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.

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.

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.

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.

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.

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.

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.

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.

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.

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:

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:

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

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:

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

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

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:

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

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:

$y=\sec x$

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

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.

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.

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.

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 $(63 t+27)^{\frac{1}{3}}$.

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 $6.93 %$.

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.

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.

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.