NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3

Exercise 9.3

1. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$.

Use trigonometric half–angle identities to simplify:

$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}$ 

$\Rightarrow \frac{dy}{dx}={\tan }^2\frac{x}{2}$ 

$\Rightarrow \frac{dy}{dx}={\sec }^2\frac{x}{2}-1$ 

Separate the differentials and integrate:

$\int dy=\int ({\sec }^2\frac{x}{2}-1)dx$ 

$\Rightarrow y=\int {\sec }^2\frac{x}{2}dx-\int dx$ 

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

Thus the general solution of given differential equation is $\text{y = 2tan}\frac{\text{x}}{\text{2}}\text{ - x + C}$.

2. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{\text{y}}^{\text{2}}}\left(\text{-2 y 2}\right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{\text{y}}^{\text{2}}}\left(\text{-2 y 2}\right)$.

Simplify the expression:

$\frac{dy}{dx}=\sqrt{4-y^2}$ 

$\Rightarrow \frac{dy}{\sqrt{4-y^2}}=dx$ 

Use standard integration:

$\int \frac{dy}{\sqrt{4-y^2}}=\int dx$ 

$\Rightarrow {\sin }^{-1}\frac{y}{2}=x+C$ 

$\Rightarrow \frac{y}{2}=\sin (x+C)$ 

$\Rightarrow y=2\sin (x+C)$ 

Thus the general solution of given differential equation is $\text{y = 2 sin}\left(\text{x + C}\right)$.

3. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{+ y =1}(\text{y}\ne \text{1})$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{+ y =1}(\text{y}\ne \text{1})$.

Simplify the expression:

$\frac{dy}{dx}+y=1$ 

$\Rightarrow \frac{dy}{dx}=1-y$ 

$\Rightarrow \frac{dy}{1-y}=dx$ 

Use standard integration:

$\int \frac{dy}{1-y}=\int dx$ 

$\Rightarrow -\log (1-y)=x+C$ 

$\Rightarrow \log (1-y)=-(x+C)$ 

$\Rightarrow 1-y=e^{-(x+C)}$ 

$y=1-A{e}^{-x}(A=e^{-C})$ 

Thus the general solution of given differential equation is $\text{y=1- A}{\text{e}}^{\text{-x}}$.

4. Find the general solution for $\text{se}{\text{c}}^{\text{2}}\text{x}\text{tany}\text{dx+se}{\text{c}}^{\text{2}}\text{y}\text{tanx}\text{dy=0}$.

Ans: The given differential equation is:

${\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy=0$.

Divide both side by $\text{tan x tan y}$:

$\frac{{\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy}{\tan x\tan y}=0$ 

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

Integrate both side:

$\int \frac{{\sec }^{2}x}{\tan x}dx+\int \frac{{\sec }^{2}y}{\tan y}dy=0$ 

$\Rightarrow \int \frac{{\sec }^{2}y}{\tan y}dy=-\int \frac{{\sec }^{2}x}{\tan x}dx$       ……(1)

Use a substitution method for integration. Substitute $\text{tan x = u}$:

For integral on RHS:

$\Rightarrow \tan x=u$ 

$\Rightarrow {\sec }^{2}xdx=du$ 

$\Rightarrow \int \frac{{\sec }^{2}x}{\tan x}dx=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{\sec }^{2}x}{\tan x}dx=\log u$ 

$\Rightarrow \int \frac{{\sec }^{2}x}{\tan x}dx=\log (\tan x)$ 

Thus evaluating result form (1):

$\Rightarrow \log (\tan y)=-\log (\tan x)+\log \left(C\right)$ 

$\Rightarrow \log (\tan y)=\log \left(\frac{C}{\tan x}\right)$ 

$\Rightarrow \tan x\tan y=C$ 

Thus the general solution of given differential equation is $\text{tan x tan y = C}$.

5. Find the general solution for $\left({\text{e}}^{\text{x}}\text{+}{\text{e}}^{\text{-x}}\right)\text{dy-}\left({\text{e}}^{\text{x}}\text{-}{\text{e}}^{\text{-x}}\right)\text{dx=0}$ .

Ans: The given differential equation is:

$(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$.

Simplify the expression:

$dy=\left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right]dx$ 

Integrate both side:

$\int dy=\int \left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right]dx$ ……(1)

Use a substitution method for integration. Substitute $${\text{e}}^{\text{x}}\text{+}{\text{e}}^{\text{-x}}\text{=t}$$:

For integral on RHS:

$\Rightarrow e^x+e^{-x}=t$ 

$\Rightarrow (e^x-e^{-x})dx=dt$ 

$\Rightarrow \int \left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right]dx=\int \frac{dt}{t}$ s

$\Rightarrow \int \left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right]dx=\ln t+C$ 

$\Rightarrow \int \left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right]dx=\log (e^x+e^{-x})+C$ 

Thus evaluating result form (1):

$y=\log (e^x+e^{-x})+C$ 

Thus the general solution of given differential equation is $\text{y=log}\left({\text{e}}^{\text{x}}\text{+}{\text{e}}^{\text{-x}}\right)\text{+C}$.

6. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\left(\text{1+}{\text{x}}^{\text{2}}\right)\left(\text{1+}{\text{y}}^{\text{2}}\right)$.

Ans: The given differential equation is:

$\frac{dy}{dx}=(1+x^2)(1+y^2)$.

Simplify the expression:

$\frac{dy}{1+y^2}=(1+x^2)dx$ 

Integrate both side:

$\int \frac{dy}{1+y^2}=\int (1+x^2)dx$

Use standard integration:

${\tan }^{-1}y=\int dx+\int x^{2}dx$ 

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

Thus the general solution of given differential equation is $\text{ta}{\text{n}}^{\text{-1}}\text{y=x+}\frac{{\text{x}}^{\text{3}}}{\text{3}}\text{+C}$.

7. Find the general solution for $\text{ylog y dx-xdy=0}$.

Ans: The given differential equation is:

$y\log ydx-xdy=0$.

Simplify the expression:

$y\log ydx=xdy$ 

$\Rightarrow \frac{dx}{x}=\frac{dy}{y\log y}$ 

Integrate both side:

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$ ……(1)

Use substitution method for integration on LHS. Substitute $$\text{log y=t}$$:

$\log y=t$ 

$\Rightarrow \frac{1}{y}dy=dt$ 

$\Rightarrow \int \frac{dy}{y\log y}=\int \frac{dt}{t}$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log t$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log (\log y)$ 

Evaluating expression (1):

$\log (\log y)=\log x+\log C$ 

$\Rightarrow \log (\log y)=\log Cx$ 

$\Rightarrow \log y=Cx$ 

$\Rightarrow y=e^{Cx}$ 

Thus the general solution of given differential equation is $\text{y=}{\text{e}}^{\text{Cx}}$.

8. Find the general solution for ${\text{x}}^{\text{5}}\frac{\text{dy}}{\text{dx}}\text{=-}{\text{y}}^{\text{5}}$.

Ans: The given differential equation is:

$x^5\frac{dy}{dx}=-y^5$.

Simplify the expression:

$\frac{dy}{y^5}=-\frac{dx}{x^5}$ 

Integrate both side:

$\int \frac{dy}{y^5}=-\int \frac{dx}{x^{-5}}$ 

$$\Rightarrow \int y^{-5}dy=-\int x^{-5}dx$$ 

$\Rightarrow \frac{y^{-5+1}}{-5+1}=-\frac{x^{-5+1}}{-5+1}+C$ 

$\Rightarrow \frac{y^{-4}}{-4}=-\frac{x^{-4}}{-4}+C$ 

$\Rightarrow x^{-4}+y^{-4}=-4C$ 

$\Rightarrow x^{-4}+y^{-4}=A(A=-4C)$ 

Thus the general solution of given differential equation is ${\text{x}}^{\text{-4}}\text{+}{\text{y}}^{\text{-4}}\text{=A}$.

9. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=si}{\text{n}}^{\text{-1}}\text{x}$.

Ans: The given differential equation is:

$\frac{dy}{dx}={\sin }^{-1}x$.

Simplify the expression:

$dy={\sin }^{-1}xdx$ 

Integrate both side:

$\int dy=\int {\sin }^{-1}xdx$ 

$$\Rightarrow y=\int 1\times {\sin }^{-1}xdx$$ 

Use product rule of integration:

$\int {\sin }^{-1}xdx={\sin }^{-1}x\int dx-\int (\frac{1}{\sqrt{1-x^2}}\int dx)dx$ 

$\Rightarrow \int {\sin }^{-1}xdx=x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx$ 

Substitute $\text{1-}{\text{x}}^{\text{2}}\text{=}{\text{t}}^{\text{2}}$ 

$1-x^2=t^2$ 

$\Rightarrow -2xdx=2tdt$ 

$\Rightarrow -xdx=tdt$ 

Evaluating the integral:

$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$ 

$\Rightarrow \int {\sin }^{-1}xdx=x{\sin }^{-1}x+\int \frac{tdt}{\sqrt{t^2}}$ 

$\Rightarrow \int {\sin }^{-1}xdx=x{\sin }^{-1}x+t+C$ 

$\Rightarrow \int {\sin }^{-1}xdx=x{\sin }^{-1}x+\sqrt{1-x^2}+C$ 

$\Rightarrow y=x{\sin }^{-1}x+\sqrt{1-x^2}+C$ 

Thus the general solution of given differential equation is $\text{y=xsi}{\text{n}}^{\text{-1}}\text{x+}\sqrt{\text{1-}{\text{x}}^{\text{2}}}\text{+C}$,

10. Find the general solution for ${\text{e}}^{\text{x}}\text{tanydx+}\left(\text{1-}{\text{e}}^{\text{x}}\right)\text{se}{\text{c}}^{\text{2}}\text{ydy=0}$.

Ans: The given differential equation is:

$e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$.

Simplify the expression:

$(1-e^x){\sec }^{2}ydy=-e^x\tan ydx$ 

$\Rightarrow \frac{{\sec }^{2}y}{\tan y}dy=-\frac{e^x}{(1-e^x)}dx$ 

Integrate both side:

$\int \frac{{\sec }^{2}y}{\tan y}dy=-\int \frac{e^x}{(1-e^x)}dx$ ……(1)

Substitute $\text{tan y=u}$ 

$\tan y=u$ 

$$\Rightarrow {\sec }^{2}y=du$$ 

Evaluating the LHS integral of (1):

$\Rightarrow \int \frac{{\sec }^{2}y}{\tan y}dy=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{\sec }^{2}y}{\tan y}dy=\log u$ 

$\Rightarrow \int \frac{{\sec }^{2}y}{\tan y}dy=\log (\tan y)$ 

Substitute $\text{1-}{\text{e}}^x\text{=v}$ 

$1-e^x=v$ 

$\Rightarrow -e^{x}dx=dv$ 

Evaluating the RHS integral of (1):

$\Rightarrow -\int \frac{e^x}{(1-e^x)}dx=\int \frac{dv}{v}$ 

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

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

Therefore the integral (1) will be:

$\log (\tan y)=\log (1-e^x)+\log C$ 

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

$\Rightarrow \tan y=C(1-e^x)$ 

Thus the general solution of given differential equation is $\text{tany=C}\left(\text{1-}{\text{e}}^{\text{x}}\right)$,

11. Find the particular solution of $\left({\text{x}}^{\text{3}}\text{+}{\text{x}}^{\text{2}}\text{+x+1}\right)\frac{\text{dy}}{\text{dx}}\text{=2}{\text{x}}^{\text{2}}\text{+x;}\text{y=1,}\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x;y=1,x=0$.

Simplify the expression:

$\frac{dy}{dx}=\frac{2x^2+x}{(x^3+x^2+x+1)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2x^2+x}{(x^3+x+x^2+1)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2x^2+x}{(x+1)(x^2+1)}$ 

$\Rightarrow dy=\frac{2x^2+x}{(x+1)(x^2+1)}dx$ 

Integrate both side:

$\int dy=\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$

$\Rightarrow 2x^2+x=A(x^2+1)+(Bx+C)(x+1)$ 

$$\Rightarrow 2x^2+x=(A+B)x^2+(B+C)x+(A+C)$$ 

By comparing coefficients:

$A+B=2$ 

$B+C=1$ 

$A+C=0$ 

Solving this we get:

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

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

Rewriting the  integral(1):

$y=\frac{1}{2}\int (\frac{1}{x+1}+\frac{3x-1}{x^2+1})dx$ 

$$\Rightarrow y=\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \frac{3x-1}{x^2+1}dx$$ 

$$\Rightarrow y=\frac{1}{2}\log (x+1)+\frac{3}{2}\int \frac{x}{x^2+1}dx-\frac{1}{2}\int \frac{1}{x^2+1}dx$$ 

$$\Rightarrow y=\frac{1}{2}\log (x+1)+\frac{3}{4}\int \frac{2x}{x^2+1}dx-\frac{1}{2}{\tan }^{-1}x$$ 

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

For $y=1$ when $\text{x=0}$. 

$$1=\frac{1}{2}\log (0+1)+\frac{3}{4}\log (0+1)-\frac{1}{2}{\tan }^{-1}0+C$$ 

$$\Rightarrow C=1$$ 

Thus the required particular solution is :

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

12. Find the particular solution of $\text{x}\left({\text{x}}^{\text{2}}\text{-1}\right)\frac{\text{dy}}{\text{dx}}\text{=1}$; $\text{y=0}$ when $\text{x=2}$ to satisfy the given condition.

Ans: The given differential equation is:

$x(x^2-1)\frac{dy}{dx}=1$; $y=0$ when $x=2$ 

Simplify the expression:

$x(x^2-1)\frac{dy}{dx}=1$ 

$\Rightarrow dy=\frac{dx}{x(x^2-1)}$ 

$\Rightarrow dy=\frac{dx}{x(x-1)(x+1)}$ 

Integrate both side:

$\int dy=\int \frac{dx}{x(x-1)(x+1)}$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

$\Rightarrow 1=A(x^2-1)+Bx(x+1)+Cx(x-1)$ 

$$\Rightarrow 1=(A+B+C)x^2+(B-C)x-A$$ 

By comparing coefficients:

$A+B+C=0$ 

$B-C=0$ 

$-A=1$ 

Solving this we get:

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

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

Rewriting the  integral(1):

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

$$\Rightarrow y=-\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{1}{x+1}dx$$ 

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

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

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

$$\Rightarrow y=\frac{1}{2}\log \left[\frac{C^2(x^2-1)}{x^2}\right]$$ 

For $y=0$ when $\text{x=2}$. 

$$0=\frac{1}{2}\log \left[\frac{C^2(2^2-1)}{2^2}\right]$$ 

$$\Rightarrow 0=\log \left[\frac{3C^2}{4}\right]$$ 

$\Rightarrow \frac{3C^2}{4}=1$ 

$\Rightarrow C^2=\frac{4}{3}$ 

Thus the required particular solution is :

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

13. Find the particular solution of $\text{cos}\left(\frac{\text{dy}}{\text{dx}}\right)\text{=a}(\text{a}\in \text{R})$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\cos \left(\frac{dy}{dx}\right)=a(a\in R)$; $y=1$ when $x=0$

Simplify the expression:

$\cos \left(\frac{dy}{dx}\right)=a$ 

$\Rightarrow \frac{dy}{dx}={\cos }^{-1}a$ 

$\Rightarrow dy={\cos }^{-1}adx$ 

Integrate both side:

$\int dy=\int {\cos }^{-1}adx$ 

$\Rightarrow y={\cos }^{-1}a\int dx$ 

$$\Rightarrow y=x{\cos }^{-1}a+C$$ 

For $y=1$ when $\text{x=0}$. 

$$1=0{\cos }^{-1}a+C$$ 

$$\Rightarrow C=1$$ 

Thus the required particular solution is:

$$y=x{\cos }^{-1}a+1$$.

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

$$\Rightarrow \cos \left(\frac{y-1}{x}\right)=a$$

14. Find the particular solution of $\frac{\text{dy}}{\text{dx}}\text{=ytanx}$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\frac{dy}{dx}=y\tan x$; $y=1$ when $x=0$

Simplify the expression:

$\frac{dy}{dx}=y\tan x$

$\Rightarrow \frac{dy}{y}=\tan xdx$ 

Integrate both side:

$\int \frac{dy}{y}=\int \tan xdx$ 

$\Rightarrow \log y=\log (\sec x)+\log C$ 

$$\Rightarrow \log y=\log (C\sec x)$$ 

$\Rightarrow y=C\sec x$ 

For $\text{y=1}$ when $\text{x=0}$. 

$$1=C\sec 0$$ 

$$\Rightarrow C=1$$ 

Thus the required particular solution is :

$$y=\sec x$$. 

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

Ans: The given differential equation is:

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

The curve passes through $(0,0)$.

Simplify the expression:

$\Rightarrow \frac{dy}{dx}=e^x\sin x$

$\Rightarrow dy=e^x\sin xdx$ 

Integrate both side:

$\int dy=\int e^x\sin xdx$ 

Use product rules for integration of RHS. Let:

$I=\int e^x\sin xdx$ 

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

$\Rightarrow I=e^x\sin x-\int e^x\cos xdx$ 

$\Rightarrow I=e^x\sin x-(\cos x\int e^{x}dx+\int (\sin x\int e^{x}dx)dx)$ 

$\Rightarrow I=e^x\sin x-e^x\cos x-\int (e^x\sin x)dx$ 

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

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

Thus integral will be:

$y=\frac{e^x}{2}(\sin x-\cos x)+C$ 

Thus as the curve passes through $\left(\text{0,0}\right)$ 

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

$\Rightarrow 0=\frac{1}{2}(0-1)+C$ 

$\Rightarrow C=\frac{1}{2}$ 

Thus the equation of the curve will be:

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

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

16. For the differential equation $\text{xy}\frac{\text{dy}}{\text{dx}}\text{=}\left(\text{x+2}\right)\left(\text{y+2}\right)$ find the solution curve passing through the point $\left(\text{1,-1}\right)$.

Ans: The given differential equation is:

$xy\frac{dy}{dx}=(x+2)(y+2)$

The curve passes through $(1,-1)$.

Simplify the expression:

$$\Rightarrow \left(\frac{y}{y+2}\right)dy=\frac{(x+2)}{x}dx$$

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

Integrate both side:

$\int (1-\frac{2}{y+2})dy=\int \frac{(x+2)}{x}dx$ 

$\Rightarrow \int dy-2\int \frac{1}{y+2}dy=\int \frac{x}{x}dx+\int \frac{2}{x}dx$ 

$\Rightarrow y-2\log (y+2)=x+2\log x+C$ 

$\Rightarrow y-x=2\log (y+2)+2\log x+C$ 

$\Rightarrow \Rightarrow y-x=2\log \left[x(y+2)\right]+C$ 

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

Thus as the curve passes through $\left(\text{1,-1}\right)$ 

$\Rightarrow -1-1=\log \left[{\left(1\right)}^2{(-1+2)}^2\right]+C$

$\Rightarrow -2=\log 1+C$ 

$\Rightarrow C=-2$ 

Thus the equation of the curve will be:

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

$\Rightarrow y-x+2=\log \left(x^2{(y+2)}^2\right)$ 

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

Ans: According to , the equation is given by:

$y\frac{dy}{dx}=x$ 

The curve passes through $(0,-2)$.

Simplify the expression:

$$\Rightarrow ydy=xdx$$ 

Integrate both side:

$\int ydy=\int xdx$ 

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

$\Rightarrow y^2-x^2=2C$ 

Thus as the curve passes through $\left(\text{0,-2}\right)$ 

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

$\Rightarrow 4=2C$ 

$\Rightarrow C=2$ 

Thus the equation of the curve will be:

$y^2-x^2=2\left(2\right)$ 

$\Rightarrow y^2-x^2=4$ .

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

Ans: Let the point of contact of the tangent be $\left(\text{x,y}\right)$.Then the slope of the segment joining point of contact and $\left(\text{-4,-3}\right)$:

$m=\frac{y+3}{x+4}$ 

According to  the for the slope of tangent $\frac{\text{dy}}{\text{dx}}$ it follows:

$\frac{dy}{dx}=2m$ 

$\Rightarrow \frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$

Simplify the expression:

$\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$ 

$\frac{dy}{y+3}=\frac{2}{x+4}dx$ 

Integrate both side:

$\int \frac{dy}{y+3}=\int \frac{2}{x+4}dx$ 

$\Rightarrow \log (y+3)=2\log (x+4)+\log C$ 

$$\Rightarrow \log (y+3)=\log {(x+4)}^2+\log C$$ 

$$\Rightarrow \log (y+3)=\log C{(x+4)}^2$$ 

$$\Rightarrow y+3=C{(x+4)}^2$$ 

Thus as the curve passes through $\left(\text{-2,1}\right)$ 

$$1+3=C{(-2+4)}^2$$

$\Rightarrow 4=4C$ 

$\Rightarrow C=1$ 

Thus the equation of the curve will be:

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

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

Ans: Let the volume of spherical balloon be $\text{V}$ and its radius $\text{r}$. Let the rate of change of volume be $\text{k}$.

$\frac{dV}{dt}=k$ 

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

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

$\Rightarrow \frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi r^2\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi r^{2}dr=kdt$ 

Integrate both side:

$\int 4\pi r^{2}dr=\int kdt$ 

$\Rightarrow 4\pi \int r^{2}dr=kt+C$