This suggests that we multiply equation (3.1) by

$\exp (\phi(x))$

and equation (3.1) goes over to

$\text { a } \frac{d}{d x}(y(x) \exp (\phi(x))-Q(x) \exp (\phi(x))$

If we integrate both sides of (3.5) we get

$y(x) \exp (\phi(x))-\int Q(x) \exp (\phi(x)) d x$

which gives us the solution of the linear differential equation.

Comments on equation (3.6)

Since the exponential function does not vanish, equations (3.1) and (3.5) are completely equivalent.

How are we going to find a $\phi(x)$ such that $\phi^{\prime}(x)=P(x)$ ? Well, if wc cannot, we are simply out of luck! All we can say is take

$\phi(x)=\int_a^x P(t) d t .$

and the final solution would have a VERY UGLY APPEARENCE WITH MANY INTEGRATION SIGNS! There is nothing one can do except to deal with it!

Comments contd

We shall assume that we can find such a $\phi(x)$ namely $\int P(x) d x$ and and rewrite the solution obtained as

$\left.y(x) \exp \left(\int P(x) d x\right)\right)-\int Q(x) \exp \left(\int P(x) d x\right) d x$

It is understood that in both occurrences of $\int P(x) d x$ are identical and so the SAME constant of integration should be assigned for both

Comments contd:

Here we strongly advise the student not to memorize the final formula but rather, follow the steps:

First compute $f P(x) d x$ ignoring the constant of integration

Multiply the differential equation by $\exp \left(\int P(x) d x\right)$.

Perform the final integration - and in case there are initial conditions given, employ definite intgrals.

Example 1: Solve the equation on the interval $(-\pi / 2, \pi / 2)$ :

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

Solution: Here $P(x)-\tan x$ and so

$\exp \int P(x) d x=\exp (\log (\sec x))=\sec x \text {. }$

Ignored the const. of integrationl Multiply the diff. eqn. by sec $x$ to get

$\frac{d}{d x}(y \sec x)=\sin x \sec x=\tan x$

Integratang (constant of integration is needed nowl) and smplifying we get

$y(x)-\cos x(\log (\sec x)+c)$

Example 2 contd

$y^{\prime}(x)-\frac{y}{x}-x$

Here $P(x)-1 / x$ and $\exp \left(\int P(x) d x\right)-1 / x$. Multiplying by $1 / x$ we get

$\frac{d}{d x}(y / x)=1$

Setting $x=1$ in (3.10) we get the initial condition $y(1)=1 / 3$. Integrating (3.11) from 1 to 2 we get

$\frac{y(2)}{2}-\frac{y(1)}{1}=1$

which gives the value $y(2)=8 / 3$.

Example from JEE 2014 (paper - 2)

Given that $f(x)$ is the solution of the differential equation and the given initial condition on $(-1,1)$.

$\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}, y(0)-0 .$

Find the value of the integral

$\int_p^y f(x) d x$

The given equation is a linear differential equation which is defined for $1<x<1$. Here wo compute

$\int P(x) d x-\frac{1}{2} \int_1^{-2 } x^2-\frac{1}{2} \log \left(1 \quad x^2\right) x d x$

Example from JEE 2014 (paper - 2)

Given that $f(x)$ is the solution of the differential equation and the given initial condition on $(-1,1)$.

$\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}, y(0)-0 .$

Find the value of the integral

$\int_p^y f(x) d x$

The given equation is a linear differential equation which is defined for $1<x<1$. Here wo compute

$\int P(x) d x-\frac{1}{2} \iint_1^{-2 x d x} x^2-\frac{1}{2} \log \left(1 \quad x^2\right) .$

Example from JEE 2014 (paper - 2) contd

Hence we should multiply the differential equation by the factor

$\exp \int P(x) d x=\sqrt{1-x^2}$

After multiplying we see that

$\frac{d}{d x}\left(y \sqrt{1-x^2}\right)-x^4+2 x$

We now integrate this from 0 to $x$ and we get

$- f(x) \sqrt{1-x^2}=f(0)+\int_0^x\left(t^4+2 t\right) d t=\frac{x^5}{5}+x^2 .$

Now what is asked is the integral of the function $f(x)$ over an interval symmetric about the origin and so the odd part of $f(x)$ namely the term coming from $x^5 / 5$ would be zero.

Example from JEE 2014 (paper - 2) contd...

Thus the desired integral is

$2 \int_0^{\sqrt{3 / 2}} \frac{x^2 d x}{\sqrt{1-x^2}}=\int_0^{\pi / 3} 2 \sin ^2 \theta d \theta$

which the student can easily compute

Exercise:

(1) Compute the value of this integral.

Example JFE 2016, paper 1

With some notational changes, the linear differential equation on $(0, \infty)$ is

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

(i) Calculate the limits as $x \rightarrow 0+$ of $y^{\prime}(1 / x), x y(1 / x)$ and $x^2 y^{\prime}(x)$

(ii) Is the function $|y(x)|$ bounded on $(0.2)$ ?

Solution: Rewriting the equation as

$y^{\prime}(x)+\frac{y(x)}{x}-2$

We see that $\exp \left(\int P(x) d x\right)=x$ Multiplying the equation by $x$ we get

$d x(x y(x))=2 x$

As such, $y(1)$ can be assigned any real value. Let us assume that $y(1)=a$

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

Integrate the last equation over $[1, x]$ :

$x y(x)-y(1)=\left(x^2-1\right)$

But $y(1)-$ a so we get

$x y(x)=a+\left(x^2-1\right) .$

It is now an easy matter for you to calculate the limits asked in the question.

This is the equation

$\frac{d y}{d x}+P(x) y=Q(x) y^n$

where $P(x)$ and $Q(x)$ are continuous on an interval $I$.

If $n=0$ or $n=1$ then the equation is linear and so we shall assume $n \neq 0,1$. Dividing by $y^n$ we get

$\frac{1}{y^n} \frac{d y}{d x}+P(x) y^{1-n}=Q(x)$

The substitution $u=y^{1-n}$ converts equation (3.15) into a linear differential equation.

Warning: If $n>0$ we are tacitly assuming that $y(x) \neq 0$ Suppose we are given initial conditions such as $y\left(x_0\right)=0$ then the method cannot be employed !

$\frac{1 d y}{y^n}+P(x) y^{1-n}=Q(x)$

We put $u-y^{1-n}$ so that

$\frac{d u}{d x}=(1-n) y^{-n} \frac{d y}{d x}$

and hence the Diff. Eqn transforms into

$\left.\frac{1 d u}{1-n d x} \right\rvert, P(x) u-Q(x)$

Multiplying through by $1-n$ gives us the desired Linear Diff. Eqn

(4) Solve the equation

$\frac{d y}{d t}=y(1-y), \quad y(0) \neq 0$

(i) By the method of separation of variables. Left for the student.

(ii) As a Bernoulli equation.

Solution (ii): The condition $y(0) \neq 0$ allows us to divide by $y^2$. The equation can be written as

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

Put $1 / y=u$ and we get the equation $u^{\prime}+u=1$ whose solution is $u=1+c \exp (-x)$ and so we get

$y(x)=\frac{1}{11 \operatorname{cexp}(x)}$

Two more exercises

(5) Solve the following homogeneous equation

$2 x y d x+\left(x^2-y^2\right) d y=0$

as a Bernoulli equation in $x$. That is write it in terms of $\frac{d x}{d y}$ and regard $x$ as a function of $y$.

Do you think solving it as a Bernoulli equation is easier?

(6) Solve the differential equation

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

Hint: Multiply by $\cos y$ and see what happens.

Discussion of Exercise (6)

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

Multiplying the Diff. Eqn. by $\cos y$ gives

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

Let us now put $\sin y=u$. Then

$\cos y \frac{d y}{d x}-d x$

The diff eqn transforms into the linear equation

$\frac{d u}{d x}+2 x u-e^{-x^2}$