}{3}$

Substituting $C^{2}=\frac{1}{3}$ in equation (2), we get:

$ \begin{aligned} & \frac{x^{2} y}{y+2 x}=\frac{1}{3} \\ & \Rightarrow y+2 x=3 x^{2} y \end{aligned} $

This is the required solution of the given differential equation.

13. $[x \sin ^{2}(\frac{y}{x})-y] d x+x d y=0 ; y=\frac{\pi}{4}$ when $x=1$

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Solution

$[x \sin ^{2}(\frac{y}{x})-y] d x+x d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}$

Let $F(x, y)=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}$.

$\therefore F(\lambda x, \lambda y)=\frac{-[\lambda x \cdot \sin ^{2}(\frac{\lambda x}{\lambda y})-\lambda y]}{\lambda x}=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x=\frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{-[x \sin ^{2} v-v x]}{x}$

$\Rightarrow v+x \frac{d v}{d x}=-[\sin ^{2} v-v]=v-\sin ^{2} v$

$\Rightarrow x \frac{d v}{d x}=-\sin ^{2} v$

$\Rightarrow \frac{d v}{\sin ^{2} v}=-\frac{d x}{d x}$

$\Rightarrow cosec^{2} v d v=-\frac{d x}{x}$

Integrating both sides, we get: $-\cot v=-\log |x|-C$

$\Rightarrow \cot v=\log |x|+C$

$\Rightarrow \cot (\frac{y}{x})=\log |x|+\log C$

$\Rightarrow \cot (\frac{y}{x})=\log |Cx|$

Now, $y=\frac{\pi}{4}$ at $x=1$

$\Rightarrow \cot (\frac{\pi}{4})=\log |C|$

$\Rightarrow 1=\log C$

$\Rightarrow C=e^{1}=e$

Substituting $C=e$ in equation (2), we get:

$\cot (\frac{y}{x})=\log |e x|$

This is the required solution of the given differential equation.

14. $\frac{d y}{d x}-\frac{y}{x}+cosec(\frac{y}{x})=0 ; y=0$ when $x=1$

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Solution

$\frac{d y}{d x}-\frac{y}{x}+cosec(\frac{y}{x})=0$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x}-cosec(\frac{y}{x})$

Let $F(x, y)=\frac{y}{x}-cosec(\frac{y}{x})$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}-cosec(\frac{\lambda y}{\lambda x})$

$\Rightarrow F(\lambda x, \lambda y)=\frac{y}{x}-cosec(\frac{y}{x})=F(x, y)=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=v-cosec v$

$\Rightarrow-\frac{d v}{cosec v}=-\frac{d x}{x}$

$\Rightarrow-\sin v d v=\frac{d x}{x}$

Integrating both sides, we get:

$\cos v=\log x+\log C=\log |C x|$

$\Rightarrow \cos (\frac{y}{x})=\log |C x|$

This is the required solution of the given differential equation.

Now, $y=0$ at $x=1$.

$\Rightarrow \cos (0)=\log C$

$\Rightarrow 1=\log C$

$\Rightarrow C=e^{1}=e$

Substituting $C=e$ in equation (2), we get:

$\cos (\frac{y}{x})=\log |(e x)|$

This is the required solution of the given differential equation.

15. $2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0 ; y=2$ when $x=1$

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Solution

$2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0$

$\Rightarrow 2 x^{2} \frac{d y}{d x}=2 x y+y^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$

Let $F(x, y)=\frac{2 x y+y^{2}}{2 x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{2(\lambda x)(\lambda y)+(\lambda y)^{2}}{2(\lambda x)^{2}}=\frac{2 x y+y^{2}}{2 x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{2 x(v x)+(v x)^{2}}{2 x^{2}}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{2 v+v^{2}}{2}$

$\Rightarrow v+x \frac{d v}{d x}=v+\frac{v^{2}}{2}$

$\Rightarrow \frac{2}{v^{2}} d v=\frac{d x}{x}$

Integrating both sides, we get:

$$ \begin{align*} & 2 \cdot \frac{v^{-2+1}}{-2+1}=\log |x|+C \\ & \Rightarrow-\frac{2}{v}=\log |x|+C \\ & \Rightarrow-\frac{2}{\frac{y}{x}}=\log |x|+C \\ & \Rightarrow-\frac{2 x}{y}=\log |x|+C \\ & \text{ Now, } y=2 \text{ at } x=1 . \tag{2}\\ & \Rightarrow-1=\log (1)+C \\ & \Rightarrow C=-1 \end{align*} $$

Substituting $C=-1$ in equation (2), we get:

$ \begin{aligned} & -\frac{2 x}{y}=\log |x|-1 \\ & \Rightarrow \frac{2 x}{y}=1-\log |x| \\ & \Rightarrow y=\frac{2 x}{1-\log |x|},(x \neq 0, x \neq e) \end{aligned} $

This is the required solution of the given differential equation.

16. A homogeneous differential equation of the from $\frac{d x}{d y}=h(\frac{x}{y})$ can be solved by making the substitution.

$\quad\quad$(A) $y=v x$

$\quad\quad$(B) $v=y x$

$\quad\quad$(C) $x=v y$

$\quad\quad$(D) $x=v$

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Solution

For solving the homogeneous equation of the form $\frac{d x}{d y}=h(\frac{x}{y})$, we need to make the substitution as $x=v y$.

Hence, the correct answer is $C$.

17. Which of the following is a homogeneous differential equation?

$\quad\quad$(A) $(4 x+6 y+5) d y-(3 y+2 x+4) d x=0$

$\quad\quad$(B) $(x y) d x-(x^{3}+y^{3}) d y=0$

$\quad\quad$(C) $(x^{3}+2 y^{2}) d x+2 x y d y=0$

$\quad\quad$(D) $y^{2} d x+(x^{2}-x y-y^{2}) d y=0$

Show Answer

Solution

Function $F(x, y)$ is said to be the homogenous function of degree $n$, if $F(\lambda x, \lambda y)=\lambda^{n} F(x, y)$ for any non-zero constant $(\lambda)$.

Consider the equation given in alternativeD:

$y^{2} d x+(x^{2}-x y-y^{2}) d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}=\frac{y^{2}}{y^{2}+x y-x^{2}}$

Let $F(x, y)=\frac{y^{2}}{y^{2}+x y-x^{2}}$.

$\Rightarrow F(\lambda x, \lambda y)=\frac{(\lambda y)^{2}}{(\lambda y)^{2}+(\lambda x)(\lambda y)-(\lambda x)^{2}}$

$=\frac{\lambda^{2} y^{2}}{\lambda^{2}(y^{2}+x y-x^{2})}$

$=\lambda^{0}(\frac{y^{2}}{y^{2}+x y-x^{2}})$

$=\lambda^{0} \cdot F(x, y)$

Hence, the differential equation given in alternative $\mathbf{D}$ is a homogenous equation.

9.4.3 Linear differential equations

$$ \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} $$

A differential equation of the from where, $P$ and $Q$ are constants or functions of $x$ only, is known as a first order linear differential equation. Some examples of the first order linear differential equation are

$ \begin{aligned} \frac{d y}{d x}+y & =\sin x \\ \frac{d y}{d x}+(\frac{1}{x}) y & =e^{x} \\ \frac{d y}{d x}+(\frac{y}{x \log x}) & =\frac{1}{x} \end{aligned} $

Another form of first order linear differential equation is

$\frac{d x}{d y}+\mathrm{P} _{1} x=\mathrm{Q} _{1}$

where, $\mathrm{P} _{1}$ and $\mathrm{Q} _{1}$ are constants or functions of $y$ only. Some examples of this type of differential equation are

$ \begin{matrix} \frac{d x}{d y}+x=\cos y \\ \frac{d x}{d y}+\frac{-2 x}{y}=y^{2} e^{-y} \end{matrix} $

To solve the first order linear differential equation of the type

$$ \begin{equation*} \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \tag{1} \end{equation*} $$

Multiply both sides of the equation by a function of $x$ say $g(x)$ to get

$$ \begin{equation*} g(x) \frac{d y}{d x}+\mathrm{P} \cdot g(x) y=\mathrm{Q} \cdot g(x) \tag{2} \end{equation*} $$

Choose $g(x)$ in such a way that R.H.S. becomes a derivative of $y . g(x)$.

$$ \text{ i.e. } \qquad g(x) \frac{d y}{d x}+\mathrm{P} \cdot g(x) y=\frac{d}{d x}[y \cdot g(x)] $$

$$ \text{ or } \qquad g(x) \frac{d y}{d x}+\mathrm{P} \cdot g(x) y=g(x) \frac{d y}{d x}+y g^{\prime}(x) $$

$\Rightarrow \quad$ P. $g(x)=g^{\prime}(x)$

$$ \text{ or } \qquad \mathrm{P}=\frac{g^{\prime}(x)}{g(x)} $$

Integrating both sides with respect to $x$, we get

$ \text{ or } \qquad \begin{aligned} or \quad\quad\quad\quad\int P d x & =\int \frac{g^{\prime}(x)}{g(x)} d x \\ \quad\quad\quad\quad\int P \cdot d x & =\log (g(x)) \\ g(x) & =e^{\int P d x} \end{aligned} $

On multiplying the equation (1) by $g(x)=e^{\int P d x}$, the L.H.S. becomes the derivative of some function of $x$ and $y$. This function $g(x)=e^{\int P d x}$ is called Integrating Factor (I.F.) of the given differential equation.

Substituting the value of $g(x)$ in equation (2), we get

$ e^{\int P d x} \frac{d y}{d x}+P e^{\int P d x} y=Q \cdot e^{\int P d x} $

or $ \frac{d}{d x}(y e^{\int P d x})=Q e^{\int P d x} $

Integrating both sides with respect to $x$, we get

or $ \begin{aligned} y \cdot e^{\int P d x} & =\int(Q \cdot e^{\int P d x}) d x \\ y & =e^{-\int P d x} \cdot \int(Q \cdot e^{\int P d x}) d x+C \end{aligned} $

which is the general solution of the differential equation.

Steps involved to solve first order linear differential equation:

(i) Write the given differential equation in the form $\frac{d y}{d x}+P y=Q$ where $P, Q$ are constants or functions of $x$ only.

(ii) Find the Integrating Factor (I.F) $=e^{\int P d x}$.

(iii) Write the solution of the given differential equation as

$$ y(I . F)=\int(Q \times I . F) d x+C $$

In case, the first order linear differential equation is in the form $\frac{d x}{d y}+P_1 x=Q_1$, where, $P_1$ and $Q_1$ are constants or functions of $y$ only. Then I.F $=e^{P_1 d y}$ and the solution of the differential equation is given by

$$ x \cdot(I \cdot F)=\int(Q_1 \times I \cdot F) d y+C $$

Example 14 Find the general solution of the differential equation $\frac{d y}{d x}-y=\cos x$.

Solution Given differential equation is of the form

$$ \frac{d y}{d x}+P y=Q \text{, where } P=-1 \text{ and } Q=\cos x $$

Therefore $\text{ I. } F=e^{\int-1 d x}=e^{-x}$

Multiplying both sides of equation by I.F, we get

$$ \begin{aligned} e^{-x} \frac{d y}{d x}-e^{-x} y & =e^{-x} \cos x \\ \frac{d y}{d x}(y e^{-x}) & =e^{-x} \cos x \end{aligned} $$

On integrating both sides with respect to $x$, we get

$$ \begin{equation*} y e^{-x}=\int e^{-x} \cos x d x+\mathrm{C} \tag{1} \end{equation*} $$

$$ \begin{aligned} \text{ Let }\qquad & =-\cos x e^{-x}-\int \sin x e^{-x} d x \\ & =-\cos x e^{-x}-[\sin x(-e^{-x})-\int \cos x(-e^{-x}) d x] \\ & =-\cos x e^{-x}+\sin x e^{-x}-\int \cos x e^{-x} d x \\ I & =-e^{-x} \cos x+\sin x e^{-x}-I \\ 2 I & =(\sin x-\cos x) e^{-x} \\ I & =\frac{(\sin x-\cos x) e^{-x}}{2} \end{aligned} $$

Substituting the value of I in equation (1), we get

$ \begin{aligned} \text{ or }\qquad y e^{-x} & =(\frac{\sin x-\cos x}{2}) e^{-x}+C \\ y & =(\frac{\sin x-\cos x}{2})+C e^{x} \end{aligned} $

which is the general solution of the given differential equation.

Example 15 Find the general solution of the differential equation $x \frac{d y}{d x}+2 y=x^{2}(x \neq 0)$.

Solution The given differential equation is

$ x \frac{d y}{d x}+2 y=x^{2} $

Dividing both sides of equation (1) by $x$, we get

$$ \frac{d y}{d x}+\frac{2}{x} y=x $$

which is a linear differential equation of the type $\frac{d y}{d x}+P y=Q$, where $P=\frac{2}{x}$ and $Q=x$.

$\text{ So } \text{ I.F }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}[\text{ as } e^{\log f(x)}=f(x)] $

Therefore, solution of the given equation is given by

$ \begin{matrix} y \cdot x^{2}=\int(x)(x^{2}) d x+C=\int x^{3} d x+C \\ y=\frac{x^{2}}{4}+C x^{-2} \end{matrix} $

which is the general solution of the given differential equation.

Example 16 Find the general solution of the differential equation $y d x-(x+2 y^{2}) d y=0$.

Solution The given differential equation can be written as

$$ \frac{d x}{d y}-\frac{x}{y}=2 y $$

This is a linear differential equation of the type $\frac{d x}{d y}+P_1 x=Q_1$, where $P_1=-\frac{1}{y}$ and $Q_1=2 y$. Therefore I.F $=e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log (y)^{-1}}=\frac{1}{y}$

Hence, the solution of the given differential equation is

$ \begin{aligned} \text{ or }\qquad x \frac{1}{y} & =\int(2 y)(\frac{1}{y}) d y+C \\ \frac{x}{y} & =\int(2 d y)+C \\ \frac{x}{y} & =2 y+C \\ x & =2 y^{2}+C y \end{aligned} $

which is a general solution of the given differential equation.

Example 17 Find the particular solution of the differential equation given that $y=0$ when $x=\frac{\pi}{2}$.

$$ \frac{d y}{d x}+y \cot x=2 x+x^{2} \cot x(x \neq 0) $$

Solution The given equation is a linear differential equation of the type $\frac{d y}{d x}+P y=Q$, where $P=\cot x$ and $Q=2 x+x^{2} \cot x$. Therefore

$ \text{ I.F }=e^{\int \cot x d x}=e^{\log \sin x}=\sin x $

Hence, the solution of the differential equation is given by

$ y \cdot \sin x=\int(2 x+x^{2} \cot x) \sin x d x+C $

$ \begin{aligned} & \text{ or } \quad y \sin x=\int 2 x \sin x d x+\int x^{2} \cos x d x+C \\ & \text{ or } \quad y \sin x=\sin x(\frac{2 x^{2}}{2})-\int \cos x(\frac{2 x^{2}}{2}) d x+\int x^{2} \cos x d x+C \\ & \text{ or } \quad y \sin x=x^{2} \sin x-\int x^{2} \cos x d x+\int x^{2} \cos x d x+C \\ & \text{ or } \quad y \sin x=x^{2} \sin x+C \end{aligned} $

Substituting $y=0$ and $x=\frac{\pi}{2}$ in equation (1), we get

$$ 0=\left(\frac{\pi}{2}\right)^{2} \sin \left(\frac{\pi}{2}\right)+C $$

$$ \text{ or }\qquad \mathrm{C}=\frac{-\pi^{2}}{4} $$ Substituting the value of $C$ in equation (1), we get

$ \begin{aligned} \text{ or }\qquad y \sin x & =x^{2} \sin x-\frac{\pi^{2}}{4} \\ y & =x^{2}-\frac{\pi^{2}}{4 \sin x}(\sin x \neq 0) \end{aligned} $

which is the particular solution of the given differential equation.

Example 18 Find the equation of a curve passing through the point $(0,1)$. If the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$ coordinate (abscissa) and the product of the $x$ coordinate and $y$ coordinate (ordinate) of that point.

Solution We know that the slope of the tangent to the curve is $\frac{d y}{d x}$.

$$ \begin{align*} \text{ Therefore, }\qquad & \frac{d y}{d x}=x+x y \\ & \frac{d y}{d x}-x y=x \tag{1} \end{align*} $$

This is a linear differential equation of the type $\frac{d y}{d x}+P y=Q$, where $P=-x$ and $Q=x$.

$ \text{ Therefore, }\qquad \text{ I. } F=e^{\int-x d x}=e^{\frac{-x^{2}}{2}} $

Hence, the solution of equation is given by

$$ \begin{equation*} \text{ Let }\qquad y \cdot e^{\frac{-x^{2}}{2}}=\int(x)\left(e^{\frac{-x^{2}}{2}}\right) d x+\mathrm{C} \tag{2} \end{equation*} $$

$$ \mathrm{I}=\int(x) e^{\frac{-x^{2}}{2}} d x $$

Let $\frac{-x^{2}}{2}=t$, then $-x d x=d t$ or $x d x=-d t$.

Therefore, $\quad I=-\int e^{t} d t=-e^{t}=-e^{\frac{-x^{2}}{2}}$

Substituting the value of I in equation (2), we get

$$ \begin{align*} \text{ or } & y e^{\frac{-x^{2}}{2}}=-e^{\frac{-x^{2}}{2}}+\mathrm{C} \\ & y=-1+\mathrm{C} e^{\frac{x^{2}}{2}} \tag{3} \end{align*} $$

Now (3) represents the equation of family of curves. But we are interested in finding a particular member of the family passing through $(0,1)$. Substituting $x=0$ and $y=1$ in equation (3) we get

$$ 1=-1+C \cdot e^{0} \text{ or } C=2 $$

Substituting the value of $C$ in equation (3), we get

$$ y=-1+2 e^{\frac{x^{2}}{2}} $$

which is the equation of the required curve.

EXERCISE 9.5

For each of the differential equations given in Exercises 1 to 12, find the general solution:

1. $\frac{d y}{d x}+2 y=\sin x$

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Solution

The given differential equation is $\frac{d y}{d x}+2 y=\sin x$.

This is in the form of $\frac{d y}{d x}+p y=Q($ where $p=2$ and $Q=\sin x)$.

Now, I.F $=e^{\int p d x}=e^{\int 2 d x}=e^{2 x}$.

The solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{2 x}=\int \sin x \cdot e^{2 x} d x+C \tag{1} \end{align*} $$

Let $I=\int \sin x \cdot e^{2 x}$.

$\Rightarrow I=\sin x \cdot \int e^{2 x} d x-\int(\frac{d}{d x}(\sin x) \cdot \int e^{2 x} d x) d x$

$\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int(\cos x \cdot \frac{e^{2 x}}{2}) d x$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \int e^{2 x}-\int(\frac{d}{d x}(\cos x) \cdot \int e^{2 x} d x) d x]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \frac{e^{2 x}}{2}-\int[(-\sin x) \cdot \frac{e^{2 x}}{2}] d x]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} \int(\sin x \cdot e^{2 x}) d x$

$\Rightarrow I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I$

$\Rightarrow \frac{5}{4} I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)$

$\Rightarrow I=\frac{e^{2 x}}{5}(2 \sin x-\cos x)$

Therefore, equation (1) becomes:

$y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+C$

$\Rightarrow y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}$

This is the required general solution of the given differential equation.

2. $\frac{d y}{d x}+3 y=e^{-2 x}$

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Solution

$ \frac{d y}{d x}+p y=Q(\text{ where } p=3 \text{ and } Q=e^{-2 x}) \text{. } $

The given differential equation is

Now, I.F $=e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$.

The solution of the given differential equation is given by the relation,

$ \begin{aligned} & y(I . F .)=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{3 x}=\int(e^{-2 x} \times e^{3 x})+C \\ & \Rightarrow y e^{3 x}=\int e^{x} d x+C \\ & \Rightarrow y e^{3 x}=e^{x}+C \\ & \Rightarrow y=e^{-2 x}+C e^{-3 x} \end{aligned} $

This is the required general solution of the given differential equation.

3. $\frac{d y}{d x}+\frac{y}{x}=x^{2}$

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Solution

The given differential equation is:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x}$ and $.Q=x^{2})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x$.

The solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(x)=\int(x^{2} \cdot x) d x+C$

$\Rightarrow x y=\int x^{3} d x+C$

$\Rightarrow x y=\frac{x^{4}}{4}+C$

This is the required general solution of the given differential equation.

4. $\frac{d y}{d x}+(\sec x) y=\tan x(0 \leq x<\frac{\pi}{2})$

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Solution

The given differential equation is:

$\frac{d y}{d x}+p y=Q($ where $p=\sec x$ and $Q=\tan x)$

Now, I.F $=e^{\int p d x}=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$.

The general solution of the given differential equation is given by the relation,

$ y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C $

$\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+C$

$\Rightarrow y(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^{2} x d x+C$

$\Rightarrow y(\sec x+\tan x)=\sec x+\int(\sec ^{2} x-1) d x+C$

$\Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C$

5. $\cos ^{2} x \frac{d y}{d x}+y=\tan x(0 \leq x<\frac{\pi}{2})$

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Solution

We have,

$\begin{aligned} & \cos ^2 x \frac{d y}{d x}+y=\tan x \ & \frac{d y}{d x}+\sec ^2 x(y)=\sec ^2 x \tan x \end{aligned}$

We know that the general equation

$\frac{d y}{d x}+P y=Q$

Here,

$P=\sec ^2 x, Q=\sec ^2 x \tan x$

Since, integrating factor I. $F=e^{\int P d x}$ I. $F=e^{\int \sec ^2 x d x}$ I. $F=e^{\tan x}$

We know that the general solution,

$\mathrm{y} \times \mathrm{I} . \mathrm{F}=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C}$

Therefore,

$y \times e^{\tan x}=\int\left(\sec ^2 x \tan x\right) e^{\tan x} d x+C$

Let $\mathrm{t}=\tan \mathrm{x}$

$\mathrm{dt}=\sec ^2 \mathrm{x} d \mathrm{x}$

Therefore,

$\begin{aligned} & y \times e^t=\int t e^t d t+C \\ & y \times e^t=t e^t-\int 1 e^t d t+C \\ & y \times e^t=t e^t-e^t+C \end{aligned}$

On putting the value of $t$, we get

$y \times e^{\tan x}=\tan x e^{\tan x}-e^{\tan x}+C$

Hence, this is the answer.

6. $x \frac{d y}{d x}+2 y=x^{2} \log x$

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Solution

The given differential equation is:

$x \frac{d y}{d x}+2 y=x^{2} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$

This equation is in the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2}{x}$ and $.Q=x \log x)$

Now, I.F $=e^{\int p d x}=e^{\int_x^{2} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}$.

The general solution of the given differential equation is given by the relation, $y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot x^{2}=\int(x \log x \cdot x^{2}) d x+C$

$\Rightarrow x^{2} y=\int(x^{3} \log x) d x+C$

$\Rightarrow x^{2} y=\log x \cdot \int x^{3} d x-\int[\frac{d}{d x}(\log x) \cdot \int x^{3} d x] d x+C$

$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int(\frac{1}{x} \cdot \frac{x^{4}}{4}) d x+C$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+C$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+C$

$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+C$

$\Rightarrow y=\frac{1}{16} x^{2}(4 \log x-1)+C x^{-2}$

7. $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

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Solution

The given differential equation is:

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}$

This equation is the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x \log x}$ and $Q=\frac{2}{x^{2}}$ )

Now, I.F $=e^{\int \rho d x}=e^{\int \frac{1}{x \log } d x}=e^{\log (\log x)}=\log x$.

The general solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(I . F .)=\int(Q \times I \text{.F. }) d x+C \\ & \Rightarrow y \log x=\int(\frac{2}{x^{2}} \log x) d x+C \tag{1} \end{align*} $$

Now, $\int(\frac{2}{x^{2}} \log x) d x=2 \int(\log x \cdot \frac{1}{x^{2}}) d x$.

$ \begin{aligned} & =2[\log x \cdot \int \frac{1}{x^{2}} d x-\int{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^{2}} d x} d x] \\ & =2[\log x(-\frac{1}{x})-\int(\frac{1}{x} \cdot(-\frac{1}{x})) d x] \\ & =2[-\frac{\log x}{x}+\int \frac{1}{x^{2}} d x] \\ & =2[-\frac{\log x}{x}-\frac{1}{x}] \\ & =-\frac{2}{x}(1+\log x) \end{aligned} $

Substituting the value of $\int(\frac{2}{x^{2}} \log x) d x$ in equation (1), we get:

$y \log x=-\frac{2}{x}(1+\log x)+C$

This is the required general solution of the given differential equation.

8. $(1+x^{2}) d y+2 x y d x=\cot x d x(x \neq 0)$

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Solution

$(1+x^{2}) d y+2 x y d x=\cot x d x$

$\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}$

This equation is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2 x}{1+x^{2}}$ and $.Q=\frac{\cot x}{1+x^{2}})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log (1+x^{2})}=1+x^{2}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(1+x^{2})=\int[\frac{\cot x}{1+x^{2}} \times(1+x^{2})] d x+C$

$\Rightarrow y(1+x^{2})=\int \cot x d x+C$

$\Rightarrow y(1+x^{2})=\log |\sin x|+C$

9. $x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)$

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Solution

$x \frac{d y}{d x}+y-x+x y \cot x=0$

$\Rightarrow x \frac{d y}{d x}+y(1+x \cot x)=x$

$\Rightarrow \frac{d y}{d x}+(\frac{1}{x}+\cot x) y=1$

This equation is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x}+\cot x$ and $.Q=1)$

Now, I.F $=e^{\int \rho d x}=e^{\int(\frac{1}{x}+\cot x) d x}=e^{\log x+\log (\sin x)}=e^{\log (x \sin x)}=x \sin x$.

The general solution of the given differential equation is given by the relation,

$ \begin{aligned} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y(x \sin x)=\int(1 \times x \sin x) d x+C \\ & \Rightarrow y(x \sin x)=\int(x \sin x) d x+C \\ & \Rightarrow y(x \sin x)=x \int \sin x d x-\int[\frac{d}{d x}(x) \cdot \int \sin x d x]+C \\ & \Rightarrow y(x \sin x)=x(-\cos x)-\int 1 \cdot(-\cos x) d x+C \\ & \Rightarrow y(x \sin x)=-x \cos x+\sin x+C \\ & \Rightarrow y=\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x} \\ & \Rightarrow y=-\cot \cdot x+\frac{1}{x}+\frac{C}{x \sin x} \end{aligned} $

10. $(x+y) \frac{d y}{d x}=1$

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Solution

$(x+y) \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}$

$\Rightarrow \frac{d x}{d y}=x+y$

$\Rightarrow \frac{d x}{d y}-x=y$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p x=Q($ where $p=-1$ and $Q=y$ )

Now, I.F $=e^{\int p d y}=e^{\int-d y}=e^{-y}$.

The general solution of the given differential equation is given by the relation, $x($ I.F. $)=\int(Q \times I . F) d y+.C$

$\Rightarrow x e^{-y}=\int(y \cdot e^{-y}) d y+C$

$\Rightarrow x e^{-y}=y \cdot \int e^{-y} d y-\int[\frac{d}{d y}(y) \int e^{-y} d y] d y+C$

$\Rightarrow x e^{-y}=y(-e^{-y})-\int(-e^{-y}) d y+C$

$\Rightarrow x e^{-y}=-y e^{-y}+\int e^{-y} d y+C$

$\Rightarrow x e^{-y}=-y e^{-y}-e^{-y}+C$

$\Rightarrow x=-y-1+Ce^{y}$

$\Rightarrow x+y+1=Ce^{y}$

11. $y d x+(x-y^{2}) d y=0$

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Solution

$y d x+(x-y^{2}) d y=0$

$\Rightarrow y d x=(y^{2}-x) d y$

$\Rightarrow \frac{d x}{d y}=\frac{y^{2}-x}{y}=y-\frac{x}{y}$

$\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p x=Q(.$ where $p=\frac{1}{y}$ and $.Q=y)$

Now, I.F $=e^{\int \rho d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y$.

The general solution of the given differential equation is given by the relation, $x(I . F)=.\int(Q \times I . F) d y+.C$

$\Rightarrow x y=\int(y \cdot y) d y+C$

$\Rightarrow x y=\int y^{2} d y+C$

$\Rightarrow x y=\frac{y^{3}}{3}+C$

$\Rightarrow x=\frac{y^{2}}{3}+\frac{C}{y}$

12. $(x+3 y^{2}) \frac{d y}{d x}=y(y>0)$.

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Solution

$(x+3 y^{2}) \frac{d y}{d x}=y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^{2}}$

$\Rightarrow \frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y$

$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p x=Q(.$ where $p=-\frac{1}{y}$ and $.Q=3 y)$

Now, I.F $=e^{\int p d y}=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log (\frac{1}{y})}=\frac{1}{y}$.

The general solution of the given differential equation is given by the relation, $x(I . F)=.\int(Q \times I . F) d y+.C$

$\Rightarrow x \times \frac{1}{y}=\int(3 y \times \frac{1}{y}) d y+C$

$\Rightarrow \frac{x}{y}=3 y+C$

$\Rightarrow x=3 y^{2}+C y$

For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition:

13. $\frac{d y}{d x}+2 y \tan x=\sin x ; y=0$ when $x=\frac{\pi}{3}$

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Solution

The given differential equation is $\frac{d y}{d x}+2 y \tan x=\sin x$.

This is a linear equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=2 \tan x$ and $Q=\sin x)$

Now, I.F $=e^{\int p d x}=e^{\int 2 tan x d x}=e^{2 \log |\sec x|}=e^{\log (\sec ^{2} x)}=\sec ^{2} x$.

The general solution of the given differential equation is given by the relation,

$y(I . F)=.\int(Q \times I . F) d x+.C$

$\Rightarrow y(\sec ^{2} x)=\int(\sin x \cdot \sec ^{2} x) d x+C$

$\Rightarrow y \sec ^{2} x=\int(\sec x \cdot \tan x) d x+C$

$\Rightarrow y \sec ^{2} x=\sec x+C$

Now, $y=0$ at $x=\frac{\pi}{3}$.

Therefore,

$0 \times \sec ^{2} \frac{\pi}{3}=\sec \frac{\pi}{3}+C$

$\Rightarrow 0=2+C$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation (1), we get: $y \sec ^{2} x=\sec x-2$

$\Rightarrow y=\cos x-2 \cos ^{2} x$

Hence, the required solution of the given differential equation is $y=\cos x-2 \cos ^{2} x$.

14. $(1+x^{2}) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0$ when $x=1$

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Solution

$(1+x^{2}) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}}$

$\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{(1+x^{2})^{2}}$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2 x}{1+x^{2}}$ and $.Q=\frac{1}{(1+x^{2})^{2}})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x d x}{1+x^{2}}}=e^{\log (1+x^{2})}=1+x^{2}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(1+x^{2})=\int[\frac{1}{(1+x^{2})^{2}} \cdot(1+x^{2})] d x+C$

$\Rightarrow y(1+x^{2})=\int \frac{1}{1+x^{2}} d x+C$

$\Rightarrow y(1+x^{2})=\tan ^{-1} x+C$

Now, $y=0$ at $x=1$.

Therefore,

$0=\tan ^{-1} 1+C$

$\Rightarrow C=-\frac{\pi}{4}$

Substituting $C=-\frac{\pi}{4}$ in equation (1), we get:

$y(1+x^{2})=\tan ^{-1} x-\frac{\pi}{4}$

This is the required general solution of the given differential equation.

15. $\frac{d y}{d x}-3 y \cot x=\sin 2 x ; y=2$ when $x=\frac{\pi}{2}$

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Solution

The given differential equation is $\frac{d y}{d x}-3 y \cot x=\sin 2 x$.

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-3 \cot x$ and $Q=\sin 2 x)$

Now, I.F $=e^{\int p d x}=e^{-3 \int \cot x d x}=e^{-3 \log |\sin x|}=e^{\log |\frac{1}{\sin ^{3} x}|}=\frac{1}{\sin ^{3} x}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot \frac{1}{\sin ^{3} x}=\int[\sin 2 x \cdot \frac{1}{\sin ^{3} x}] d x+C$

$\Rightarrow y cosec^{3} x=2 \int(\cot x cosec x) d x+C$

$\Rightarrow y cosec^{3} x=2 cosec x+C$

$\Rightarrow y=-\frac{2}{cosec^{2} x}+\frac{3}{cosec^{3} x}$

$\Rightarrow y=-2 \sin ^{2} x+C \sin ^{3} x$

Now, $y=2$ at $x=\frac{\pi}{2}$.

Therefore, we get:

$2=-2+C$

$\Rightarrow C=4$

Substituting $C=4$ in equation (1), we get:

$y=-2 \sin ^{2} x+4 \sin ^{3} x$

$\Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x$

This is the required particular solution of the given differential equation.

16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

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Solution

Let $F(x, y)$ be the curve passing through the origin.

At point $(x, y)$, the slope of the curve will be $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}=x+y$

$\Rightarrow \frac{d y}{d x}-y=x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x)$

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{-x}=\int x e^{-x} d x+C \tag{1} \end{align*} $$

Now, $\int x e^{-x} d x=x \int e^{-x} d x-\int[\frac{d}{d x}(x) \cdot \int e^{-x} d x] d x$.

$ =-x e^{-x}-\int-e^{-x} d x $

$ \begin{aligned} & =-x e^{-x}+(-e^{-x}) \\ & =-e^{-x}(x+1) \end{aligned} $

Substituting in equation (1), we get:

$$ \begin{align*} & y e^{-x}=-e^{-x}(x+1)+C \\ & \Rightarrow y=-(x+1)+C e^{x} \\ & \Rightarrow x+y+1=C e^{x} \tag{2} \end{align*} $$

The curve passes through the origin.

Therefore, equation (2) becomes:

$1=C$

Substituting $C=1$ in equation (2), we get:

$\Rightarrow \quad x+y+1=e^{x}$

Hence, the required equation of curve passing through the origin is $x+y+1=e^{x}$.

17. Find the equation of a curve passing through the point $(0,2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 .

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Solution

Let $F(x, y)$ be the curve and let $(x, y)$ be a point on the curve. The slope of the tangent

to the curve at $(x, y)$ is $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}+5=x+y$

$\Rightarrow \frac{d y}{d x}-y=x-5$

This is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x-5)$

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general equation of the curve is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot e^{-x}=\int(x-5) e^{-x} d x+C$

Now, $\int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int[\frac{d}{d x}(x-5) \cdot \int e^{-x} d x] d x$.

$ \begin{aligned} & =(x-5)(-e^{-x})-\int(-e^{-x}) d x \\ & =(5-x) e^{-x}+(-e^{-x}) \\ & =(4-x) e^{-x} \end{aligned} $

Therefore, equation (1) becomes:

$$ \begin{align*} & y e^{-x}=(4-x) e^{-x}+C \\ & \Rightarrow y=4-x+C e^{x} \\ & \Rightarrow x+y-4=C e^{x} \tag{2} \end{align*} $$

The curve passes through point $(0,2)$.

Therefore, equation ( 2 ) becomes:

$0+2-4=Ce^{0}$

$\Rightarrow-2=C$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation (2), we get:

$ \begin{aligned} & x+y-4=-2 e^{x} \\ & \Rightarrow y=4-x-2 e^{x} \end{aligned} $

This is the required equation of the curve.

18. The Integrating Factor of the differential equation $x \frac{d y}{d x}-y=2 x^{2}$ is

$\quad\quad$(A) $e^{-x}$

$\quad\quad$(B) $e^{-y}$

$\quad\quad$(C) $\frac{1}{x}$

$\quad\quad$(D) $x$

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Solution

The given differential equation is:

$x \frac{d y}{d x}-y=2 x^{2}$

$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=-\frac{1}{x}$ and $.Q=2 x)$

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore$ I.F $=e^{\int \frac{1}{x} d x}=e^{-\log x}=e^{\log (x^{-1})}=x^{-1}=\frac{1}{x}$

Hence, the correct answer is $C$.

19. The Integrating Factor of the differential equation $\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y(-1<y<1)$

(A) $\frac{1}{y^{2}-1}$

(B) $\frac{1}{\sqrt{y^{2}-1}}$

(C) $\frac{1}{1-y^{2}}$

(D) $\frac{1}{\sqrt{1-y^{2}}}$

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Solution

The given differential equation is:

$(1-y^{2}) \frac{d x}{d y}+y x=a y$

$\Rightarrow \frac{d y}{d x}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p y=Q(.$ where $p=\frac{y}{1-y^{2}}$ and $Q=\frac{a y}{1-y^{2}}$ )

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore I . F=e^{\int p d y}=e^{\int \frac{y}{1-y^{2}} d y}=e^{-\frac{1}{2} \log (1-y^{2})}=e^{\log [\frac{1}{\sqrt{1-y^{2}}}]}=\frac{1}{\sqrt{1-y^{2}}}$

Hence, the correct answer is D.

Miscellaneous Examples

Example 19 Verify that the function $y=c_1 e^{a x} \cos b x+c_2 e^{a x} \sin b x$, where $c_1, c_2$ are arbitrary constants is a solution of the differential equation

$$ \frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+(a^{2}+b^{2}) y=0 $$

Solution The given function is

$$ \begin{equation*} y=e^{a x}\left[c _{1} \cos b x+c _{2} \sin b x\right] \tag{1} \end{equation*} $$

Differentiating both sides of equation (1) with respect to $x$, we get

$$ \begin{align*} & \frac{d y}{d x}=e^{a x}\left[-b c _{1} \sin b x+b c _{2} \cos b x\right]+\left[c _{1} \cos b x+c _{2} \sin b x\right] e^{a x} \cdot a \\ & \frac{d y}{d x}=e^{a x}\left[\left(b c _{2}+a c _{1}\right) \cos b x+\left(a c _{2}-b c _{1}\right) \sin b x\right] \tag{2} \end{align*} $$

Differentiating both sides of equation (2) with respect to $x$, we get

$ \begin{aligned} \frac{d^{2} y}{d x^{2}}= & e^{a x}[(b c_2+a c_1)(-b \sin b x)+(a c_2-b c_1)(b \cos b x)] \\ & +[(b c_2+a c_1) \cos b x+(a c_2-b c_1) \sin b x] e^{a x} \cdot a \\ = & e^{a x}[(a^{2} c_2-2 a b c_1-b^{2} c_2) \sin b x+(a^{2} c_1+2 a b c_2-b^{2} c_1) \cos b x] \end{aligned} $

Substituting the values of $\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ and $y$ in the given differential equation, we get

$ \begin{aligned} \text{ L.H.S. }= & .e^{a x}[a^{2} c_2-2 a b c_1-b^{2} c_2) \sin b x+(a^{2} c_1+2 a b c_2-b^{2} c_1) \cos b x] \\ & -2 a e^{a x}[(b c_2+a c_1) \cos b x+(a c_2-b c_1) \sin b x] \\ & +(a^{2}+b^{2}) e^{a x}[c_1 \cos b x+c_2 \sin b x] \\ = & e^{a x} \begin{bmatrix} (a^{2} c_2-2 a b c_1-b^{2} c_2-2 a^{2} c_2+2 a b c_1+a^{2} c_2+b^{2} c_2) \sin b x \\ +(a^{2} c_1+2 a b c_2-b^{2} c_1-2 a b c_2-2 a^{2} c_1+a^{2} c_1+b^{2} c_1) \cos b x \end{bmatrix} \\ = & e^{a x}[0 \times \sin b x+0 \cos b x]=e^{a x} \times 0=0=\text{ R.H.S. } \end{aligned} $

Hence, the given function is a solution of the given differential equation.

Example 20 Find the particular solution of the differential equation $\log (\frac{d y}{d x})=3 x+4 y$ given that $y=0$ when $x=0$.

Solution The given differential equation can be written as

$ \frac{d y}{d x}=e^{(3 x+4 y)} $

$ \frac{d y}{d x}=e^{3 x} \cdot e^{4 y} $

Separating the variables, we get

$$ \frac{d y}{e^{4 y}}=e^{3 x} d x $$

$$\text{ Therefore } \int e^{-4 y} d y=\int e^{3 x} d x $$

$$\text{ or } \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C $$

$$\text{ or } 4 e^{3 x}+3 e^{-4 y}+12 C=0 \qquad\text{(2)} $$

Substituting $x=0$ and $y=0$ in (2), we get

$$ 4+3+12 C=0 \text{ or } C=\frac{-7}{12} $$

Substituting the value of $C$ in equation (2), we get

$$ 4 e^{3 x}+3 e^{-4 y}-7=0 $$

which is a particular solution of the given differential equation.

Example 21 Solve the differential equation

$$ (x d y-y d x) y \sin (\frac{y}{x})=(y d x+x d y) x \cos (\frac{y}{x}) . $$

Solution The given differential equation can be written as

$$ [x y \sin (\frac{y}{x})-x^{2} \cos (\frac{y}{x})] d y=[x y \cos (\frac{y}{x})+y^{2} \sin (\frac{y}{x})] d x $$

$$\text{ or } \frac{d y}{d x}=\frac{x y \cos (\frac{y}{x})+y^{2} \sin (\frac{y}{x})}{x y \sin (\frac{y}{x})-x^{2} \cos (\frac{y}{x})} $$

Dividing numerator and denominator on RHS by $x^{2}$, we get

$$ \frac{d y}{d x}=\frac{\frac{y}{x} \cos (\frac{y}{x})+(\frac{y^{2}}{x^{2}}) \sin (\frac{y}{x})}{\frac{y}{x} \sin (\frac{y}{x})-\cos (\frac{y}{x})} \tag{1} $$

Clearly, equation (1) is a homogeneous differential equation of the form $\frac{d y}{d x}=g(\frac{y}{x})$. To solve it,

we make the substitution

$$ \begin{aligned} & y=v x \\ & \text{ or } \\ & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & v+x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v} \\ & x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\ & (\frac{v \sin v-\cos v}{v \cos v}) d v=\frac{2 d x}{x} \\ & \text{ Therefore } \quad \int(\frac{v \sin v-\cos v}{v \cos v}) d v=2 \int \frac{1}{x} d x \\ & \text{ or } \\ & \int \tan v d v-\int \frac{1}{v} d v=2 \int \frac{1}{x} d x \\ & \text{ or } \\ & \text{ or } \\ & \log |\sec v|-\log |v|=2 \log |x|+\log |C_1| \\ & \log |\frac{\sec v}{v x^{2}}|=\log |C_1| \\ & \text{ or } \\ & \frac{\sec v}{v x^{2}}= \pm C_1 \end{aligned} $$

Replacing $v$ by $\frac{y}{x}$ in equation (3), we get

$ \begin{aligned} & \frac{\sec (\frac{y}{x})}{(\frac{y}{x})(x^{2})}=C \text{ where, } C= \pm C_1 \\ & \sec (\frac{y}{x})=C x y \end{aligned} $

which is the general solution of the given differential equation.

Example 22 Solve the differential equation

$$ (\tan ^{-1} y-x) d y=(1+y^{2}) d x \text{. } $$

Solution The given differential equation can be written as

$ \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}} $

Now (1) is a linear differential equation of the form $\frac{d x}{d y}+P_1 x=Q_1$,

where, $\quad P_1=\frac{1}{1+y^{2}}$ and $Q_1=\frac{\tan ^{-1} y}{1+y^{2}}$. Therefore, $\quad I . F=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$

Thus, the solution of the given differential equation is

$$ \begin{equation*} x e^{\tan ^{-1} y}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} d y+\mathrm{C} \tag{2} \end{equation*} $$

$ \text{ Let } \quad I=\int(\frac{\tan ^{-1} y}{1+y^{2}}) e^{\tan ^{-1} y} d y $

Substituting $\tan ^{-1} y=t$ so that $(\frac{1}{1+y^{2}}) d y=d t$, we get

$ I=\int t e^{t} d t=t e^{t}-\int 1 \cdot e^{t} d t=t e^{t}-e^{t}=e^{t}(t-1) $

$$\text{ or }\qquad I=e^{\tan ^{-1} y}(\tan ^{-1} y-1) $$

Substituting the value of $I$ in equation (2), we get

$$ \begin{aligned} \text{ or }\qquad & x \cdot e^{\tan ^{-1} y}=e^{\tan ^{-1} y}(\tan ^{-1} y-1)+C \\ x= & (\tan ^{-1} y-1)+C e^{-\tan ^{-1} y} \end{aligned} $$

which is the general solution of the given differential equation.

Miscellaneous Exercise on Chapter 9

1. For each of the differential equations given below, indicate its order and degree (if defined).

$\quad\quad$(i) $\frac{d^{2} y}{d x^{2}}+5 x(\frac{d y}{d x})^{2}-6 y=\log x$

$\quad\quad$(ii) $(\frac{d y}{d x})^{3}-4(\frac{d y}{d x})^{2}+7 y=\sin x$

$\quad\quad$(iii) $\frac{d^{4} y}{d x^{4}}-\sin (\frac{d^{3} y}{d x^{3}})=0$

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Solution

(i) The differential equation is given as:

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}+5 x(\frac{d y}{d x})^{2}-6 y=\log x \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}+5 x(\frac{d y}{d x})^{2}-6 y-\log x=0 \end{aligned} $

The highest order derivative present in the differential equation is $\frac{d^{2} y}{d x^{2}}$. Thus, its order is two. The highest power raised to $\frac{d^{2} y}{d x^{2}}$ is one. Hence, its degree is one.

(ii) The differential equation is given as:

$ \begin{aligned} & (\frac{d y}{d x})^{3}-4(\frac{d y}{d x})^{2}+7 y=\sin x \\ & \Rightarrow(\frac{d y}{d x})^{3}-4(\frac{d y}{d x})^{2}+7 y-\sin x=0 \end{aligned} $

The highest order derivative present in the differential equation is $\frac{d y}{d x}$. Thus, its order is one. The highest power raised to $\frac{d y}{d x}$ is three. Hence, its degree is three.

(iii) The differential equation is given as:

$\frac{d^{4} y}{d x^{4}}-\sin (\frac{d^{3} y}{d x^{3}})=0$

The highest order derivative present in the differential equation is $\frac{d^{4} y}{d x^{4}}$. Thus, its order is four.

However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$\quad\quad$(i) $x y=a e^{x}+b e^{-x}+x^{2}$

$\quad\quad: x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0$

$\quad\quad$(ii) $y=e^{x}(a \cos x+b \sin x) \quad: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

$\quad\quad$(iii) $y=x \sin 3 x$

$\quad\quad: \frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x=0$

$\quad\quad$(iv) $x^{2}=2 y^{2} \log y$

$ \quad\quad:(x^{2}+y^{2}) \frac{d y}{d x}-x y=0 $

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Solution

(i) $y=a e^{x}+b e^{-x}+x^{2}$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=a \frac{d}{d x}(e^{x})+b \frac{d}{d x}(e^{-x})+\frac{d}{d x}(x^{2})$ $\Rightarrow \frac{d y}{d x}=a e^{x}-b e^{-x}+2 x$

Again, differentiating both sides with respect to $x$, we get:

$\frac{d^{2} y}{d x^{2}}=a e^{x}+b e^{-x}+2$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S.

$x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2$

$=x(a e^{x}+b e^{-x}+2)+2(a e^{x}-b e^{-x}+2 x)-x(a e^{x}+b e^{-x}+x^{2})+x^{2}-2$

$=(a x e^{x}+b x e^{-x}+2 x)+(2 a e^{x}-2 b e^{-x}+4 x)-(a x e^{x}+b x e^{-x}+x^{3})+x^{2}-2$

$=2 a e^{x}-2 b e^{-x}+x^{2}+6 x-2$

$\neq 0$

$\Rightarrow$ L.H.S. $\neq$ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii) $y=e^{x}(a \cos x+b \sin x)=a e^{x} \cos x+b e^{x} \sin x$

Differentiating both sides with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=a \cdot \frac{d}{d x}(e^{x} \cos x)+b \cdot \frac{d}{d x}(e^{x} \sin x) \\ & \Rightarrow \frac{d y}{d x}=a(e^{x} \cos x-e^{x} \sin x)+b \cdot(e^{x} \sin x+e^{x} \cos x) \\ & \Rightarrow \frac{d y}{d x}=(a+b) e^{x} \cos x+(b-a) e^{x} \sin x \end{aligned} $

Again, differentiating both sides with respect to $x$, we get:

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}=(a+b) \cdot \frac{d}{d x}(e^{x} \cos x)+(b-a) \frac{d}{d x}(e^{x} \sin x) \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=(a+b) \cdot[e^{x} \cos x-e^{x} \sin x]+(b-a)[e^{x} \sin x+e^{x} \cos x] \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}[(a+b)(\cos x-\sin x)+(b-a)(\sin x+\cos x)] \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}[a \cos x-a \sin x+b \cos x-b \sin x+b \sin x+b \cos x-a \sin x-a \cos x] \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=[2 e^{x}(b \cos x-a \sin x)] \end{aligned} $

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get:

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y \\ & =2 e^{x}(b \cos x-a \sin x)-2 e^{x}[(a+b) \cos x+(b-a) \sin x]+2 e^{x}(a \cos x+b \sin x) \\ & =e^{x} \begin{bmatrix} (2 b \cos x-2 a \sin x)-(2 a \cos x+2 b \cos x) \\ -(2 b \sin x-2 a \sin x)+(2 a \cos x+2 b \sin x) \end{bmatrix} \\ & =e^{x}[(2 b-2 a-2 b+2 a) \cos x]+e^{x}[(-2 a-2 b+2 a+2 b) \sin x] \\ & =0 \end{aligned} $

Hence, the given function is a solution of the corresponding differential equation.

(iii) $y=x \sin 3 x$

Differentiating both sides with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(x \sin 3 x)=\sin 3 x+x \cdot \cos 3 x \cdot 3 \\ & \Rightarrow \frac{d y}{d x}=\sin 3 x+3 x \cos 3 x \end{aligned} $

Again, differentiating both sides with respect to $x$, we get: $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\sin 3 x)+3 \frac{d}{d x}(x \cos 3 x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=3 \cos 3 x+3[\cos 3 x+x(-\sin 3 x) \cdot 3]$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=6 \cos 3 x-9 x \sin 3 x$

Substituting the value of $\frac{d^{2} y}{d x^{2}}$ in the L.H.S. of the given differential equation, we get:

$\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x$

$=(6 \cdot \cos 3 x-9 x \sin 3 x)+9 x \sin 3 x-6 \cos 3 x$

$=0$

Hence, the given function is a solution of the corresponding differential equation.

(iv) $x^{2}=2 y^{2} \log y$

Differentiating both sides with respect to $x$, we get:

$2 x=2 \cdot \frac{d}{d x}=[y^{2} \log y]$

$\Rightarrow x=[2 y \cdot \log y \cdot \frac{d y}{d x}+y^{2} \cdot \frac{1}{y} \cdot \frac{d y}{d x}]$

$\Rightarrow x=\frac{d y}{d x}(2 y \log y+y)$

$\Rightarrow \frac{d y}{d x}=\frac{x}{y(1+2 \log y)}$

Substituting the value of $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get: $(x^{2}+y^{2}) \frac{d y}{d x}-x y$

$=(2 y^{2} \log y+y^{2}) \cdot \frac{x}{y(1+2 \log y)}-x y$

$=y^{2}(1+2 \log y) \cdot \frac{x}{y(1+2 \log y)}-x y$

$=x y-x y$

$=0$

Hence, the given function is a solution of the corresponding differential equation.

3. Prove that $x^{2}-y^{2}=c(x^{2}+y^{2})^{2}$ is the general solution of differential equation $(x^{3}-3 x y^{2}) d x=(y^{3}-3 x^{2} y) d y$, where $c$ is a parameter.

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Solution

$(x^{3}-3 x y^{2}) d x=(y^{3}-3 x^{2} y) d y$

$\Rightarrow \frac{d y}{d x}=\frac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}$

This is a homogeneous equation. To simplify it, we need to make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d v}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x^{3}-3 x(v x)^{2}}{(v x)^{3}-3 x^{2}(v x)}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{1-3 v^{2}}{v^{3}-3 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1-3 v^{2}}{v^{3}-3 v}-v$

$\Rightarrow x \frac{d v}{d x}=\frac{1-3 v^{2}-v(v^{3}-3 v)}{v^{3}-3 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1-v^{4}}{v^{3}-3 v}$

$\Rightarrow(\frac{v^{3}-3 v}{1-v^{4}}) d v=\frac{d x}{x}$

Integrating both sides, we get:

$\int(\frac{v^{3}-3 v}{1-v^{4}}) d v=\log x+\log C^{\prime}$

Now, $\int(\frac{v^{3}-3 v}{1-v^{4}}) d v=\int \frac{v^{3} d v}{1-v^{4}}-3 \int \frac{v d v}{1-v^{4}}$

$\Rightarrow \int(\frac{v^{3}-3 v}{1-v^{4}}) d v=I_1-3 I_2$, where $I_1=\int \frac{v^{3} d v}{1-v^{4}}$ and $I_2=\int \frac{v d v}{1-v^{4}}$

Let $1-v^{4}=t$.

$\therefore \frac{d}{d v}(1-v^{4})=\frac{d t}{d v}$

$\Rightarrow-4 v^{3}=\frac{d t}{d v}$

$\Rightarrow v^{3} d v=-\frac{d t}{4}$

Now, $I_1=\int \frac{-d t}{4 t}=-\frac{1}{4} \log t=-\frac{1}{4} \log (1-v^{4})$

And, $I_2=\int \frac{v d v}{1-v^{4}}=\int \frac{v d v}{1-(v^{2})^{2}}$

Let $v^{2}=p$.

$\therefore \frac{d}{d v}(v^{2})=\frac{d p}{d v}$

$\Rightarrow 2 v=\frac{d p}{d v}$

$\Rightarrow v d v=\frac{d p}{2}$

$\Rightarrow I_2=\frac{1}{2} \int \frac{d p}{1-p^{2}}=\frac{1}{2 \times 2} \log |\frac{1+p}{1-p}|=\frac{1}{4} \log |\frac{1+v^{2}}{1-v^{2}}|$

Substituting the values of $I_1$ and $I_2$ in equation (3), we get:

$\int(\frac{v^{3}-3 v}{1-v^{4}}) d v=-\frac{1}{4} \log (1-v^{4})-\frac{3}{4} \log |\frac{1-v^{2}}{1+v^{2}}|$

Therefore, equation (2) becomes:

$ \begin{aligned} & \frac{1}{4} \log (1-v^{4})-\frac{3}{4} \log |\frac{1+v^{2}}{1-v^{2}}|=\log x+\log C^{\prime} \\ & \Rightarrow-\frac{1}{4} \log [(1-v^{4})(\frac{1+v^{2}}{1-v^{2}})^{3}]=\log C^{\prime} x \\ & \Rightarrow \frac{(1+v^{2})^{4}}{(1-v^{2})^{2}}=(C^{\prime} x)^{-4} \\ & \Rightarrow \frac{(1+\frac{y^{2}}{x^{2}})^{4}}{(1-\frac{y^{2}}{x^{2}})^{2}}=\frac{1}{C^{\prime 4} x^{4}} \\ & \Rightarrow \frac{(x^{2}+y^{2})^{4}}{x^{4}(x^{2}-y^{2})^{2}}=\frac{1}{C^{\prime 4} x^{4}} \\ & \Rightarrow(x^{2}-y^{2})^{2}=C^{\prime 4}(x^{2}+y^{2})^{4} \\ & \Rightarrow(x^{2}-y^{2})=C^{\prime 2}(x^{2}+y^{2})^{2} \\ & \Rightarrow x^{2}-y^{2}=C(x^{2}+y^{2})^{2}, \text{ where } C=C^{\prime 2} \end{aligned} $

Hence, the given result is proved.

4. Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$.

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Solution

$\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\frac{-d x}{\sqrt{1-x^{2}}}$

Integrating both sides, we get:

$\sin ^{-1} y=-\sin ^{-1} x+C$

$\Rightarrow \sin ^{-1} x+\sin ^{-1} y=C$

5. Show that the general solution of the differential equation $\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$ is given by $(x+y+1)=A(1-x-y-2 x y)$, where $A$ is parameter.

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Solution

$ \begin{aligned} & \frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0 \\ & \Rightarrow \frac{d y}{d x}=-\frac{(y^{2}+y+1)}{x^{2}+x+1} \\ & \Rightarrow \frac{d y}{y^{2}+y+1}=\frac{-d x}{x^{2}+x+1} \\ & \Rightarrow \frac{d y}{y^{2}+y+1}+\frac{d x}{x^{2}+x+1}=0 \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & \int \frac{d y}{y^{2}+y+1}+\int \frac{d x}{x^{2}+x+1}=C \\ & \Rightarrow \int \frac{d y}{(y+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}+\int \frac{d x}{(x+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}=C \\ & \Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}[\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}]+\frac{2}{\sqrt{3}} \tan ^{-1}[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}]=C \\ & \Rightarrow \tan ^{-1}[\frac{2 y+1}{\sqrt{3}}]+\tan ^{-1}[\frac{2 x+1}{\sqrt{3}}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{(2 y+1)}{\sqrt{3}} \cdot \frac{(2 x+1)}{\sqrt{3}}}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{\frac{2 x+2 y+2}{\sqrt{3}}}{1-(\frac{4 x y+2 x+2 y+1}{3})}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}=\tan (\frac{\sqrt{3} C}{2})=B, \text{ where } B=\tan (\frac{\sqrt{3} C}{2}) \\ & \Rightarrow x+y+1=\frac{2 B}{\sqrt{3}}(1-x y-2 x y) \\ & \Rightarrow x+y+1=A(1-x-y-2 x y) \text{, where } A=\frac{2 B}{\sqrt{3}} \end{aligned} $

Hence, the given result is proved.

6. Find the equation of the curve passing through the point $(0, \frac{\pi}{4})$ whose differential equation is $\sin x \cos y d x+\cos x \sin y d y=0$.

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Solution

The differential equation of the given curve is:

$\sin x \cos y d x+\cos x \sin y d y=0$

$\Rightarrow \frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0$

$\Rightarrow \tan x d x+\tan y d y=0$

Integrating both sides, we get:

$\log (\sec x)+\log (\sec y)=\log C$

$\log (\sec x \cdot \sec y)=\log C$

$\Rightarrow \sec x \cdot \sec y=C$

The curve passes through point $(0, \frac{\pi}{4})$.

$\therefore 1 \times \sqrt{2}=C$

$\Rightarrow C=\sqrt{2}$

On substituting in equation (1), we get:

$\sec x \cdot \sec y=\sqrt{2}$

$\Rightarrow \sec x \cdot \frac{1}{\cos y}=\sqrt{2}$

$\Rightarrow \cos y=\frac{\sec x}{\sqrt{2}}$

Hence, the required equation of the curve is $\cos y=\frac{\sec x}{\sqrt{2}}$.

7. Find the particular solution of the differential equation $(1+e^{2 x}) d y+(1+y^{2}) e^{x} d x=0$, given that $y=1$ when $x=0$.

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Solution

$(1+e^{2 x}) d y+(1+y^{2}) e^{x} d x=0$

$\Rightarrow \frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}=0$

Integrating both sides, we get:

$\tan ^{-1} y+\int \frac{e^{x} d x}{1+e^{2 x}}=C$

Let $e^{x}=t \Rightarrow e^{2 x}=t^{2}$.

$\Rightarrow \frac{d}{d x}(e^{x})=\frac{d t}{d x}$

$\Rightarrow e^{x}=\frac{d t}{d x}$

$\Rightarrow e^{x} d x=d t$

Substituting these values in equation (1), we get:

$\tan ^{-1} y+\int \frac{d t}{1+t^{2}}=C$

$\Rightarrow \tan ^{-1} y+\tan ^{-1} t=C$

$\Rightarrow \tan ^{-1} y+\tan ^{-1}(e^{x})=C$

Now, $y=1$ at $x=0$.

Therefore, equation (2) becomes:

$\tan ^{-1} 1+\tan ^{-1} 1=C$

$\Rightarrow \frac{\pi}{4}+\frac{\pi}{4}=C$

$\Rightarrow C=\frac{\pi}{2}$

Substituting $C=\frac{\pi}{2}$ in equation (2), we get:

$\tan ^{-1} y+\tan ^{-1}(e^{x})=\frac{\pi}{2}$

This is the required particular solution of the given differential equation.

8. Solve the differential equation $y e^{\frac{x}{y}} d x=(x e^{\frac{x}{y}}+y^{2}) d y(y \neq 0)$.

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Solution

$y e^{\frac{x}{y}} d x=(x e^{\frac{x}{y}}+y^{2}) d y$

$\Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y}=x e^{\frac{x}{y}}+y^{2}$

$\Rightarrow e^{\frac{x}{y}}[y \cdot \frac{d x}{d y}-x]=y^{2}$

$\Rightarrow e^{\frac{x}{y}} \frac{[y \cdot \frac{d x}{d y}-x]}{y^{2}}=1$

Let $e^{\frac{x}{y}}=z$.

Differentiating it with respect to $y$, we get:

$\frac{d}{d y}(e^{\frac{x}{y}})=\frac{d z}{d y}$

$\Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}(\frac{x}{y})=\frac{d z}{d y}$

$\Rightarrow e^{\frac{x}{y}} \cdot[\frac{y \cdot \frac{d x}{d y}-x}{y^{2}}]=\frac{d z}{d y}$

From equation (1) and equation (2), we get:

$\frac{d z}{d y}=1$

$\Rightarrow d z=d y$

Integrating both sides, we get: $z=y+C$

$\Rightarrow e^{\frac{x}{y}}=y+C$

9. Find a particular solution of the differential equation $(x-y)(d x+d y)=d x-d y$, given that $y=-1$, when $x=0$. (Hint: put $x-y=t$ )

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Solution

$(x-y)(d x+d y)=d x-d y$

$\Rightarrow(x-y+1) d y=(1-x+y) d x$

$\Rightarrow \frac{d y}{d x}=\frac{1-x+y}{x-y+1}$

$\Rightarrow \frac{d y}{d x}=\frac{1-(x-y)}{1+(x-y)}$

Let $x-y=t$.

$\Rightarrow \frac{d}{d x}(x-y)=\frac{d t}{d x}$

$\Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow 1-\frac{d t}{d x}=\frac{d y}{d x}$

Substituting the values of $x-y$ and $\frac{d y}{d x}$ in equation (1), we get: $1-\frac{d t}{d x}=\frac{1-t}{1+t}$

$\Rightarrow \frac{d t}{d x}=1-(\frac{1-t}{1+t})$

$\Rightarrow \frac{d t}{d x}=\frac{(1+t)-(1-t)}{1+t}$

$\Rightarrow \frac{d t}{d x}=\frac{2 t}{1+t}$

$\Rightarrow(\frac{1+t}{t}) d t=2 d x$

$\Rightarrow(1+\frac{1}{t}) d t=2 d x$

Integrating both sides, we get:

$t+\log |t|=2 x+C$

$\Rightarrow(x-y)+\log |x-y|=2 x+C$

$\Rightarrow \log |x-y|=x+y+C$

Now, $y=-1$ at $x=0$.

Therefore, equation (3) becomes:

$\log 1=0-1+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (3) we get:

$\log |x-y|=x+y+1$

This is the required particular solution of the given differential equation.

10. Solve the differential equation $[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}] \frac{d x}{d y}=1(x \neq 0)$.

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Solution

$[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}] \frac{d x}{d y}=1$

$\Rightarrow \frac{d y}{d x}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$

$\Rightarrow \frac{d y}{d x}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$

This equation is a linear differential equation of the form

$\frac{d y}{d x}+P y=Q$, where $P=\frac{1}{\sqrt{x}}$ and $Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$.

Now, I.F $=e^{\int P d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}$

The general solution of the given differential equation is given by,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y e^{2 \sqrt{x}}=\int(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \times e^{2 \sqrt{x}}) d x+C$

$\Rightarrow y e^{2 \sqrt{x}}=\int \frac{1}{\sqrt{x}} d x+C$

$\Rightarrow y e^{2 \sqrt{x}}=2 \sqrt{x}+C$

11. Find a particular solution of the differential equation $\frac{d y}{d x}+y \cot x=4 x cosec x$ $(x \neq 0)$, given that $y=0$ when $x=\frac{\pi}{2}$.

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Solution

The given differential equation is:

$\frac{d y}{d x}+y \cot x=4 x cosec x$

This equation is a linear differential equation of the form $\frac{d y}{d x}+p y=Q$, where $p=\cot x$ and $Q=4 x cosec x$.

Now, I.F $=e^{\int p p d x}=e^{\int \cot x d x}=e^{\log |\sin x|}=\sin x$

The general solution of the given differential equation is given by,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y \sin x=\int(4 x cosec x \cdot \sin x) d x+C$

$\Rightarrow y \sin x=4 \int x d x+C$

$\Rightarrow y \sin x=4 \cdot \frac{x^{2}}{2}+C$

$\Rightarrow y \sin x=2 x^{2}+C$

Now, $y=0$ at $x=\frac{\pi}{2}$.

Therefore, equation (1) becomes:

$0=2 \times \frac{\pi^{2}}{4}+C$

$\Rightarrow C=-\frac{\pi^{2}}{2}$

Substituting $C=-\frac{\pi^{2}}{2}$ in equation (1), we get:

$y \sin x=2 x^{2}-\frac{\pi^{2}}{2}$

This is the required particular solution of the given differential equation.

12. Find a particular solution of the differential equation $(x+1) \frac{d y}{d x}=2 e^{-y}-1$, given that $y=0$ when $x=0$.

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Solution

$ \begin{aligned} & (x+1) \frac{d y}{d x}=2 e^{-y}-1 \\ & \Rightarrow \frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1} \\ & \Rightarrow \frac{e^{y} d y}{2-e^{y}}=\frac{d x}{x+1} \end{aligned} $

Integrating both sides, we get:

$$ \begin{equation*} \int \frac{e^{y} d y}{2-e^{y}}=\log |x+1|+\log C \tag{1} \end{equation*} $$

Let $2-e^{y}=t$.

$\therefore \frac{d}{d y}(2-e^{y})=\frac{d t}{d y}$

$\Rightarrow-e^{y}=\frac{d t}{d y}$

$\Rightarrow e^{y} d t=-d t$

Substituting this value in equation (1), we get:

$$ \begin{align*} & \int \frac{-d t}{t}=\log |x+1|+\log C \\ & \Rightarrow-\log |t|=\log |C(x+1)| \\ & \Rightarrow-\log |2-e^{y}|=\log |C(x+1)| \\ & \Rightarrow \frac{1}{2-e^{y}}=C(x+1) \\ & \Rightarrow 2-e^{y}=\frac{1}{C(x+1)} \tag{2} \end{align*} $$

Now, at $x=0$ and $y=0$, equation (2) becomes:

$\Rightarrow 2-1=\frac{1}{C}$

$\Rightarrow C=1$

Substituting $C=1$ in equation (2), we get: $2-e^{y}=\frac{1}{x+1}$

$\Rightarrow e^{y}=2-\frac{1}{x+1}$

$\Rightarrow e^{y}=\frac{2 x+2-1}{x+1}$

$\Rightarrow e^{y}=\frac{2 x+1}{x+1}$

$\Rightarrow y=\log |\frac{2 x+1}{x+1}|,(x \neq-1)$

This is the required particular solution of the given differential equation.

13. The general solution of the differential equation $\frac{y d x-x d y}{y}=0$ is

$\quad\quad$(A) $x y=C$

$\quad\quad$(B) $x=C y^{2}$

$\quad\quad$(C) $y=C x$

$\quad\quad$(D) $y=C x^{2}$

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Solution

The given differential equation is:

$ \begin{aligned} & \frac{y d x-x d y}{y}=0 \\ & \Rightarrow \frac{y d x-x d y}{x y}=0 \\ & \Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0 \end{aligned} $

Integrating both sides, we get:

$\log |x|-\log |y|=\log k$

$\Rightarrow \log |\frac{x}{y}|=\log k$

$\Rightarrow \frac{x}{y}=k$

$\Rightarrow y=\frac{1}{k} x$

$\Rightarrow y=C x$ where $C=\frac{1}{k}$

Hence, the correct answer is C.

14. The general solution of a differential equation of the type $\frac{d x}{d y}+P_1 x=Q_1$ is

$\quad\quad$(A) $y e^{\int P_1 d y}=\int(Q_1 e^{\int P_1 d y}) d y+C$

$\quad\quad$(B) $y \cdot e^{\int P_1 d x}=\int(Q_1 e^{\int P_1 d x}) d x+C$

$\quad\quad$(C) $x e^{\int P_1 d y}=\int(Q_1 e^{\int P_1 d y}) d y+C$

$\quad\quad$(D) $x e^{\int P_1 d x}=\int(Q_1 e^{\int P_1 d x}) d x+C$

Show Answer

Solution

The integrating factor of the given differential equation $\frac{d x}{d y}+P_1 x=Q_1$ is $e^{\int P_P d y}$.

The general solution of the differential equation is given by,

$ \begin{aligned} & x(\text{ I.F. })=\int(Q \times \text{ I.F. }) d y+C \\ & \Rightarrow x \cdot e^{\int P_1 d y}=\int(Q_1 e^{\int P_1 d y}) d y+C \end{aligned} $

Hence, the correct answer is C.

15. The general solution of the differential equation $e^{x} d y+(y e^{x}+2 x) d x=0$ is

$\quad\quad$(A) $x e^{y}+x^{2}=C$

$\quad\quad$(B) $x e^{y}+y^{2}=C$

$\quad\quad$(C) $y e^{x}+x^{2}=C$

$\quad\quad$(D) $y e^{y}+x^{2}=C$

Show Answer

Solution

The given differential equation is:

$ \begin{aligned} & e^{x} d y+(y e^{x}+2 x) d x=0 \\ & \Rightarrow e^{x} \frac{d y}{d x}+y e^{x}+2 x=0 \\ & \Rightarrow \frac{d y}{d x}+y=-2 x e^{-x} \end{aligned} $

This is a linear differential equation of the form

$\frac{d y}{d x}+P y=Q$, where $P=1$ and $Q=-2 x e^{-x}$.

Now, I.F $=e^{\int P d t}=e^{\int d x}=e^{x}$

The general solution of the given differential equation is given by,

$ \begin{aligned} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{x}=\int(-2 x e^{-x} \cdot e^{x}) d x+C \\ & \Rightarrow y e^{x}=-\int 2 x d x+C \\ & \Rightarrow y e^{x}=-x^{2}+C \\ & \Rightarrow y e^{x}+x^{2}=C \end{aligned} $

Hence, the correct answer is $C$.

Summary

  • An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation.

  • Order of a differential equation is the order of the highest order derivative occurring in the differential equation.

  • Degree of a differential equation is defined if it is a polynomial equation in its derivatives.

  • Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it.

  • A function which satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution.

  • Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing $y$ should remain with $d y$ and terms containing $x$ should remain with $d x$.

  • A differential equation which can be expressed in the form $\frac{d y}{d x}=f(x, y)$ or $\frac{d x}{d y}=g(x, y)$ where, $f(x, y)$ and $g(x, y)$ are homogenous functions of degree zero is called a homogeneous differential equation.

A differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P$ and $Q$ are constants or functions of $x$ only is called a first order linear differential equation.

Historical Note

One of the principal languages of Science is that of differential equations. Interestingly, the date of birth of differential equations is taken to be November, 11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black and white the identity $\int y d y=\frac{1}{2} y^{2}$, thereby introducing both the symbols $\int$ and $d y$. Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed. This led him to discover the ‘method of separation of variables’ 1691. A year later he formulated the ‘method of solving the homogeneous differential equations of the first order’. He went further in a very short time to the discovery of the ‘method of solving a linear differential equation of the first-order’. How surprising is it that all these methods came from a single man and that too within 25 years of the birth of differential equations!

In the old days, what we now call the ‘solution’ of a differential equation, was used to be referred to as ‘integral’ of the differential equation, the word being coined by James Bernoulli (1654 - 1705) in 1690. The word ‘solution was first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred years since the birth of differential equations. It was Jules Henri Poincare (1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the word ‘solution’ has found its deserved place in modern terminology. The name of the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), a younger brother of James Bernoulli. Application to geometric problems were also considered. It was again John Bernoulli who first brought into light the intricate nature of differential equations. In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the differential equation

$$ x^{2} y^{\prime \prime}=2 y $$

which led to three types of curves, viz., parabolas, hyperbolas and a class of cubic curves. This shows how varied the solutions of such innocent looking differential equation can be. From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading ‘qualitative analysis of differential equations’. Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations.

He who seeks fअथवा methods without having a definite problem in mind seeks fअथवा the most part in vain - D. HILBERT

9.1 भूमिका (Introduction )

कक्षा XI एवं इस पुस्तक के अध्याय 5 में हमने चर्चा की थी, कि एक स्वतंत्र चर के सापेक्ष किसी फलन $f$ का अवकलज कैसे ज्ञात किया जाता है अर्थात् किसी फलन $f$ की परिभाषित प्रांत के प्रत्येक $x$ के लिए, $f^{\prime}(x)$ कैसे ज्ञात किया जाता है। इसके अतिरिक्त समाकल गणित के अध्याय में हमने चर्चा की थी, कि यदि किसी फलन $f$ का अवकलज फलन $g$ है तो फलन $f$ कैसे ज्ञात किया जाए। इसको निम्न रूप में सूत्रबद्ध किया जा सकता है:

किसी दिए हुए फलन $g$ के लिए फलन $f$ ज्ञात कीजिए ताकि

$$ \frac{d y}{d x}=g(x) \text { जहाँ } y=f(x) $$

Henri Poincare $(1854-1912)$

समीकरण (1) के रूप वाले समीकरण को अवकल समीकरण कहते हैं। इसकी औपचारिक परिभाषा बाद में दी जाएगी।

अवकल समीकरणों का उपयोग मुख्य रूप से भौतिकी, रसायन विज्ञान, जीव विज्ञान, मानव विज्ञान, भूविज्ञान, अर्थशास्त्र आदि विभिन्न क्षेत्रों में किया जाता है। अतः सभी अत्याधुनिक वैज्ञानिक अन्वेषणों के लिए अवकल समीकरणों के गहन अध्ययन की अत्यंत आवश्यकता है।

इस अध्याय में, हम अवकल समीकरण की कुछ आधारभूत संकल्पनाओं, अवकल समीकरण के व्यापक एवं विशिष्ट हल, अवकल समीकरण का निर्माण, प्रथम कोटि एवं प्रथम घात के अवकल समीकरण को हल करने की कुछ विधियाँ और विभिन्न क्षेत्रों में अवकल समीकरणों के कुछ उपयोगों के बारे में अध्ययन करेंगे।

9.2 आधारभूत संकल्पनाएँ (Basic Concepts)

हम पहले से ही निम्नलिखित प्रकार के समीकरणों से परिचित हैं

$$ \begin{align*} x^{2}-3 x+3=0 \tag{1} \\ \sin x+\cos x=0 \tag{2} \\ x+y=7 \tag{3} \end{align*} $$

आइए निम्नलिखित समीकरण पर विचार करें

$$ \begin{equation*} x \frac{d y}{d x}+y=0 \tag{4} \end{equation*} $$

हम पाते हैं कि समीकरणों (1), (2) एवं (3) में केवल स्वतंत्र और/अथवा आश्रित चर (एक या अधिक) शामिल हैं जब कि समीकरण (4) में चर के साथ-साथ स्वतंत्र चर $(x)$ के सापेक्ष आश्रित चर $(y)$ का अवकलज भी शामिल है। इस प्रकार का समीकरण अवकल समीकरण कहलाता है।

सामान्यतः एक ऐसा समीकरण, जिसमें स्वतंत्र चर (चरों) के सापेक्ष आश्रित चर के अवकलज सम्मिलित हों, अवकल समीकरण कहलाता है।

एक ऐसा अवकल समीकरण, जिसमें केवल एक स्वतंत्र चर के सापेक्ष, आश्रित चर के अवकलज सम्मिलित हों, सामान्य अवकल समीकरण कहलाता है। उदाहरणतया

$ 2 \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}=0 \tag{5} \text{ एक सामान्य अवकल समीकरण है। } $

निःसन्देह ऐसे भी अवकल समीकरण होते हैं जिनमें एक से अधिक स्वतंत्र चरों के सापेक्ष अवकलज शामिल होते हैं, इस प्रकार के अवकल समीकरण आंशिक अवकल समीकरण कहलाते हैं। लेकिन इस स्तर पर हम अपने आप को केवल सामान्य अवकल समीकरणों के अध्ययन तक सीमित रखेंगे। इससे आगे हम सामान्य अवकल समीकरण के लिए अवकल समीकरण शब्द का ही उपयोग करेंगे।

टिप्पणी

1. हम अवकलजों के लिए निम्नलिखित संकेतों के उपयोग को वरीयता देंगे

$$ \frac{d y}{d x}=y^{\prime}, \frac{d^{2} y}{d x^{2}}=y^{\prime \prime}, \frac{d^{3} y}{d x^{3}}=y^{\prime \prime \prime} $$

2. उच्च कोटि वाले अवकलजों के लिए, इतने अधिक डैशों (dashes) को उच्च प्रत्यय के रूप में प्रयुक्त करना असुविधाजनक होगा इसलिए $n$ वें कोटि वाले अवकलज $\frac{d^{n} y}{d x^{n}}$ के लिए हम संकेत $y _{n}$ का उपयोग करेंगे।

9.2.1 अवकल समीकरण की कोटि (अथवाder of a differential equation)

किसी अवकल समीकरण की कोटि उस अवकल समीकरण में सम्मिलित स्वतंत्र चर के सापेक्ष आश्रित चर के उच्चतम कोटि के अवकलज की कोटि द्वारा परिभाषित होती है।

निम्नलिखित अवकल समीकरणों पर विचार कीजिए:

$$ \begin{align*} & \frac{d y}{d x}=e^{x} \tag{6}\\ & \frac{d^{2} y}{d x^{2}}+y=0 \tag{7}\\ & \frac{d^{3} y}{d x^{3}}+x^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=0 \tag{8} \end{align*} $$

समीकरण (6), (7) एवं (8) में क्रमशः प्रथम, द्वितीय एवं तृतीय कोटि के उच्चतम अवकलज उपस्थित हैं इसलिए इन समीकरणों की कोटि क्रमशः 1,2 एवं 3 है।

9.2.2 अवकल समीकरण की घात (Degree of a differential equation)

किसी अवकल समीकरण की घात का अध्ययन करने के लिए मुख्य बिंदु यह है कि वह अवकल समीकरण, अवकलजों $y^{\prime}, y^{\prime \prime}, y^{\prime \prime \prime}$ इत्यादि में बहुपद समीकरण होना चाहिए। निम्नलिखित समीकरणों पर विचार कीजिए:

$ \begin{aligned} \frac{d^{3} y}{d x^{3}}+2(\frac{d^{2} y}{d x^{2}})^{2}-\frac{d y}{d x}+y & =0 \\ (\frac{d y}{d x})^{2}+(\frac{d y}{d x})-\sin ^{2} y & =0 \\ \frac{d y}{d x}+\sin (\frac{d y}{d x}) & =0 \end{aligned} $

हम प्रेक्षित करते हैं कि समीकरण (9) $y^{\prime \prime \prime}, y^{\prime \prime}$ एवं $y^{\prime}$ में बहुपद समीकरण है। समीकरण (10) $y^{\prime}$ में बहुपद समीकरण है (यद्यपि यह $y$ में बहुपद नहीं है) इस प्रकार के अवकल समीकरणों की घात को परिभाषित किया जा सकता है। परंतु समीकरण (11) $y^{\prime}$ में बहुपद समीकरण नहीं है और इस प्रकार के अवकल समीकरण की घात को परिभाषित नहीं किया जा सकता है।

यदि एक अवकल समीकरण अवकलजों का बहुपद समीकरण है तो उस अवकल समीकरण की घात से हमारा तात्पर्य है उस अवकल समीकरण में उपस्थित उच्चतम कोटि के अवकलज की उच्चतम घात (धनात्मक पूर्णांक)

उपरोक्त परिभाषा के संदर्भ में हम प्रेक्षित कर सकते हैं कि समीकरणों (6), (7), (8) एवं (9) में से प्रत्येक की घात 1 है, समीकरण (10) की घात 2 है जब कि अवकल समीकरण (11) की घात परिभाषित नहीं है।

टिप्पणी किसी अवकल समीकरण की कोटि एवं घात (यदि परिभाषित हो) हमेशा धनात्मक पूर्णांक होते हैं।

उदाहरण 1 निम्नलिखित अवकल समीकरणों में से प्रत्येक की कोटि एवं घात (यदि परिभाषित हो) ज्ञात कीजिए:

(i) $\frac{d y}{d x}-\cos x=0$

(ii) $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$

(iii) $y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0$

हल

(i) इस अवकल समीकरण में उपस्थित उच्चतम कोटि अवकलज $\frac{d y}{d x}$ है। इसलिए इसकी कोटि 1 है। यह $y^{\prime}$ में बहुपद समीकरण है एवं $\frac{d y}{d x}$ की अधिकतम घातांक 1 है, इसलिए इस अवकल समीकरण की घात 1 है।

(ii) इस अवकल समीकरण में उपस्थित उच्चतम कोटि अवकलज $\frac{d^{2} y}{d x^{2}}$ है । इसलिए इसकी कोटि 2 है। यह अवकल समीकरण $\frac{d^{2} y}{d x^{2}}$ एवं $\frac{d y}{d x}$ में बहुपद समीकरण है और $\frac{d^{2} y}{d x^{2}}$ की अधिकतम घातांक 1 है, इसलिए इस अवकल समीकरण की घात 1 है।

(iii) इस अवकल समीकरण में उपस्थित उच्चतम कोटि अवकलज $y^{\prime \prime \prime}$ है। इसलिए इसकी कोटि 3 है। इस समीकरण का बायाँ पक्ष अवकलजों में बहुपद नहीं है इसलिए इसकी घात परिभाषित नहीं है।

प्रश्नावली 9.1

1 से 10 तक के प्रश्नों में प्रत्येक अवकल समीकरण की कोटि एवं घात (यदि परिभाषित हो) ज्ञात कीजिए।

1. $\frac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0$

Show Answer #missing

2. $y^{\prime}+5 y=0$

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3. $\left(\frac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0$

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4. $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$

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5. $\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x$

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6. $\left(y^{\prime \prime \prime}\right)^{2}+\left(y^{\prime \prime}\right)^{3}+\left(y^{\prime}\right)^{4}+y^{5}=0$

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7. $y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$

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8. $y^{\prime}+y=e^{x}$

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9. $y^{\prime \prime}+\left(y^{\prime}\right)^{2}+2 y=0$

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10. $y^{\prime \prime}+2 y^{\prime}+\sin y=0$

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11. अवकल समीकरण

$\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0$ की घात है:

(A) 3

(B) 2

(C) 1

(D) परिभाषित नहीं है

Show Answer #missing

12. अवकल समीकरण

$2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$ की कोटि है:

(A) 2

(B) 1

(C) 0

(D) परिभाषित नहीं है

Show Answer #missing

9.3. अवकल समीकरण का व्यापक एवं विशिष्ट हल (General and Particular Solutions of a Differential Equation)

पिछली कक्षाओं में हमने निम्नलिखित प्रकार के समीकरणों को हल किया है:

$$ \begin{align*} x^{2}+1=0 \tag{1} \\ \sin ^{2} x-\cos x=0 \tag{2} \end{align*} $$

समीकरणों (1) तथा (2) का हल एक ऐसी वास्तविक अथवा सम्मिश्र संख्या है जो दिए हुए समीकरण को संतुष्ट करती है अर्थात् जब इस संख्या को समीकरण में अज्ञात $x$ के स्थान पर प्रतिस्थापित कर दिया जाता है तो दायाँ पक्ष और बायाँ पक्ष आपस में बराबर हो जाते हैं।

अब अवकल समीकरण पर विचार करते हैं।

$\frac{d^{2} y}{d x^{2}}+y=0$

प्रथम दो समीकरणों के विपरीत इस अवकल समीकरण का हल एक ऐसा फलन $\phi$ है जो इस समीकरण को संतुष्ट करेगा अर्थात् जब इस फलन $\phi$ को अवकल समीकरण में अज्ञात $y$ (आश्रित चर) के स्थान पर प्रतिस्थापित कर दिया जाता है तो बायाँ पक्ष और दायाँ पक्ष बराबर हो जाते हैं।

वक्र $y=\phi(x)$ अवकल समीकरण का हल वक्र (समाकलन वक्र) कहलाता है। निम्नलिखित फलन पर विचार कीजिए

$$ \begin{equation*} y=\phi(x)=a \sin (x+b) \tag{4} \end{equation*} $$

जहाँ $a, b \in \mathbf{R}$. यदि इस फलन और इसके अवकलजों को समीकरण (3) में प्रतिस्थापित कर दिया जाए तो बायाँ पक्ष और दायाँ पक्ष बराबर हो जाते हैं। इसलिए यह फलन अवकल समीकरण (3) का हल है।

मान लीजिए कि $a$ और $b$ को कुछ विशिष्ट मान $a=2$ एवं $b=\frac{\pi}{4}$ दे दिए जाते हैं तो हमें निम्नलिखित फलन प्राप्त होता है:

$$ \begin{equation*} y=\phi _{1}(x)=2 \sin \left(x+\frac{\pi}{4}\right) \tag{5} \end{equation*} $$

यदि इस फलन और इसके अवकलजों को समीकरण (3) में प्रतिस्थापित कर दिया जाए तो पुन: बायाँ पक्ष और दायाँ पक्ष बराबर हो जाते हैं। इसलिए $\phi _{1}$ भी समीकरण (3) का एक हल है।

फलन $\phi$ में दो स्वेच्छ अचर (प्राचल) $a, b$ सम्मिलित हैं तथा यह फलन दिए हुए अवकल समीकरण का व्यापक हल कहलाता है। जबकि फलन $\phi _{1}$ में कोई भी स्वेच्छ अचर सम्मिलित नहीं है लेकिन प्राचलों $a$ तथा $b$ के विशिष्ट मान उपस्थित हैं और इसलिए इसको अवकल समीकरण का विशिष्ट हल कहा जाता है।

ऐसा हल, जिसमें स्वेच्छ अचर उपस्थित हो अवकल समीकरण का व्यापक हल कहलाता है। ऐसा हल, जो स्वेच्छ अचरों से मुक्त है अर्थात् व्यापक हल में स्वेच्छ अचरों को विशिष्ट मान देने पर प्राप्त हल, अवकल समीकरण का विशिष्ट हल कहलाता है।

उदाहरण 2 सत्यापित कीजिए कि फलन $y=e^{-3 x}$, अवकल समीकरण $\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-6 y=0$ का एक हल है।

हल दिया हुआ फलन $y=e^{-3 x}$ है। इसके दोनों पक्षों का $x$ के सापेक्ष अवकलन करने पर हम प्राप्त करते है:

$$ \begin{equation*} \frac{d y}{d x}=3 e^{-3 x} \tag{1} \end{equation*} $$

अब समीकरण (1) का $x$ के सापेक्ष पुनः अवकलन करने पर हम देखते हैं कि

$$ \frac{d^{2} y}{d x^{2}}=9 e^{-3 x} $$

$\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ और $y$ का मान, दिए गए अवकल समीकरण में प्रतिस्थापित करने पर पर हम पाते हैं

एल.एच.एस. $=9 e^{-3 x}+(-3 e^{-3 x})-6 . e^{-3 x}=9 e^{-3 x}-9 e^{-3 x}=0=$ आर.एच.एस.

दायाँ पक्ष इसलिए दिया हुआ फलन दिए हुए अवकल समीकरण का एक हल है।

उदाहारण 3सत्यापित कीजिए कि फलन $y=a \cos x+b \sin x$, जिसमें $a, b \in \mathbf{R}$, अवकल समीकरण $\frac{d^{2} y}{d x^{2}}+y=0$ का हल है।

हल दिया हुआ फलन है

$$ \begin{equation*} y=a \cos x+b \sin x \tag{1} \end{equation*} $$

समीकरण (1) के दोनों पक्षों का $x$, के सापेक्ष उत्तरोत्तर अवकलन करने पर हम देखते हैं:

$$ \begin{aligned} \frac{d y}{d x} & =-a \sin x+b \cos x \\ \frac{d^{2} y}{d x^{2}} & =-a \cos x-b \sin x \end{aligned} $$

$\frac{d^{2} y}{d x^{2}}$ एवं $y$ का मान दिए हुए अवकल समीकरण में प्रतिस्थापित करने पर प्राप्त करते हैं:

बायाँ पक्ष $=(-a \cos x-b \sin x)+(a \cos x+b \sin x)=0=$ दायाँ पक्ष

इसलिए दिया हुआ फलन, दिए हुए अवकल समीकरण का हल है।

प्रश्नावली 9.2

1 से 10 तक प्रत्येक प्रश्न में सत्यापित कीजिए कि दिया हुआ फलन (स्पष्ट अथवा अस्पष्ट) संगत अवकल समीकरण का हल है:

1. $y=e^{x}+1 \quad: \quad y^{\prime \prime}-y^{\prime}=0$

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2. $y=x^{2}+2 x+\mathrm{C} \quad: \quad y^{\prime}-2 x-2=0$

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3. $y=\cos x+\mathrm{C} \quad: \quad y^{\prime}+\sin x=0$

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4. $y=\sqrt{1+x^{2}} \quad: \quad y^{\prime}=\frac{x y}{1+x^{2}}$

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5. $y=\mathrm{A} x \quad: \quad x y^{\prime}=y(x \neq 0)$

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6. $y=x \sin x \quad: \quad x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0$ और $x>y$ अथवा $x<-y)$

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7. $x y=\log y+\mathrm{C} \quad: \quad y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)$

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8. $y-\cos y=x \quad: \quad(y \sin y+\cos y+x) y^{\prime}=y$

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9. $x+y=\tan ^{-1} y \quad: \quad y^{2} y^{\prime}+y^{2}+1=0$

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10. $y=\sqrt{a^{2}-x^{2}} x \in(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)$

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11. चार कोटि वाले किसी अवकल समीकरण के व्यापक हल में उपस्थित स्वेच्छ अचरों की संख्या है:

(A) 0

(B) 2

(C) 3

(D) 4

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12. तीन कोटि वाले किसी अवकल समीकरण के विशिष्ट हल में उपस्थित स्वेच्छ अचरों की संख्या है:

(A) 3

(B) 2

(C) 1

(D) 0

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9.4. प्रथम कोटि एवं प्रथम घात के अवकल समीकरणों को हल करने की विधियाँ (Methods of Solving First अथवाder, First Degree Differential Equations)

इस परिच्छेद में हम प्रथम कोटि एवं प्रथम घात के अवकल समीकरणों को हल करने की तीन विधियों की चर्चा करेंगे।

9.4.1 पृथक्करणीय चर वाले अवकल समीकरण (Differential equations with variables separable)

प्रथम कोटि एवं प्रथम घात का अवकल समीकरण निम्नलिखित रूप का होता है:

$$ \begin{equation*} \frac{d y}{d x}=\mathrm{F}(x, y) \tag{1} \end{equation*} $$

यदि $\mathrm{F}(x, y)$ को गुणनफल $\mathrm{g}(x), h(y)$ के रूप में अभिव्यक्त किया जा सकता है जहाँ $g(x), x$ का फलन है और $h(y), y$ का एक फलन है तो समीकरण (1) पृथक्करणीय चर वाला समीकरण कहलाता है। ऐसा होने पर समीकरण (1) को निम्नलिखित रूप में लिखा जा सकता है:

$$ \begin{equation*} \frac{d y}{d x}=h(y) \cdot g(x) \tag{2} \end{equation*} $$

यदि $h(y) \neq 0$, तो चरों को पृथक् करते हुए समीकरण (2) को

$$ \begin{equation*} \frac{1}{h(y)} d y=g(x) d x \tag{3} \end{equation*} $$

के रूप में लिखा जा सकता है। समीकरण (3) के दोनों पक्षों का समाकलन करने पर हम प्राप्त करते हैं:

$$ \begin{equation*} \int \frac{1}{h(y)} d y=\int g(x) d x \tag{4} \end{equation*} $$

इस प्रकार समीकरण (4), दिए हुए अवकल समीकरण का हल निम्नलिखित रूप में प्रदान करता है:

$$ \begin{equation*} \mathrm{H}(y)=\mathrm{G}(x)+\mathrm{C} \tag{5} \end{equation*} $$

यहाँ $\mathrm{H}(y)$ एवं $\mathrm{G}(x)$ क्रमशः $\frac{1}{h(y)}$ एवं $g(x)$ के प्रतिअवकलज हैं और $\mathrm{C}$ स्वेच्छ अचर है।

उदाहरण 4 अवकल समीकरण $\frac{d y}{d x}=\frac{x+1}{2-y},(y \neq 2)$ का व्यापक हल ज्ञात कीजिए।

हल दिया गया है कि

$$ \begin{equation*} \frac{d y}{d x}=\frac{x+1}{2-y}(y \neq 2) \tag{1} \end{equation*} $$

समीकरण (1) में चरों को पृथक् करने पर हम प्राप्त करते हैं :

$$ \begin{equation*} (2-y) d y=(x+1) d x \tag{2} \end{equation*} $$

समीकरण (2) के दोनों पक्षों का समाकलन करने पर हम प्राप्त करते हैं :

$$ \int(2-y) d y=\int(x+1) d x $$

$$ \text{ अथवा } \qquad 2 y-\frac{y^{2}}{2}=\frac{x^{2}}{2}+x+\mathrm{C} _{1} $$

$$ \begin{equation*} \text{ अथवा } \qquad x^{2}+y^{2}+2 x-4 y+2 \mathrm{C} _{1}=0 \tag{3} \end{equation*} $$

$\text{ अथवा } \qquad x^{2}+y^{2}+2 x-4 y+\mathrm{C}=0 \text { जहां } \mathrm{C}=2 \mathrm{C} _{1}$

समीकरण (3) अवकल समीकरण (1) का व्यापक हल है।

उदाहरण 5 अवकल समीकरण $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$ का व्यापक हल ज्ञात कीजिए।

हल चूँकि $1+y^{2} \neq 0$, इसलिए चरों को पृथक् करते हुए दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है:

$$ \begin{equation*} \frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}} \tag{1} \end{equation*} $$

समीकरण (1) के दोनों पक्षों का समाकलन करते हुए हम पाते हैं:

$$ \int \frac{d y}{1+y^{2}}=\int \frac{d x}{1+x^{2}} $$

$$\text{ अथवा }\qquad \tan ^{-1} y=\tan ^{-1} x+\mathrm{C} $$

यह समीकरण (1) का व्यापक हल है।

उदाहरण 6 अवकल समीकरण $\frac{d y}{d x}=-4 x y^{2}$ का विशिष्ट हल ज्ञात कीजिए, यदि $y=1$ जब $x=0$ हो

हल यदि $y \neq 0$, दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है:

$$ \begin{equation*} \frac{d y}{y^{2}}=-4 x d x \tag{1} \end{equation*} $$

समीकरण (1) के दोनों पक्षों का समाकलन करने पर हम पाते हैं:

$$ \begin{align*} & \int \frac{d y}{y^{2}}=-4 \int x d x \\ & -\frac{1}{y}=-2 x^{2}+\mathrm{C} \\ & \text{ अथवा } \qquad y=\frac{1}{2 x^{2}-\mathrm{C}} \end{align*} $$

समीकरण (2) में $y=1$ और $x=0$ प्रतिस्थापित करने पर हमें $\mathrm{C}=-1$ प्राप्त होता है।

$\mathrm{C}$ का मान समीकरण (2) में प्रतिस्थापित करने पर दिए हुए अवकल समीकरण का विशिष्ट हल $y=\frac{1}{2 x^{2}+1}$ प्राप्त होता है।

उदाहरण 7 बिंदु $(1,1)$ से गुजरने वाले एक ऐसे वक्र का समीकरण कीजिए जिसका अवकल समीकरण $x d y=(2 x^{2}+1) d x(x \neq 0)$ है।

हल दिए हुए अवकल समीकरण को निम्नलिखित रूप मे अभिव्यक्त किया जा सकता है:

$\text{ अथवा } \qquad dy $ $ =(\frac{2x^2+1}{x}) dx \\ dy =(2 x+\frac{1}{x}) d x $

समीकरण (1) के दोनों पक्षों का समाकलन करने पर हम प्राप्त करते हैं:

$$ \int d y=\int\left(2 x+\frac{1}{x}\right) d x $$

$ \begin{equation*} \text{ अथवा }\qquad y=x^{2}+\log |x|+\mathrm{C} \tag{2} \end{equation*} $

समीकरण (2) दिए हुए अवकल समीकरण के हल वक्रों के कुल को निरूपित करता है परंतु हम इस कुल के एक ऐसे विशिष्ट सदस्य का समीकरण ज्ञात करना चाहते हैं जो बिंदु $(1,1)$ से गुजरता हो।

इसलिए समीकरण (2) में $x=1, y=1$ प्रतिस्थापित करने पर हमें $\mathrm{C}=0$ प्राप्त होता है। $\mathrm{C}$ का मान समीकरण (2) में प्रतिस्थापित करने पर हमें अभीष्ट वक्र का समीकरण $y=x^{2}+\log |x|$ के रूप में प्राप्त होता है।

उदाहरण 8 बिंदु $(-2,3)$, से गुजरने वाले ऐसे वक्र का समीकरण ज्ञात कीजिए जिसके किसी बिंदु $(x, y)$ पर स्पर्श रेखा की प्रवणता $\frac{2 x}{y^{2}}$ है।

हल हम जानते हैं कि किसी वक्र की स्पर्श रेखा की प्रवणता $\frac{d y}{d x}$ के बराबर होती है। इसलिए

$$ \begin{equation*} \frac{d y}{d x}=\frac{2 x}{y^{2}} \tag{1} \end{equation*} $$

चरों को पृथक् करते हुए समीकरण (1) को निम्नलिखित रूप में लिखा जा सकता है :

$$ \begin{equation*} y^{2} d y=2 x d x \tag{2} \end{equation*} $$

समीकरण (2) के दोनों पक्षों का समाकलन करने पर हम प्राप्त करते हैं :

$$ \int y^{2} d y=\int 2 x d x $$

$$ \begin{equation*} \text{ अथवा } \qquad \frac{y^{3}}{3}=x^{2}+\mathrm{C} \tag{3} \end{equation*} $$

समीकरण (3) में $x=-2, y=3$ प्रतिस्थापित करने पर हमें $\mathrm{C}=5$ प्राप्त होता है।

$\mathrm{C}$ का मान समीकरण (3) में प्रतिस्थापित करने पर हमें अभीष्ट वक्र का समीकरण के रूप में प्राप्त होता है।

$$ \frac{y^{3}}{3}=x^{2}+5 \text { अथवा } y=\left(3 x^{2}+15\right)^{\frac{1}{3}} $$

उदाहरण 9 किसी बैंक में मूलधन की वृद्धि $5 \%$ वार्षिक की दर से होती है। कितने वर्षों में Rs 1000 की राशि दुगुनी हो जाएगी?

हल मान लीजिए किसी समय $t$ पर मूलधन $\mathrm{P}$ है। दी हुई समस्या के अनुसार

$$ \begin{align*} & \frac{d \mathrm{P}}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P} \\ & \frac{d \mathrm{P}}{d t}=\frac{\mathrm{P}}{20} \tag{1} \end{align*} $$

समीकरण (1) में चरों को पृथक् करने पर, हम प्राप्त करते हैं :

$$ \begin{equation*} \frac{d \mathrm{P}}{\mathrm{P}}=\frac{d t}{20} \tag{2} \end{equation*} $$

समीकरण (2) के दोनों पक्षों का समाकलन करने पर हम प्राप्त करते हैं :

$$ \text{ अथवा } \qquad \log \mathrm{P}=\frac{t}{20}+\mathrm{C} _{1} $$ $$ \mathrm{P}=e^{\frac{t}{20}} \cdot e^{c _{1}} $$

$$ \begin{equation*} \text{ अथवा } \qquad \mathrm{P}=\mathrm{C} e^{\frac{t}{20}} \quad\left(\text { जहाँ } e^{\mathrm{C} _{1}}=\mathrm{C}\right) \tag{3} \end{equation*} $$

अब $\qquad\mathrm{P}=1000, \quad \text { जब } t=0$

$\mathrm{P}$ और $t$ का मान समीकरण (3) में रखने पर हम $\mathrm{C}=1000$ प्राप्त करते हैं।

इसलिए समीकरण (3) से हम प्राप्त करते हैं :

$$ \mathrm{P}=1000 e^{\frac{t}{20}} $$

मान लीजिए $t$ वर्षों में मूलधन दुगुना हो जाता है, तब

$$ 2000=1000 e^{\frac{t}{20}} \Rightarrow t=20 \log _{e} 2 $$

प्रश्नावली 9.3

1 से 10 तक के प्रश्नों में, प्रत्येक अवकल समीकरण का व्यापक हल ज्ञात कीजिए।

1. $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x} \quad$

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2.$\frac{d y}{d x}=\sqrt{4-y^{2}}(-2<y<2)$

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3. $\frac{d y}{d x}+y=1(y \neq 1)$

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4. $\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

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5. $(e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0$

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6. $\frac{d y}{d x}=(1+x^{2})(1+y^{2})$

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7. $y \log y d x-x d y=0$

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8. $x^{5} \frac{d y}{d x}=-y^{5}$

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9. $\frac{d y}{d x}=\sin ^{-1} x$

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10. $e^{x} \tan y d x+(1-e^{x}) \sec ^{2} y d y=0$

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11 से 14 तक के प्रश्नों में, प्रत्येक अवकल समीकरण के लिए दिए हुए प्रतिबंध को संतुष्ट करने वाला विशिष्ट हल ज्ञात कीजिए।

11. $(x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x ; y=1$ when $x=0$

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12. $x\left(x^{2}-1\right) \frac{d y}{d x}=1 ; y=0$ यदि $x=2$

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13. $\cos \left(\frac{d y}{d x}\right)=a(a \in \mathrm{R}) ; y=1$ यदि $x=0$

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14. $\frac{d y}{d x}=y \tan x ; y=2$ यदि $x=0$

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15. बिंदु $(0,0)$ से गुजरने वाले एक ऐसे वक्र का समीकरण ज्ञात कीजिए जिसका अवकल समीकरण $y^{\prime}=e^{x} \sin x$ है।

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16. अवकल समीकरण $x y \frac{d y}{d x}=(x+2)(y+2)$ के लिए बिंदु $(1,-1)$ से गुजरने वाला वक्र ज्ञात कीजिए।

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17. बिंदु $(0,-2)$ से गुजरने वाले एक ऐसे वक्र का समीकरण ज्ञात कीजिए जिसके किसी बिंदु $(x, y)$ पर स्पर्श रेखा की प्रवणता और उस बिंदु के $y$ निर्देशांक का गुणनफल उस बिंदु के $x$ निर्देशांक के बराबर है।

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18. एक वक्र के किसी बिंदु $(x, y)$ पर स्पर्श रेखा की प्रवणता, स्पर्श बिंदु को, बिंदु $(-4,-3)$. से मिलाने वाले रेखाखंड की प्रवणता की दुगुनी है। यदि यह वक्र बिंदु $(-2,1)$ से गुज़रता हो तो इस वक्र का समीकरण ज्ञात कीजिए।

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19. एक गोलाकार गुब्बारे का आयतन, जिसे हवा भरकर फुलाया जा रहा है, स्थिर गति से बदल रहा है यदि आरंभ में इस गुब्बारे की त्रिज्या 3 ईकाई है और 3 सेकेंड बाद 6 ईकाई है, तो $t$ सेकेंड बाद उस गुब्बारे की त्रिज्या ज्ञात कीजिए।

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20. किसी बैंक में मूलधन की वृद्धि $r \%$ वार्षिक की दर से होती है। यदि 100 रुपये 10 वर्षों में दुगुने हो जाते हैं, तो $r$ का मान ज्ञात कीजिए। $\left(\log _{e} 2=0.6931\right)$.

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21. किसी बैंक में मूलधन की वृद्धि $5 \%$ वार्षिक की दर से होती है। इस बैंक में Rs 1000 जमा कराए जाते हैं। ज्ञात कीजिए कि 10 वर्ष बाद यह राशि कितनी हो जाएगी? $\left(e^{0.5}=1.648\right)$

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22. किसी जीवाणु समूह में जीवाणुओं की संख्या $1,00,000$ है। 2 घंटो में इनकी संख्या में $10 \%$ की वृद्धि होती है। कितने घंटों में जीवाणुओं की संख्या $2,00,000$ हो जाएगी, यदि जीवाणुओं के वृद्धि की दर उनके उपस्थित संख्या के समानुपाती है।

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23. अवकल समीकरण $\frac{d y}{d x}=e^{x+y}$ का व्यापक हल है:

(A) $e^{x}+e^{-y}=\mathrm{C}$

(B) $e^{x}+e^{y}=\mathrm{C}$

(C) $e^{-x}+e^{y}=\mathrm{C}$

(D) $e^{-x}+e^{-y}=\mathrm{C}$

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9.4.2 समघातीय अवकल समीकरण (Homogenous differential equations)

$x$ एवं $y$ के निम्नलिखित फलनों पर विचार कीजिए

$$ \begin{array}{ll} \mathrm{F} _{1}(x, y)=y^{2}+2 x y, & \mathrm{~F} _{2}(x, y)=2 x-3 y, \\ \mathrm{~F} _{3}(x, y)=\cos \left(\frac{y}{x}\right), & \mathrm{F} _{4}(x, y)=\sin x+\cos y \end{array} $$

यदि उपरोक्त फलनों में $x$ और $y$ को किसी शून्येतर अचर $\lambda$ के लिए क्रमशः $\lambda x$ एवं $\lambda y$ से प्रतिस्थापित कर दिया जाए तो हम प्राप्त करते हैं:

$$ \begin{aligned} & \mathrm{F} _{1}(\lambda x, \lambda y)=\lambda^{2}\left(y^{2}+2 x y\right)=\lambda^{2} \mathrm{~F} _{1}(x, y) \\ & \mathrm{F} _{2}(\lambda x, \lambda y)=\lambda(2 x-3 y)=\lambda \mathrm{F} _{2}(x, y) \\ & \mathrm{F} _{3}(\lambda x, \lambda y)=\cos \left(\frac{\lambda y}{\lambda x}\right)=\cos \left(\frac{y}{x}\right)=\lambda^{0} \mathrm{~F} _{3}(x, y) \\ & \mathrm{F} _{4}(\lambda x, \lambda y)=\sin \lambda x+\cos \lambda y \neq \lambda^{n} \mathrm{~F} _{4}(x, y), \text { किसी भी } n \text { के लिए } \end{aligned} $$

यहाँ हम प्रेक्षित करते हैं कि फलनों $\mathrm{F} _{1}, \mathrm{~F} _{2}, \mathrm{~F} _{3}$ को $\mathrm{F}(\lambda x, \lambda y)=\lambda^{n} \mathrm{~F}(x, y)$ के रूप में लिखा जा सकता है परंतु फलन $\mathrm{F} _{4}$ को इस रूप में नहीं लिखा जा सकता है। इससे हम निम्नलिखित परिभाषा प्राप्त करते हैं।

फलन $\mathrm{F}(x, y), n$ घात वाला समघातीय फलन कहलाता है। यदि किसी शून्येतर अचर $\lambda$ के लिए $\mathrm{F}(\lambda x, \lambda y)=\lambda^{n} \mathrm{~F}(x, y)$

हम नोट करते हैं कि उपरोक्त उदाहरणों में $\mathrm{F} _{1}, \mathrm{~F} _{2}, \mathrm{~F} _{3}$ क्रमशः $2,1,0$ घात वाले समघातीय फलन हैं जबकि $\mathrm{F} _{4}$ समघातीय फलन नहीं है।

हम यह भी प्रेक्षित करते हैं कि

$ \begin{aligned} & \qquad F_1(x, y)=x^{2}(\frac{y^{2}}{x^{2}}+\frac{2 y}{x})=x^{2} h_1(\frac{y}{x}) \\ & \qquad F_1(x, y)=y^{2}(1+\frac{2 x}{y})=y^{2} h_2(\frac{x}{y}) \\ & \text{अथवा}\qquad F_2(x, y)=x^{1}(2-\frac{3 y}{x})=x^{1} h_3(\frac{y}{x}) \\ & \qquad F_2(x, y)=y^{1}(2 \frac{x}{y}-3)=y^{1} h_4(\frac{x}{y}) \\ & \qquad F_3(x, y)=x^{0} \cos (\frac{y}{x})=x^{0} h_5(\frac{y}{x}) \\ & \qquad F_4(x, y) \neq x^{n} h_6(\frac{y}{x}), \text{ किसी के लिए } n \in \mathbf{N} \\ & .\qquad F_4(x, y) \neq y^{n} h_7(\frac{x}{y}), \text{ किसी के लिए } n \in \mathbf{N} \end{aligned} $

$$ \begin{aligned} & \text{अथवा}\qquad \mathrm{F} _{4}(x, y) \neq x^{n} h _{6}\left(\frac{y}{x}\right), n \in \mathbf{N} \text { के किसी भी मान के लिए } \\ & \qquad \mathrm{F} _{4}(x, y) \neq y^{n} h _{7}\left(\frac{x}{y}\right), n \in \mathbf{N} \end{aligned} $$

इसलिए एक फलन $\mathrm{F}(x, y), n$ घात वाला समघातीय फलन कहलाता है यदि $\frac{d y}{d x}=\mathrm{F}(x, y)$ के रूप वाला अवकल समीकरण समघातीय कहलाता है यदि $\mathrm{F}(x, y)$ शून्य घात वाला समघातीय फलन है।

के रूप वाले समघातीय अवकल समीकरण को हल करने के लिए हम $\frac{y}{x}=v$ अर्थात्

$$ \mathrm{F}(x, y)=x^{n} g\left(\frac{y}{x}\right) \quad \text { अथवा } \quad y^{n} h\left(\frac{x}{y}\right) $$

$$ \begin{equation*} \frac{d y}{d x}=\mathrm{F}(x, y)=g\left(\frac{y}{x}\right) \tag{1} \end{equation*} $$

$\text{ प्रतिस्थापित करते हैं }\qquad y=v x \tag{2} $

समीकरण (2) का $x$ के सापेक्ष अवकलन करने पर हम प्राप्त करते हैं :

$$ \begin{equation*} \frac{d y}{d x}=v+x \frac{d v}{d x} \tag{3} \end{equation*} $$

समीकरण (3) से $\frac{d y}{d x}$ का मान समीकरण (1) में प्रतिस्थापित करने पर हम प्राप्त करते हैं :

$$ v+x \frac{d v}{d x}=g(v) $$

$$ \begin{equation*} x \frac{d v}{d x}=g(v)-v \tag{4} \end{equation*} $$

समीकरण (4) में चरों को पृथक् करने पर हम प्राप्त करते हैं :

$$ \begin{equation*} \frac{d v}{g(v)-v}=\frac{d x}{x} \tag{5} \end{equation*} $$

समीकरण (5) के दोनों पक्षों का समाकलन करने पर हमें प्राप्त होता है:

$$ \begin{equation*} \int \frac{d v}{g(v)-v}=\int \frac{1}{x} d x+\mathrm{C} \tag{6} \end{equation*} $$

यदि $v$ को $\frac{y}{x}$ से प्रतिस्थापित कर दिया जाए तो समीकरण (6), अवकल समीकरण (1) का व्यापक हल प्रदान करता है।

2 टिप्पणी यदि समघातीय अवकल समीकरण $\frac{d x}{d y}=\mathrm{F}(x, y)$ के रूप में है। जहाँ $\mathrm{F}(x, y)$ शून्य घात वाला समघातीय फलन है तो हम $\frac{x}{y}=v$ अर्थात, $x=v y$ प्रतिस्थापित करते हैं और फिर उपरोक्त चर्चा के अनुसार $\frac{d x}{d y}=\mathrm{F}(x, y)=h\left(\frac{x}{y}\right)$ के रूप में लिखकर व्यापक हल ज्ञात करने के लिए आगे बढ़ते हैं।

उदाहरण 10 दर्शाइए कि अवकल समीकरण $(x-y) \frac{d y}{d x}=x+2 y$ समघातीय है और इसका हल ज्ञात कीजिए।

हल दिए गए अवकल समीकरण को निम्नलिखित रूप में अभिव्यक्त किया जा सकता है :

$$ \begin{equation*} \frac{d y}{d x}=\frac{x+2 y}{x-y} \tag{1} \end{equation*} $$

मान लीजिए $$ \mathrm{F}(x, y)=\frac{x+2 y}{x-y} $$

अब $$ \mathrm{F}(\lambda x, \lambda y)=\frac{\lambda(x+2 y)}{\lambda(x-y)}=\lambda^{\circ} \cdot \mathrm{F}(x, y) $$

इसलिए $\mathrm{F}(x, y)$ शून्य घात वाला समघातीय फलन है। अतः दिया हुआ अवकल समीकरण एक समघातीय अवकल समीकरण है।

विकल्पत:

$$ \begin{equation*} \frac{d y}{d x}=\left(\frac{1+\frac{2 y}{x}}{1-\frac{y}{x}}\right)=g\left(\frac{y}{x}\right) \tag{2} \end{equation*} $$

समीकरण (2) का दायाँ पक्ष $g\left(\frac{y}{x}\right)$ के रूप में है इसलिए यह शून्य घात वाला एक समघातीय फलन है। इसलिए समीकरण (1) एक समघातीय अवकल समीकरण है। इसको हल करने के लिए हम प्रतिस्थापन करते है:

$$ \begin{equation*} y=v x \tag{3} \end{equation*} $$

समीकरण (3) का $x$ के सापेक्ष अवकलन करने पर हम प्राप्त करते हैं :

$$ \begin{equation*} \frac{d y}{d x}=v+x \frac{d v}{d x} \tag{4} \end{equation*} $$

समीकरण (1) में $y$ एवं $\frac{d y}{d x}$ का मान प्रतिस्थापित करने पर हम प्राप्त करते हैं :

$$ v+x \frac{d v}{d x}=\frac{1+2 v}{1-v} $$

$\text{ अर्थात् }\qquad x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v $

$\text{ अर्थात् }\qquad x \frac{d v}{d x}=\frac{v^{2}+v+1}{1-v} $

$\text{ अर्थात् }\qquad \frac{v-1}{v^{2}+v+1} d v=\frac{-d x}{x} $

समीकरण (5) के दोनों पक्षों का समाकलन करने पर हम प्राप्त करते हैं :

$$ \int \frac{v-1}{v^{2}+v+1} d v=-\int \frac{d x}{x} $$

$$\text{ अर्थात् }\qquad \frac{1}{2} \int \frac{2 v+1-3}{v^{2}+v+1} d v=-\log |x|+\mathrm{C} $$

$$\text{ अर्थात् }\qquad \frac{1}{2} \int \frac{2 v+1}{v^{2}+v+1} d v-\frac{3}{2} \int \frac{1}{v^{2}+v+1} d v=-\log |x|+\mathrm{C} $$

$$\text{ अर्थात् }\qquad \frac{1}{2} \log \left|v^{2}+v+1\right|-\frac{3}{2} \int \frac{1}{\left(v+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d v=-\log |x|+\mathrm{C} $$

$$\text{ अर्थात् }\qquad \frac{1}{2} \log \left|v^{2}+v+1\right|-\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)=-\log |x|+\mathrm{C} $$

$v$ को $\frac{y}{x}$, से प्रतिस्थापित करने पर हम प्राप्त करते हैं :

अथवा $$ \frac{1}{2} \log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|+\frac{1}{2} \log x^{2}=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+C $$

$$ \frac{1}{2} \log \left|\left(\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right) x^{2}\right|=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+\mathrm{C} _{1} $$

अथवा $$ \log \left|\left(y^{2}+x y+x^{2}\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+2 \mathrm{C} _{1} $$

$$ \log \left|\left(x^{2}+x y+y^{2}\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+\mathrm{C} $$

यह अवकल समीकरण (1) का व्यापक हल है।

उदाहरण 11 दर्शाइए कि अवकल समीकरण $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ समघातीय है और इसका हल ज्ञात कीजिए।

हल दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है :

$$ \begin{equation*} \frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)} \tag{1} \end{equation*} $$

यहाँ $\frac{d y}{d x}=\mathrm{F}(x, y)$ के रूप का अवकल समीकरण है।

यहाँ $\mathrm{F}(x, y)=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}$ है।

$x$ को $\lambda x$ से एवं $y$ को $\lambda y$ से प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$$ \mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left[y \cos \left(\frac{y}{x}\right)+x\right]}{\lambda\left(x \cos \frac{y}{x}\right)}=\lambda^{0}[\mathrm{~F}(x, y)] $$

$\mathrm{F}(x, y)$ शून्य घात वाला समघातीय फलन है, इसलिए दिया हुआ अवकल समीकरण एक समघातीय अवकल समीकरण है। इसको हल करने के लिए हम प्रतिस्थापन करते हैं:

$$ \begin{equation*} y=v x \tag{2} \end{equation*} $$

समीकरण (2) का $x$ के सापेक्ष अवकलन करने पर हम प्राप्त करते हैं :

$$ \begin{equation*} \frac{d y}{d x}=v+x \frac{d v}{d x} \tag{3} \end{equation*} $$

समीकरण (1) में $y$ एवं $\frac{d y}{d x}$ का मान प्रतिस्थापित करने पर हम प्राप्त करते हैं :

$$ v+x \frac{d v}{d x}=\frac{v \cos v+1}{\cos v} $$

$$\text{ अर्थात् }\qquad x \frac{d v}{d x}=\frac{v \cos v+1}{\cos v}-v $$

$$\text{ अर्थात् }\qquad x \frac{d v}{d x}=\frac{1}{\cos v} $$

$$\text{ अर्थात् }\qquad \cos v d v=\frac{d x}{x} $$

$$\text{ इसलिए }\qquad \int \cos v d v=\int \frac{1}{x} d x $$

$\text{ अर्थात् }\qquad \sin v=\log |x|+\log |\mathrm{C}| $ $ \sin v=\log |\mathrm{C} x| $

$v$ को $\frac{y}{x}$ प्रतिस्थापित करने पर हम प्राप्त करते हैं।

$$ \sin \left(\frac{y}{x}\right)=\log |C x| $$

यह अवकल समीकरण (1) का व्यापक हल है।

उदाहरण 12 दर्शाइए कि अवकल समीकरण $2 y e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0$ समघातीय है और यदि, $x=0$ जब $y=1$ दिया हुआ हो तो इस समीकरण का विशिष्ट हल ज्ञात कीजिए।

हल दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है:

$$\text{ मान लीजिए }\qquad \begin{equation*} \frac{d x}{d y}=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}} \tag{1} \end{equation*} $$

$$ \mathrm{F}(x, y)=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}} \text { तब } \mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left(2 x e^{\frac{x}{y}}-y\right)}{\lambda\left(2 y e^{\frac{x}{y}}\right)}=\lambda^{\circ}[\mathrm{F}(x, y)] $$

अत: $\mathrm{F}(x, y)$ शून्य घात वाला समघातीय फलन है।

इसलिए, दिया हुआ अवकल समीकरण एक समघातीय अवकल समीकरण है।

इसका हल ज्ञात करने के लिए, हम $x=v y$ प्रतिस्थापन करते हैं।

समीकरण (2) का $y$ के सापेक्ष अवकलन करने पर हम प्राप्त करते हैं :

$$ \frac{d x}{d y}=v+y \frac{d v}{d y} \tag{2} $$

समीकरण (1) में $x$ एवं $\frac{d x}{d y}$ का मान प्रतिस्थापित करने पर हम प्राप्त करते हैं :

अथवा $ \begin{aligned} v+y \frac{d v}{d y} & =\frac{2 v e^{v}-1}{2 e^{v}} \\ y \frac{d v}{d y} & =\frac{2 v e^{v}-1}{2 e^{v}}-v \\ y \frac{d v}{d y} & =-\frac{1}{2 e^{v}} \\ 2 e^{v} d v & =\frac{-d y}{y} \\ \int 2 e^{v} \cdot d v & =-\int \frac{d y}{y} \\ 2 e^{v} & =-\log |y|+C \end{aligned} $

अथवा $v$ को $\frac{x}{y}$ से प्रतिस्थापित करने पर हम प्राप्त करते हैं :

$$ \begin{equation*} 2 e^{\frac{x}{y}}+\log |y|=\mathrm{C} \tag{3} \end{equation*} $$

समीकरण (3) में, $x=0$ एवं $y=1$ प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$$ 2 e^{0}+\log |1|=\mathrm{C} \Rightarrow \mathrm{C}=2 $$

$\mathrm{C}$ का मान समीकरण (3) में प्रतिस्थापित करने पर हम प्राप्त करते हैं :

$$ 2 e^{\frac{x}{y}}+\log |y|=2 $$

यह दिए हुए अवकल समीकरण का एक विशिष्ट हल है।

उदाहरण 13 दर्शाइए कि वक्रों का कुल, जिनके किसी बिंदु $(x, y)$ पर स्पर्श रेखा की प्रवणता $\frac{x^{2}+y^{2}}{2 x y}$ है, $x^{2}-y^{2}=c x$ द्वारा प्रदत्त है।

हल हम जानते हैं कि एक वक्र के किसी बिंदु पर स्पर्श रेखा की प्रवणता $\frac{d y}{d x}$ के बराबर होती है।

$$ \begin{equation*} \text{ इसलिए } \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y} \text { या } \frac{d y}{d x}=\frac{1+\frac{y^{2}}{x^{2}}}{\frac{2 y}{x}} \tag{1} \end{equation*} $$

स्पष्टतः समीकरण (1) समघातीय अवकल समीकरण है। इसको हल करने के लिए हम $y=v x$ प्रतिस्थापन करते हैं। $y=v x$ का $x$ के सापेक्ष अवकलन करने पर हम पाते हैं:

$$ \text{ अत: } \qquad \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \text { या } v+x \frac{d v}{d x}=\frac{1+v^{2}}{2 v} \\ & x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \text { या } \frac{2 v}{1-v^{2}} d v=\frac{d x}{x} \text { या } \frac{2 v}{v^{2}-1} d v=-\frac{d x}{x} \end{aligned} $$

$$ \text{ इसलिए } \qquad \int \frac{2 v}{v^{2}-1} d v=-\int \frac{1}{x} d x $$

$$ \text{ अथवा } \qquad \log \left|v^{2}-1\right|=-\log |x|+\log \left|\mathrm{C} _{1}\right| $$

$$ \text{ अथवा } \qquad \log \left|\left(v^{2}-1\right)(x)\right|=\log \left|\mathrm{C} _{1}\right| $$

$$ \text{ अथवा } \qquad \left(v^{2}-1\right) x= \pm \mathrm{C} _{1} $$

$v$ को $\frac{y}{x}$ से प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$$ \left(\frac{y^{2}}{x^{2}}-1\right) x= \pm \mathrm{C} _{1} $$

$$ \text{ अथवा } \qquad \left(y^{2}-x^{2}\right)= \pm \mathrm{C} _{1} x \text { या } x^{2}-y^{2}=\mathrm{C} x $$

प्रश्नावली 9.4

1 से 10 तक के प्रत्येक प्रश्न में दर्शाइए कि दिया हुआ अवकल समीकरण समघातीय है और इनमें से प्रत्येक को हल कीजिए:

1. $\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x$

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2. $y^{\prime}=\frac{x+y}{x}$

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3. $(x-y) d y-(x+y) d x=0$

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4. $\left(x^{2}-y^{2}\right) d x+2 x y d y=0$

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5. $x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$

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6. $x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

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7. ${x \cos (\frac{y}{x})+y \sin (\frac{y}{x})} y d x={y \sin (\frac{y}{x})-x \cos (\frac{y}{x})} x d y$

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8. $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$

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9. $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

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10. $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$

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11 से 15 तक के प्रश्नों में प्रत्येक अवकल समीकरण के लिए दिए हुए प्रतिबंध को संतुष्ट करने वाला विशिष्ट हल ज्ञात कीजिए।

11. $(x+y) d y+(x-y) d x=0 ; y=1$ यदि $x=1$

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12. $x^{2} d y+\left(x y+y^{2}\right) d x=0 ; y=1$ यदि $x=1$

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13. $\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0 ; y=\frac{\pi}{4}$ यदि $x=1$

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14. $\frac{d y}{d x}-\frac{y}{x}+\operatअथवाname{cosec}\left(\frac{y}{x}\right)=0 ; y=0$ यदि $x=1$

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15. $2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0 ; y=2$ यदि $x=1$

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16. $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$ के रूप वाले समघातीय अवकल समीकरण को हल करने के लिए निम्नलिखित में से कौन सा प्रतिस्थापन किया जाता है:

(A) $y=v x$

(B) $v=y x$

(C) $x=v y$

(D) $x=v$

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17. निम्नलिखित में से कौन सा समघातीय अवकल समीकरण है?

(A) $(4 x+6 y+5) d y-(3 y+2 x+4) d x=0$

(B) $(x y) d x-\left(x^{3}+y^{3}\right) d y=0$

(C) $\left(x^{3}+2 y^{2}\right) d x+2 x y d y=0$

(D) $y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0$

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9.4.3 रैखिक अवकल समीकरण (Linear differential equations)

$$ \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} $$

के रूप वाला अवकल समीकरण, जिसमें $\mathrm{P}$ एवं $\mathrm{Q}$ अचर अथवा केवल $x$ के फलन हैं, प्रथम कोटि का रैखिक अवकल समीकरण कहलाता है। प्रथम कोटि के रैखिक अवकल समीकरण के कुछ उदाहरण इस प्रकार हैं:

$ \begin{aligned} \frac{d y}{d x}+y & =\sin x \\ \frac{d y}{d x}+(\frac{1}{x}) y & =e^{x} \\ \frac{d y}{d x}+(\frac{y}{x \log x}) & =\frac{1}{x} \end{aligned} $

प्रथम कोटि के रैखिक अवकल समीकरण का दूसरा रूप सेकेंड

$\frac{d x}{d y}+\mathrm{P} _{1} x=\mathrm{Q} _{1}$ है,

जिसमें $\mathrm{P} _{1}$ और $\mathrm{Q} _{1}$ अचर अथवा केवल $y$ के फलन हैं। इस प्रकार के अवकल समीकरण के कुछ उदाहरण निम्नलिखित हैं:

$ \begin{matrix} \frac{d x}{d y}+x=\cos y \\ \frac{d x}{d y}+\frac{-2 x}{y}=y^{2} e^{-y} \end{matrix} $

प्रथम कोटि के रैखिक अवकल समीकरण

$$ \begin{equation*} \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \tag{1} \end{equation*} $$

को हल करने के लिए समीकरण के दोनों पक्षों को $x$ के फलन $g(x)$ से गुणा करने पर हम प्राप्त करते हैं:

$$ \begin{equation*} g(x) \frac{d y}{d x}+\mathrm{P} \cdot g(x) y=\mathrm{Q} \cdot g(x) \tag{2} \end{equation*} $$

$g(x)$ का चयन इस प्रकार कीजिए ताकि समीकरण का बायाँ पक्ष $y \cdot g(x)$ का अवकलज बन जाए:

$$ \text{ अर्थात् } \qquad g(x) \frac{d y}{d x}+\mathrm{P} \cdot g(x) y=\frac{d}{d x}[y \cdot g(x)] $$

$$\text{ अथवा } \qquad g(x) \frac{d y}{d x}+\mathrm{P} \cdot g(x) y=g(x) \frac{d y}{d x}+y g^{\prime}(x) $$

$\Rightarrow \quad$ P. $g(x)=g^{\prime}(x)$

$$\text{ अथवा } \qquad \mathrm{P}=\frac{g^{\prime}(x)}{g(x)} $$

दोनों पक्षों का $x$ के सापेक्ष समाकलन करने पर हम प्राप्त करते हैं:

$ \text{ अथवा } \qquad \begin{aligned} अथवा \quad\quad\quad\quad\int P d x & =\int \frac{g^{\prime}(x)}{g(x)} d x \\ \quad\quad\quad\quad\int P \cdot d x & =\log (g(x)) \\ g(x) & =e^{\int P d x} \end{aligned} $

समीकरण (1) को $g(x)=e^{\int p d x}$ से गुणा करने पर उस समीकरण का बायाँ पक्ष $x$ तथा $y$ के किसी फलन का अवकलज बन जाता है। यह फलन $g(x)=e^{\int p d x}$ दिए हुए अवकल समीकरण का समाकलन गुणक (I.F.) कहलाता है।

समीकरण (2) में $g(x)$ का मान प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$ e^{\int P d x} \frac{d y}{d x}+P e^{\int P d x} y=Q \cdot e^{\int P d x} $

अथवा $ \frac{d}{d x}(y e^{\int P d x})=Q e^{\int P d x} $

दोनों पक्षों का $x$, के सापेक्ष समाकलन करने पर हम प्राप्त करते हैं:

अथवा $ \begin{aligned} y \cdot e^{\int P d x} & =\int(Q \cdot e^{\int P d x}) d x \\ y & =e^{-\int P d x} \cdot \int(Q \cdot e^{\int P d x}) d x+C \end{aligned} $

यह अवकल समीकरण का व्यापक हल है।

प्रथम कोटि के रैखिक अवकल समीकरण को हल करने के लिए सम्मिलित चरण:

(i) दिए हुए अवकल समीकरण को $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ के रूप में लिखिए जिसमें $\mathrm{P}, \mathrm{Q}$ अचर अथवा केवल $x$ के फलन हैं।

(ii) समाकलन गुणक (I.F.) $=e^{\int \mathrm{P} d x}$ ज्ञात कीजिए।

(iii) दिए हुए अवकल समीकरण का हल निम्नलिखित रूप में लिखिए:

$$ y \cdot(\text { I.F. })=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d x+\mathrm{C} $$

यदि प्रथम कोटि का रैखिक अवकल समीकरण $\frac{d x}{d y}+\mathrm{P} _{1} x=\mathrm{Q} _{1}$ के रूप में है जिसमें $\mathrm{P} _{1}$ और $\mathrm{Q} _{1}$ अचर अथवा केवल $y$ के फलन हैं, तब I.F. $=e^{\int \mathrm{P} _{1} d y}$ और

$$ x . \text { (I.F.) }=\int\left(\mathrm{Q} _{1} \times \text { I.F. }\right) d y+\mathrm{C} \text { अवकल समीकरण का हल है। } $$

उदाहरण 14 अवकल समीकरण $\frac{d y}{d x}-y=\cos x$ का व्यापक हल ज्ञात कीजिए।

हल दिया हुआ अवकल समीकरण

$$ \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \text { है, जहाँ } \mathrm{P}=-1 \text { और } \mathrm{Q}=\cos x $$

इसलिए I.F. $=e^{\int-1 d x}=e^{-x}$

समीकरण के दोनों पक्षों को I.F. से गुणा करने पर हम प्राप्त करते हैं:

$$ \begin{aligned} & e^{-x} \frac{d y}{d x}-e^{-x} y=e^{-x} \cos x \\ & \frac{d}{d x}\left(y e^{-x}\right)=e^{-x} \cos x \end{aligned} $$

दोनों पक्षों का $x$ के सापेक्ष समाकलन करने पर हम प्राप्त करते हैं:

$$ \begin{equation*} y e^{-x}=\int e^{-x} \cos x d x+\mathrm{C} \tag{1} \end{equation*} $$

$$ \begin{aligned} \text{ मान लीजिए कि }\qquad & =-\cos x e^{-x}-\int \sin x e^{-x} d x \\ & =-\cos x e^{-x}-[\sin x(-e^{-x})-\int \cos x(-e^{-x}) d x] \\ & =-\cos x e^{-x}+\sin x e^{-x}-\int \cos x e^{-x} d x \\ I & =-e^{-x} \cos x+\sin x e^{-x}-I \\ 2 I & =(\sin x-\cos x) e^{-x} \\ I & =\frac{(\sin x-\cos x) e^{-x}}{2} \end{aligned} $$

समीकरण (1) में I का मान प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$ \begin{aligned} \text{ अथवा }\qquad y e^{-x} & =(\frac{\sin x-\cos x}{2}) e^{-x}+C \\ y & =(\frac{\sin x-\cos x}{2})+C e^{x} \end{aligned} $

यह दिए हुए अवकल समीकरण का व्यापक हल है।

उदाहरण 15 अवकल समीकरण $x \frac{d y}{d x}+2 y=x^{2}(x \neq 0)$ का व्यापक हल ज्ञात कीजिए।

हल दिया हुआ अवकल समीकरण है:

$ x \frac{d y}{d x}+2 y=x^{2} $

समीकरण (1) के दोनों पक्षों को $x$ से भाग देने पर हम प्राप्त करते हैं:

$$ \frac{d y}{d x}+\frac{2}{x} y=x $$

यह, $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, के रूप का रैखिक अवकल समीकरण है। यहाँ $\mathrm{P}=\frac{2}{x}$ एवं $\mathrm{Q}=x$ है।

$\text{ इसलिए } \text{ I.F }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}[\text{ जैसा कि } e^{\log f(x)}=f(x)] $

इसलिए दिए हुए समीकरण का हल है:

$ \begin{matrix} y \cdot x^{2}=\int(x)(x^{2}) d x+C=\int x^{3} d x+C \\ y=\frac{x^{2}}{4}+C x^{-2} \end{matrix} $

यह दिए हुए अवकल समीकरण का व्यापक हल है।

उदाहरण 16 अवकल समीकरण $y d x-\left(x+2 y^{2}\right) d y=0$ का व्यापक हल ज्ञात कीजिए।

हल दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है:

$$ \frac{d x}{d y}-\frac{x}{y}=2 y $$

यह, $\frac{d x}{d y}+\mathrm{P} _{1} x=\mathrm{Q} _{1}$, के रूप वाला रैखिक अवकल समीकरण है। यहाँ $\mathrm{P} _{1}=-\frac{1}{y}$ एवं $\mathrm{Q} _{1}=2 y$ है। इसलिए I.F. $=e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log (y)^{-1}}=\frac{1}{y}$

अतः दिए हुए अवकल समीकरण का हल है:

$ \begin{aligned} \text{ अथवा }\qquad x \frac{1}{y} & =\int(2 y)(\frac{1}{y}) d y+C \\ \frac{x}{y} & =\int(2 d y)+C \\ \frac{x}{y} & =2 y+C \\ x & =2 y^{2}+C y \end{aligned} $

यह दिए हुए अवकल समीकरण का व्यापक हल है। का विशिष्ट हल ज्ञात कीजिए, दिया हुआ है कि $y=0$ यदि $x=\frac{\pi}{2}$

उदाहरण 17 अवकल समीकरण

$$ \frac{d y}{d x}+y \cot x=2 x+x^{2} \cot x(x \neq 0) $$

हल दिया हुआ अवकल समीकरण $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, के रूप का रैखिक अवकल समीकरण है। यहाँ $\mathrm{P}=\cot x$ और $\mathrm{Q}=2 x+x^{2} \cot x$ है। इसलिए

$$ \text { I.F. }=e^{\int \cot x d x}=e^{\log \sin x}=\sin x $$

अत: अवकल समीकरण का हल है:

$$ y \cdot \sin x=\int\left(2 x+x^{2} \cot x\right) \sin x d x+\mathrm{C} $$

$ \begin{aligned} & \text{ अथवा } \quad y \sin x=\int 2 x \sin x d x+\int x^{2} \cos x d x+C \\ & \text{ अथवा } \quad y \sin x=\sin x(\frac{2 x^{2}}{2})-\int \cos x(\frac{2 x^{2}}{2}) d x+\int x^{2} \cos x d x+C \\ & \text{ अथवा } \quad y \sin x=x^{2} \sin x-\int x^{2} \cos x d x+\int x^{2} \cos x d x+C \\ & \text{ अथवा } \quad y \sin x=x^{2} \sin x+C \end{aligned} $

समीकरण (1) में $y=0$ एवं $x=\frac{\pi}{2}$ प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$$ 0=\left(\frac{\pi}{2}\right)^{2} \sin \left(\frac{\pi}{2}\right)+C $$

$$\text{ अथवा }\qquad \mathrm{C}=\frac{-\pi^{2}}{4} $$

$$ \text{ अथवा }\qquad y \sin x=x^{2} \sin x-\frac{\pi^{2}}{4} $$ $$ y=x^{2}-\frac{\pi^{2}}{4 \sin x}(\sin x \neq 0) $$

यह दिए हुए अवकल समीकरण का विशिष्ट हल है।

उदाहरण 18 बिन्दु $(0,1)$ से गुजरने वाले एक वक्र का समीकरण ज्ञात कीजिए, यदि इस वक्र के किसी बिंदु $(x, y)$ पर स्पर्श रेखा की प्रवणता, उस बिंदु के $x$ निर्देशांक ( भुज) तथा $x$ निर्देशांक और $y$ निर्देशांक (कोटि) के गुणनफल के योग के बराबर है।

हल हम जानते हैं कि वक्र की स्पर्श रेखा की प्रवणता $\frac{d y}{d x}$ के बराबर होती है। इसलिए

$$ \begin{align*} \text{ अथवा }\qquad & \frac{d y}{d x}=x+x y \\ & \frac{d y}{d x}-x y=x \tag{1} \end{align*} $$

समीकरण (1), $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ के रूप का रैखिक अवकल समीकरण है। यहाँ $\mathrm{P}=-x$ एवं $\mathrm{Q}=x$ है।

$ \text{ इसलिए }\qquad \text { I.F. }=e^{\int-x d x}=e^{\frac{-x^{2}}{2}} $

अतः दिए हुए समीकरण का हल है:

$$ \begin{equation*} \text{ मान लीजिए }\qquad y \cdot e^{\frac{-x^{2}}{2}}=\int(x)\left(e^{\frac{-x^{2}}{2}}\right) d x+\mathrm{C} \tag{2} \end{equation*} $$

$$ \mathrm{I}=\int(x) e^{\frac{-x^{2}}{2}} d x $$

मान लीजिए $ \frac{-x^{2}}{2}=t, \text { तब }-x d x=d t \text { या } x d x=-d t $

इसलिए $ \mathrm{I}=-\int e^{t} d t=-e^{t}=-e^{\frac{-x^{2}}{2}} $

समीकरण (2) में I का मान प्रतिस्थापित करने पर, हम पाते हैं:

$$ \begin{align*} \text{ अथवा } & y e^{\frac{-x^{2}}{2}}=-e^{\frac{-x^{2}}{2}}+\mathrm{C} \\ & y=-1+\mathrm{C} e^{\frac{x^{2}}{2}} \tag{3} \end{align*} $$

समीकरण (3) वक्रों के कुल का समीकरण है परंतु हम इस कुल के ऐसे सदस्य का समीकरण ज्ञात करना चाहते हैं जो बिंदु $(0,1)$ से गुजरता हो। समीकरण (3) में $x=0$ एवं $y=1$ प्रतिस्थापित करने पर हम पाते हैं:

$$ 1=-1+\mathrm{C} \cdot e^{\mathrm{o}} \text { अथवा } \mathrm{C}=2 $$

समीकरण (3) में $\mathrm{C}$ का मान प्रतिस्थापित करने पर हम प्राप्त करते हैं:

$$ y=-1+2 e^{\frac{x^{2}}{2}} $$

यह वक्र का अभीष्ट समीकरण है।

प्रश्नावली 9.5

1 से 12 तक के प्रश्नों में, प्रत्येक अवकल समीकरण का व्यापक हल ज्ञात कीजिए:

1. $\frac{d y}{d x}+2 y=\sin x$

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2. $\frac{d y}{d x}+3 y=e^{-2 x} $

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3. $\frac{d y}{d x}+\frac{y}{x}=x^{2}$

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4. $\frac{d y}{d x}+(\sec x) y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

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5. $\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

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6. $x \frac{d y}{d x}+2 y=x^{2} \log x \quad$

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7. $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

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8. $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)$

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9. $x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0) \quad$

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10. $(x+y) \frac{d y}{d x}=1$

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11. $y d x+\left(x-y^{2}\right) d y=0$

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12. $\left(x+3 y^{2}\right) \frac{d y}{d x}=y(y>0)$.

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13 से 15 तक के प्रश्नों में प्रत्येक अवकल समीकरण के लिए दिए हुए प्रतिबंध को संतुष्ट करने वाला विशिष्ट हल ज्ञात कीजिए:

13. $\frac{d y}{d x}+2 y \tan x=\sin x ; y=0$ यदि $x=\frac{\pi}{3}$

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14. $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0$ यदि $x=1$

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15. $\frac{d y}{d x}-3 y \cot x=\sin 2 x ; y=2$ यदि $x=\frac{\pi}{2}$

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16. मूल बिंदु से गुज़रने वाले एक वक्र का समीकरण ज्ञात कीजिए यदि इस वक्र के किसी बिंदु $(x, y)$ पर स्पर्श रेखा की प्रवणता उस बिंदु के निर्देशांकों के योग के बराबर है।

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17. बिंदु $(0,2)$ से गुजरने वाले वक्र का समीकरण ज्ञात कीजिए यदि इस वक्र के किसी बिंदु के निर्देशांकों का योग उस बिंदु पर खींची गई स्पर्श रेखा की प्रवणता के परिमाण से 5 अधिक है।

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18. अवकल समीकरण $x \frac{d y}{d x}-y=2 x^{2}$ का समाकलन गुणक है:

(A) $e^{-x}$

(B) $e^{-y}$

(C) $\frac{1}{x}$

(D) $x$

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19. अवकल समीकरण $\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y(-1<y<1)$ का समाकलन गुणक है:

(A) $\frac{1}{y^{2}-1}$

(B) $\frac{1}{\sqrt{y^{2}-1}}$

(C) $\frac{1}{1-y^{2}}$

(D) $\frac{1}{\sqrt{1-y^{2}}}$

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विविध उदाहरण

उदाहरण 19 सत्यापित कीजिए कि फलन $y=c _{1} e^{a x} \cos b x+c _{2} e^{a x} \sin b x$, जहाँ $c _{1}, c _{2}$ स्वेच्छ अचर है, अवकल समीकरण

$$ \frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0 \text { का हल है। } $$

हल दिया हुआ फलन है:

$$ \begin{equation*} y=e^{a x}\left[c _{1} \cos b x+c _{2} \sin b x\right] \tag{1} \end{equation*} $$

समीकरण (1) के दोनों पक्षों का $x$ के सापेक्ष अवकलन करने पर हम पाते हैं कि

$$ \begin{align*} & \frac{d y}{d x}=e^{a x}\left[-b c _{1} \sin b x+b c _{2} \cos b x\right]+\left[c _{1} \cos b x+c _{2} \sin b x\right] e^{a x} \cdot a \\ & \frac{d y}{d x}=e^{a x}\left[\left(b c _{2}+a c _{1}\right) \cos b x+\left(a c _{2}-b c _{1}\right) \sin b x\right] \tag{2} \end{align*} $$

समीकरण (2) के दोनों पक्षों का $x$, के सापेक्ष अवकलन करने पर हम पाते हैं कि

$$ \begin{aligned} \frac{d^{2} y}{d x^{2}}= & e^{a x}\left[\left(b c _{2}+a c _{1}\right)(-\sin b x \cdot b)+\left(a c _{2}-b c _{1}\right)(\cos b x \cdot b)\right] \\ & +\left[\left(b c _{2}+a c _{1}\right) \cos b x+\left(a c _{2}-b c _{1}\right) \sin b x\right] e^{a x} \cdot a \\ = & e^{a x}\left[\left(a^{2} c _{2}-2 a b c _{1}-b^{2} c _{2}\right) \sin b x+\left(a^{2} c _{1}+2 a b c _{2}-b^{2} c _{1}\right) \cos b x\right] \end{aligned} $$

दिए गए अवकल समीकरण में $\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ एवं $y$ का मान प्रतिस्थापित करने पर हम पाते हैं:

बायाँ पक्ष $$ \begin{aligned} = & \left.e^{a x}\left[a^{2} c _{2}-2 a b c _{1}-b^{2} c _{2}\right) \sin b x+\left(a^{2} c _{1}+2 a b c _{2}-b^{2} c _{1}\right) \cos b x\right] \\ & -2 a e^{a x}\left[\left(b c _{2}+a c _{1}\right) \cos b x+\left(a c _{2}-b c _{1}\right) \sin b x\right] \\ & +\left(a^{2}+b^{2}\right) e^{a x}\left[c _{1} \cos b x+c _{2} \sin b x\right] \\ = & e^{a x}\left[\begin{array}{l} \left(a^{2} c _{2}-2 a b c _{1}-b^{2} c _{2}-2 a^{2} c _{2}+2 a b c _{1}+a^{2} c _{2}+b^{2} c _{2}\right) \sin b x \\ +\left(a^{2} c _{1}+2 a b c _{2}-b^{2} c _{1}-2 a b c _{2}-2 a^{2} c _{1}+a^{2} c _{1}+b^{2} c _{1}\right) \cos b x \end{array}\right] \\ = & e^{a x}[0 \times \sin b x+0 \cos b x]=e^{a x} \times 0=0=\text { दायाँ पक्ष } \end{aligned} $$

इसलिए दिया हुआ फलन दिए हुए अवकल समीकरण का हल है।

उदाहरण 20 अवकल समीकरण $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ का विशिष्ट हल ज्ञात कीजिए। दिया हुआ है कि $y=0$ यदि $x=0$

हल दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है:

$ \frac{d y}{d x}=e^{(3 x+4 y)} $

$ \frac{d y}{d x}=e^{3 x} \cdot e^{4 y} $

चरों को पृथक् करने पर हम पाते हैं,

$$ \frac{d y}{e^{4 y}}=e^{3 x} d x $$

$$\text{ इसलिए } \int e^{-4 y} d y=\int e^{3 x} d x $$

$$\text{ अथवा } \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+\mathrm{C} $$

$$\text{ अथवा } 4 e^{3 x}+3 e^{-4 y}+12 \mathrm{C}=0 \tag{2} $$

समीकरण (2) में $x=0$ एवं $y=0$ प्रतिस्थापित करने पर हम पाते हैं:

$$ 4+3+12 \mathrm{C}=0 \text { अथवा } \mathrm{C}=\frac{-7}{12} $$

समीकरण (2) में $\mathrm{C}$ का मान प्रतिस्थापित करने पर हम,

$$ 4 e^{3 x}+3 e^{-4 y}-7=0 \text {, प्राप्त करते हैं } $$

यह दिए हुए अवकल समीकरण का एक विशिष्ट हल है।

उदाहरण 21 अवकल समीकरण

$$ (x d y-y d x) y \sin \left(\frac{y}{x}\right)=(y d x+x d y) x \cos \left(\frac{y}{x}\right) \text { को हल कीजिए। } $$

हल दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है।

$$ \left[x y \sin \left(\frac{y}{x}\right)-x^{2} \cos \left(\frac{y}{x}\right)\right] d y=\left[x y \cos \left(\frac{y}{x}\right)+y^{2} \sin \left(\frac{y}{x}\right)\right] d x $$

$$\text{ अथवा } \frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^{2} \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^{2} \cos \left(\frac{y}{x}\right)} $$

दायें पक्ष पर अंश एवं हर दोनों को $x^{2}$ से भाग देने पर हम पाते हैं:

$$ \begin{equation*} \frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y^{2}}{x^{2}}\right) \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)} \tag{1} \end{equation*} $$

स्पष्टतः समीकरण (1), $\frac{d y}{d x}=g\left(\frac{y}{x}\right)$ के रूप का समघातीय अवकल समीकरण है, इसलिए इस समीकरण को हल करने के लिए हम

प्रतिस्थापित करते हैं। अथवा इसलिए

$$ \begin{aligned} & y=v x \\ & \text{ अथवा } \\ & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & v+x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v} \\ & x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\ & (\frac{v \sin v-\cos v}{v \cos v}) d v=\frac{2 d x}{x} \\ & \text{ Therefअथवाe } \quad \int(\frac{v \sin v-\cos v}{v \cos v}) d v=2 \int \frac{1}{x} d x \\ & \text{ अथवा } \\ & \int \tan v d v-\int \frac{1}{v} d v=2 \int \frac{1}{x} d x \\ & \text{ अथवा } \\ & \text{ } \\ & \log |\sec v|-\log |v|=2 \log |x|+\log |C_1| \\ & \log |\frac{\sec v}{v x^{2}}|=\log |C_1| \\ & \text{अथवा } \\ & \frac{\sec v}{v x^{2}}= \pm C_1 \end{aligned} $$

समीकरण (3) में $v$ को $\frac{y}{x}$ से प्रतिस्थापित करने पर हम पाते हैं कि

$ \begin{aligned} & \frac{\sec (\frac{y}{x})}{(\frac{y}{x})(x^{2})}=C \text{ जहाँ, } C= \pm C_1 \\ & \sec (\frac{y}{x})=C x y \end{aligned} $

यह दिए हुए अवकल समीकरण का व्यापक हल है।

उदाहरण 22 अवकल समीकरण

$$ \left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x \text { का हल ज्ञात कीजिए। } $$

हल दिया हुआ अवकल समीकरण निम्नलिखित रूप में लिखा जा सकता है:

$ \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}} $

समीकरण (1), $\frac{d x}{d y}+\mathrm{P} _{1} x=\mathrm{Q} _{1}$, के रूप का रैखिक अवकल समीकरण है। यहाँ

$$ \begin{aligned} & \mathrm{P} _{1}=\frac{1}{1+y^{2}} \text { एवं } \mathrm{Q} _{1}=\frac{\tan ^{-1} y}{1+y^{2}} \text { है। इसलिए } \\ & \text { I.F. }=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y} \end{aligned} $$

इसलिए दिए हुए अवकल समीकरण का हल है:

$$ \begin{equation*} x e^{\tan ^{-1} y}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} d y+\mathrm{C} \tag{2} \end{equation*} $$

मान लीजिए $\quad \mathrm{I}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} d y$

$\tan ^{-1} y=t$ प्रतिस्थापित करने पर हम पाते हैं कि $\left(\frac{1}{1+y^{2}}\right) d y=d t$

$$ \mathrm{I}=\int t e^{t} d t, \mathrm{I}=t e^{t}-\int 1 . e^{t} e t, \mathrm{I}=t e^{t}-e^{t}=e^{t}(t-1) $$

$$\text{ अथवा }\qquad \mathrm{I}=e^{\tan ^{-1} y\left(\tan ^{-1} y-1\right)} $$

समीकरण (2) में I का मान प्रतिस्थापित करने पर हम

$$ \begin{aligned} \text{ अथवा }\qquad & x \cdot e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+\mathrm{C} \text { पाते हैं } \\ & x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{-\tan ^{-1} y} \end{aligned} $$

यह दिए हुए अवकल समीकरण का व्यापक हल है।

अध्याय 9 पर विविध प्रश्नावली

1. निम्नलिखित अवकल समीकरणों में से प्रत्येक की कोटि एवं घात (यदि परिभाषित हो) ज्ञात कीजिए।

(i) $\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$

(ii) $\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x$

(iii) $\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$

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2. निम्नलिखित प्रश्नों में प्रत्येक के लिए सत्यापित कीजिए कि दिया हुआ फलन (अस्पष्ट अथवा स्पष्ट) संगत अवकल समीकरण का हल है।

$\quad\quad$(i) $x y=a e^{x}+b e^{-x}+x^{2}$

$\quad\quad: x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0$

$\quad\quad$(ii) $y=e^{x}(a \cos x+b \sin x) \quad: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

$\quad\quad$(iii) $y=x \sin 3 x$

$\quad\quad: \frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x=0$

$\quad\quad$(iv) $x^{2}=2 y^{2} \log y$

$ \quad\quad:(x^{2}+y^{2}) \frac{d y}{d x}-x y=0 $

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3. सिद्ध कीजिए कि $x^{2}-y^{2}=c\left(x^{2}+y^{2}\right)^{2}$ जहाँ $c$ एक प्राचल है, अवकल समीकरण $\left(x^{3}-3 x y^{2}\right) d x=\left(y^{3}-3 x^{2} y\right) d y$ का व्यापक हल है।

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4. अवकल समीकरण $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$, जबकि $\mathrm{x} \neq 1$ का व्यापक हल ज्ञात कीजिए।

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5. दर्शाइए कि अवकल समीकरण $\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$ का व्यापक हल $(x+y+1)=\mathrm{A}(1-x-y-2 x y)$ है, जिसमें $\mathrm{A}$ एक प्राचल है।

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6. बिंदु $\left(0, \frac{\pi}{4}\right)$ से गुजरने वाले एक ऐसे वक्र का समीकरण ज्ञात कीजिए जिसका अवकल समीकरण $\sin x \cos y d x+\cos x \sin y d y=0$ है।

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7. अवकल समीकरण $\left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0$ का एक विशिष्ट हल ज्ञात कीजिए, दिया हुआ है कि $y=1$ यदि $x=0$.

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8. अवकल समीकरण $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y(y \neq 0)$ का हल ज्ञात कीजिए।

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9. अवकल समीकरण $(x-y)(d x+d y)=d x-d y$ का एक विशिष्ट हल ज्ञात कीजिए, दिया हुआ है कि $y=-1$, यदि $x=0$ (संकेत: $x-y=t$ रखें)।

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10. अवकल समीकरण $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0)$ का हल ज्ञात कीजिए।

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11. अवकल समीकरण $\frac{d y}{d x}+y \cot x=4 x अथवा {cosec} x(x \neq 0)$ का एक विशिष्ट हल ज्ञात कीजिए, दिया हुआ है कि $y=0$ यदि $x=\frac{\pi}{2}$.

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12. अवकल समीकरण $(x+1) \frac{d y}{d x}=2 e^{-y}-1$ का एक विशिष्ट हल ज्ञात कीजिए, दिया हुआ है कि $y=0$ यदि $x=0$.

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13. अवकल समीकरण $\frac{y d x-x d y}{y}=0$ का व्यापक हल है:

(A) $x y=\mathrm{C}$

(B) $x=\mathrm{C} y^{2}$

(C) $y=\mathrm{C} x$

(D) $y=\mathrm{C} x^{2}$

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14. $\frac{d x}{d y}+\mathrm{P} _{1} x=\mathrm{Q} _{1}$ के रूप वाले अवकल समीकरण का व्यापक हल है:

(A) $y e^{\int \mathrm{P} _{1} d y}=\int\left(\mathrm{Q} _{1} e^{\int \mathrm{P} _{1} d y}\right) d y+\mathrm{C}$

(B) $y \cdot e^{\int \mathrm{P} _{1} d x}=\int\left(\mathrm{Q} _{1} e^{\int \mathrm{P} _{1} d x}\right) d x+\mathrm{C}$

(C) $x e^{\int \mathrm{P} _{1} d y}=\int\left(\mathrm{Q} _{1} e^{\int \mathrm{P} _{1} d y}\right) d y+\mathrm{C}$

(D) $x e^{\int \mathrm{P} _{1} d x}=\int\left(\mathrm{Q} _{1} e^{\int \mathrm{P} _{1} d x}\right) d x+\mathrm{C}$

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15. अवकल समीकरण $e^{x} d y+\left(y e^{x}+2 x\right) d x=0$ का व्यापक हल है:

(A) $x e^{y}+x^{2}=\mathrm{C}$

(B) $x e^{y}+y^{2}=\mathrm{C}$

(C) $y e^{x}+x^{2}=\mathrm{C}$

(D) $y e^{y}+x^{2}=\mathrm{C}$

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सारांश

  • एक ऐसा समीकरण जिसमें स्वतंत्र चर (चरों) के सापेक्ष आश्रित चर के अवकलज (अवकलजों) सम्मिलित हों, अवकल समीकरण कहलाता है।

  • किसी अवकल समीकरण में सम्मिलित उच्चतम अवकलज की कोटि, उस अवकल समीकरण की कोटि कहलाती है।

  • यदि कोई अवकल समीकरण अवकलजों में बहुपद समीकरण हैं तो उस अवकल समीकरण की घात परिभाषित होती है।

  • किसी अवकल समीकरण की घात (यदि परिभाषित हो) उस अवकल समीकरण में सम्मिलित उच्चतम कोटि अवकलज की उच्चतम घात (केवल धनात्मक पूर्णांक) होती है।

  • एक दिए हुए अवकल समीकरण को संतुष्ट करने वाला फलन उस अवकल समीकरण का हल कहलाता है। एक ऐसा हल जिसमें उतने ही स्वेच्छ अचर हों, जितनी उस अवकल समीकरण की कोटि है, व्यापक हल कहलाता है और स्वेच्छ अचरों से मुक्त हल विशिष्ट हल कहलाता है।

  • एक ऐसा अवकल समीकरण, जिसको $\frac{d y}{d x}=f(x, y)$ अथवा $\frac{d x}{d y}=g(x, y)$ के रूप में अभिव्यक्त किया जा सकता है, जहाँ $f(x, y)$ एवं $g(x, y)$ शून्य घात वाले समघातीय फलन हैं, समघातीय अवकल समीकरण कहलाता है।

$\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, के रूप वाला अवकल समीकरण, जिसमें $\mathrm{P}$ तथा $\mathrm{Q}$ अचर अथवा केवल $x$ के फलन हैं, प्रथम कोटि रैखिक अवकल समीकरण कहलाता है।

ऐतिहासिक पृष्ठभूमि

अवकल समीकरण विज्ञान की प्रमुख भाषाओं में से एक है। रोचक तथ्य यह है कि अवकल समीकरणों का अस्तित्व नवंबर 11, 1675 Gottfried Wilthelm Freiherr Leibnitz (1646-1716) ने सर्वप्रथम सर्वसमिका, $\int y d y=\frac{1}{2} y^{2}$, को लिखित रूप में प्रस्तुत किया तथा उनसे दोनों प्रतीकों $\int$ और $d y$ से परिचित कराया। वस्तुतः Leibnitz ऐसी वक्र को ज्ञात करने की समस्या में मग्न थे जिसकी स्पर्श रेखा निर्दिष्ट हों, इस समस्या ने सन् 1691 में उन्हें ‘चरों के पृथक्करणीय विधि’ के अन्वेषण का मार्गदर्शन कराया। एक वर्ष पश्चात् उन्होंने ‘प्रथम कोटि के समघातीय समीकरणों के हल करने की विधि’ का सूत्रीकरण किया। वे आगे बढ़े और अल्प समय में उन्होंने ‘प्रथम कोटि के रैखिक अवकल समीकरणों को हल करने की विधि’ का अन्वेषण किया। कितना आश्चर्यजनक है कि उपर्युक्त सभी विधियों की खोज अकेले एक व्यक्ति द्वारा अवकल समीकरणों के जन्म के पच्चीस वर्षों के अल्पावधि के अंतर्गत संपन्न हुई।

प्रारंभ में केवल समीकरणों के ‘हल’ करने की प्रविधि को अवकल समीकरणों के ‘समाकलन’ के रूप में निर्दिश्ट किया गया था। यह शब्द सन् 1690 में प्रथमतः James Bernoulli, (1654-1705) द्वारा प्रचलन में लाया गया। शब्द ‘हल’ का सर्वप्रथम प्रयोग Joseph Louis Lagrange (1736-1813), द्वारा सन् 1774 में किया गया। यह घटना अवकल समीकरणों के जन्म से लगभग 100 वर्षों बाद घटित हुई। ये Jules Henri Poincare (1854-1912), थे, जिन्होंने शब्द ‘हल’ के प्रयोग के लिए अकाट्य तर्क प्रस्तुत किया, फलतः आधुनिक शब्दावली में शब्द हल को अपना उचित स्थान प्राप्त हुआ। ‘चरों के पृथ्क्करणीय विधि का नामकरण John Bernoulli (1667-1748), James Bernoulli के अनुज द्वारा किया गया। मई 20,1715 को Leibnitz को लिखे अपने पत्र में, उन्होने निम्नलिखित अवकल समीकरण के हल की खोज किए

$$ x^{2} y^{\prime \prime}=2 y $$

के हल तीन प्रकार की वक्रों नामतः परवलय, अतिपरवलय और घनीय वक्रों के एक समूह का मार्गदर्शन कराते हैं। यह दर्शाता है कि ऐसे सरल दिखाई पड़ने वाले अवकल समीकरणों के हल कैसे नाना रूप धारण करते हैं। 20 वीं शताब्दी के उतरार्ध में ‘अवकल समीकरणों के गुणात्मक विश्लेषण’ शीर्षक के अंतर्गत अवकल समीकरणों के हलों की जटिल प्रकृति के आविष्कार हेतु ध्यान आकर्षित किया गया। आजकल इसने लगभग सभी अविष्कारों हेतु अत्यंत प्रविधि के रूप में प्रमुख स्थान प्राप्त कर लिया है।



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