# NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.6

Go back to  'Differential Equations'

## Chapter 9 Ex.9.6 Question 1

Find the general solution for the differential equation given below

$$\frac{{dy}}{{dx}} + 2y = \sin x$$

### Solution

Differential equation is $$\frac{{dy}}{{dx}} + 2y = \sin x$$

This is in the form $$\frac{{dy}}{{dx}} + py = Q$$ where $$p = 2$$ and $$Q = \sin x$$

\begin{align}I.F.& = {e^{\int {pdx} }} = {e^{\int {2dx} }} = {e^{2x}}\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F} \right).dx + C} \\ \Rightarrow \; y{e^{2x}} &= \int {\sin x.} {e^{2x}}dx + C\end{align}

Let, $$I = \int {\sin x.{e^{2x}}}$$

\begin{align} &\Rightarrow \; I = \int {\sin x.\int {{e^{2x}}dx - \int {\left( {\frac{d}{{dx}}\left( {\sin x} \right).\int {{e^{2x}}dx} } \right)} } } dx\\ &\Rightarrow \; I = \sin x.\frac{{{e^{2x}}}}{2} - \int {\left( {\cos x.\frac{{{e^{2x}}}}{2}} \right)} dx\\& \Rightarrow \; I = \frac{{{e^{2x}}\sin x}}{2} - \frac{1}{2}\left[ {\cos x.\int {{e^{2x}}} - \int {\left( {\frac{d}{{dx}}\left( {\cos x} \right).\int {{e^{2x}}dx} } \right)dx} } \right]\\& \Rightarrow \; I = \frac{{{e^{2x}}\sin x}}{2} - \frac{1}{2}\left[ {\cos x.\int {\frac{{{e^{2x}}}}{2}} - \int {\left[ {\left( { - \sin x} \right).\frac{{{e^{2x}}}}{2}} \right]dx} } \right]\\& \Rightarrow \; I = \frac{{{e^{2x}}\sin x}}{2} - \frac{{{e^{2x}}\cos x}}{4} = \frac{1}{4}\int {\left( {\sin x.{e^{2x}}} \right)} dx\\& \Rightarrow \; I = \frac{{{e^{2x}}}}{4}\left( {2\sin x - \cos x} \right) - \frac{1}{4}t\\ &\Rightarrow \; \frac{5}{4}I = \frac{{{e^{2x}}}}{4}\left( {2\sin x - \cos x} \right)\\& \Rightarrow \; I = \frac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right)\\&y{e^{2x}} = \frac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + C\\& \Rightarrow \; y = \frac{1}{5}\left( {2\sin x - \cos x} \right) + C{e^{ - 2x}}\end{align}

## Chapter 9 Ex.9.6 Question 2

Find the general solution for the differential equation given below

$$\frac{{dy}}{{dx}} + 3y = {e^{ - 2x}}$$

### Solution

Differential equation is $$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = 3{\text{ and }}Q = {e^{ - 2x}}} \right)$$

\begin{align}&I.F. = {e^{\int {pdx} }} = {e^{\int {3dx} }} = {e^{3x}}\\&y\left( {I.F} \right) = \int {\left( {Q \times I.F.} \right)dx + C} \\ &\Rightarrow \; y{e^{3x}} = \int {\left( {{e^{ - 2x}} \times {e^{3x}}} \right)} + C\\& \Rightarrow \; y{e^{3x}} = \int {{e^x}} dx + C\\ &\Rightarrow \; y{e^{3x}} = {e^x} + C\\ &\Rightarrow \; y = {e^{ - 2x}} + C{e^{ - 3x}}\end{align}

## Chapter 9 Ex.9.6 Question 3

Find the general solution for the differential equation given below

$$\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}$$

### Solution

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = \frac{1}{x}{\text{ and }}Q = {x^2}} \right)$$

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{\int {\frac{1}{x}dx} }} = {e^{\log x}} = x\\y\left( {I.F} \right)&= \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; yx = \int {\left( {{x^2}.x} \right)dx} + C\\& \Rightarrow \; yx = \int {{x^3}} dx + C\\& \Rightarrow \; yx = \frac{{{x^4}}}{4} + C\\& \Rightarrow \; xy = \frac{{{x^4}}}{4} + C\end{align}

## Chapter 9 Ex.9.6 Question 4

Find the general solution for the differential equation given below

$$\frac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x\left( {0 \le x \le \frac{\pi }{2}} \right)$$

### Solution

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = \sec x{\text{ and }}Q = \tan x} \right)$$

\begin{align}&I.F. = {e^{\int {pdx} }} = {e^{\int {\sec xdx} }} = {e^{\log \left( {\sec x + \tan x} \right)}} = \sec x + \tan x\\&y\left( {I.F} \right) = \int {\left( {Q \times I.F.} \right)dx + C} \\ &\Rightarrow \; y\left( {\sec x + \tan x} \right) = \int {\tan x\left( {\sec x + \tan x} \right)dx} + C\\& \Rightarrow \; y\left( {\sec x + \tan x} \right) = \int {\sec x\tan xdx} + \int {{{\tan }^2}dx} + C\\ &\Rightarrow \; y\left( {\sec x + \tan x} \right) = \sec x + \int {\left( {{{\sec }^2}x - 1} \right)} dx + C\\ &\Rightarrow \; y\left( {\sec x + \tan x} \right) = \sec x + \tan x - x + C\end{align}

## Chapter 9 Ex.9.6 Question 5

Find the general solution for the differential equation given below

$${\cos ^2}x\frac{{dy}}{{dx}} + y = \tan x\left( {0 \le x < \frac{\pi }{2}} \right)$$

### Solution

\begin{align}{\cos ^2}x\frac{{dy}}{{dx}} + y& = \tan x\left( {0 \le x < \frac{\pi }{2}} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} + \frac{y}{{{{\cos }^2}x}} = \frac{{\tan x}}{{{{\cos }^2}x}}\\& \Rightarrow \; \frac{{dy}}{{dx}} + \left( {{{\sec }^2}x} \right)y = {\sec ^2}x.\tan x\end{align}

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = {{\sec }^2}x{\text{ and }}Q = {{\sec }^2}x.\tan x} \right)$$

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{\int {{{\sec }^2}xdx} }} = {e^{\tan x}}\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; y{e^{\tan x}} = \int {\left( {{{\sec }^2}x.\tan x.{e^{\tan x}}} \right)dx} + C\\ &\Rightarrow \; y{e^{\tan x}} = {e^{\tan x}}\left( {\tan x - 1} \right) + C\\& \Rightarrow \; y = \tan x - 1 + C{e^{\tan x}}\end{align}

## Chapter 9 Ex.9.6 Question 6

Find the general solution for the differential equation given below

$$x\frac{{dy}}{{dx}} + 2y = {x^2}\log x$$

### Solution

\begin{align}&x\frac{{dy}}{{dx}} + 2y = {x^2}\log x\\ &\Rightarrow \; \frac{{dy}}{{dx}} + \frac{2}{x}y = x\log x\end{align}

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = \frac{2}{x}{\text{ and }}Q = x\log x} \right)$$

\begin{align}I.F &= {e^{\int {pdx} }} = {e^{\int {\frac{2}{x}dx} }} = {e^{2\log x}} = {e^{\log {x^2}}} = {x^2}\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F} \right)dx + C} \\& \Rightarrow \; y.{x^2} = \int {\left( {x\log x.{x^2}} \right)dx + C} \\ &\Rightarrow \; {x^2}y = \int {\left( {{x^3}\log x} \right)dx + C} \\& \Rightarrow \; {x^2}y = \log x\int {{x^3}dx} - \int {\left[ {\frac{d}{{dx}}\left( {\log x} \right).\int {{x^3}dx} } \right]dx + C} \\& \Rightarrow \; {x^2}y = \log x.\frac{{{x^4}}}{4} - \int {\left( {\frac{1}{x}.\frac{{{x^4}}}{4}} \right)} dx + C\\ &\Rightarrow \; {x^2}y = \frac{{{x^4}\log x}}{4} - \frac{1}{4}\int {{x^3}dx} + C\\ &\Rightarrow \; {x^2}y = \frac{{{x^4}\log x}}{4} - \frac{1}{4}\frac{{{x^4}}}{4} + C\\& \Rightarrow \; {x^2}y = \frac{1}{{16}}{x^4}\left( {4\log x - 1} \right) + C\\& \Rightarrow \; y = \frac{1}{{16}}{x^2}\left( {4\log x - 1} \right) + C{x^{ - 2}}\end{align}

## Chapter 9 Ex.9.6 Question 7

Find the general solution for the differential equation given below

$$x\log x\frac{{dy}}{{dx}} + y = \frac{2}{x}\log x$$

### Solution

$$x\log x\frac{{dy}}{{dx}} + y = \frac{2}{x}\log x$$

\begin{align} &\Rightarrow \; \frac{{dy}}{{dx}} + \frac{y}{{x\log x}} = \frac{2}{{{x^2}}}\\&\frac{{dy}}{{dx}} + py = Q\end{align}

$$\left( {{\text{where, }}p = \frac{1}{{x\log x}}{\text{ and }}Q = \frac{2}{{{x^2}}}} \right)$$

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{\int {\frac{1}{{x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; y\log x = \int {\left( {\frac{2}{{{x^2}}}\log x} \right)dx} + C\\\int {\left( {\frac{2}{{{x^2}}}\log x} \right)dx} &= 2\int {\left( {\log x.\frac{1}{{{x^2}}}} \right)} dx\\& = 2\left[ {\log x.\int {\frac{1}{{{x^2}}}dx - \int {\left\{ {\frac{d}{{dx}}\left( {\log x} \right).\int {\frac{1}{{{x^2}}}dx} } \right\}dx} } } \right]\\ &= 2\left[ {\log x\left( { - \frac{1}{x}} \right) - \int {\left( {\frac{1}{x}\left( { - \frac{1}{x}} \right)} \right)dx} } \right]\\ &= 2\left[ { - \frac{{\log x}}{x} + \int {\frac{1}{{{x^2}}}dx} } \right]\\ &= 2\left[ { - \frac{{\log x}}{x} - \frac{1}{x}} \right]\\ &= - \frac{2}{x}\left( {1 + \log x} \right)\\y\log x &= - \frac{2}{x}\left( {1 + \log x} \right) + C\end{align}

## Chapter 9 Ex.9.6 Question 8

Find the general solution for the differential equation given below

$$\left( {1 + {x^2}} \right)dy + 2xydx = \cot xdx\left( {x \ne 0} \right)$$

### Solution

\begin{align}&\left( {1 + {x^2}} \right)dy + 2xydx = \cot xdx\\& \Rightarrow \; \frac{{dy}}{{dx}} + \frac{{2xy}}{{\left( {1 + {x^2}} \right)}} = \frac{{\cot x}}{{\left( {1 + {x^2}} \right)}}\\&\frac{{dy}}{{dx}} + py = Q{\text{ }}\left( {{\text{where, }}p = \frac{{2xy}}{{\left( {1 + {x^2}} \right)}}{\text{ and }}Q = \frac{{\cot x}}{{\left( {1 + {x^2}} \right)}}} \right)\end{align}

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}dx} }} = {e^{\log \left( {1 + {x^2}} \right)}} = 1 + {x^2}\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F.} \right)dx + C} \\ &\Rightarrow \; y\left( {1 + {x^2}} \right) = \int {\left( {\frac{{\cot x}}{{1 + {x^2}}} \times \left( {1 + {x^2}} \right)} \right)dx} + C\\&\Rightarrow \; y\left( {1 + {x^2}} \right) = \int {\cot x} dx + C\\ &\Rightarrow \; y\left( {1 + {x^2}} \right) = \log \left| {\sin x} \right| + C\end{align}

## Chapter 9 Ex.9.6 Question 9

Find the general solution for the differential equation given below

$$x\frac{{dy}}{{dx}} + y - x + xy\cot x = 0\left( {x \ne 0} \right)$$

### Solution

\begin{align}&x\frac{{dy}}{{dx}} + y - x + xy\cot x = 0\left( {x \ne 0} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} + \frac{y}{x} - 1 + y\cot x = 0\\& \Rightarrow \; \frac{{dy}}{{dx}} + y\left( {\frac{1}{x} + \cot x} \right) - 1 = 0\\ &\Rightarrow \; \frac{{dy}}{{dx}} + \left( {\frac{1}{x} + \cot x} \right)y = 1\\&\frac{{dy}}{{dx}} + py = Q{\text{ }}\left( {{\text{where, }}p = \left( {\frac{1}{x} + \cot x} \right){\text{ and }}Q = 1} \right)\end{align}

\begin{align}I.F.& = {e^{\int {pdx} }} = {e^{\int {\left( {\frac{1}{x} + \cot x} \right)dx} }} = {e^{\log + \log \left( {\sin x} \right)}} = {e^{\log \left( {x\sin x} \right)}} = x\sin x\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F.} \right)dx + C} \\ &\Rightarrow \; y\left( {x\sin x} \right) = \int {\left( {1 \times x\sin x} \right)dx} + C\\ &\Rightarrow \; y\left( {x\sin x} \right) = \int {\left( {x\sin x} \right)} dx + C\\& \Rightarrow \; y\left( {x\sin x} \right) = x\int {\sin xdx - \int {\left[ {\frac{d}{{dx}}\left( x \right).\int {\sin xdx} } \right] + C} } \\& \Rightarrow \; y\left( {x\sin x} \right) = x\left( { - \cos x} \right) - \int {1.\left( { - \cos x} \right)dx + C} \\& \Rightarrow \; y\left( {x\sin x} \right) = - 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 x + \frac{1}{x} + \frac{C}{{x\sin x}}\\ &\Rightarrow \; y = \frac{1}{x} - \cot x + \frac{C}{{x\sin x}}\end{align}

## Chapter 9 Ex.9.6 Question 10

Find the general solution for the differential equation given below

$$\left( {x + y\frac{{dy}}{{dx}}} \right) = 1$$

### Solution

\begin{align}&\left( {x + y} \right)\frac{{dy}}{{dx}} = 1\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{1}{{x + y}}\\ &\Rightarrow \; \frac{{dx}}{{dy}} = x + y\\&\Rightarrow \; \frac{{dx}}{{dy}} - x = y\end{align}

Put in form $$\frac{{dx}}{{dy}} + {p_1}x = {Q_1}$$

$$\left( {{\text{where, }}{p_1} = - 1{\text{ and }}{Q_1} = y} \right)$$

\begin{align}I.F. &= {e^{\int {{p_1}dy} }} = \int {{e^{ - dy}}} = {e^{ - y}}\\y\left( {I.F} \right) &= \int {\left( {{Q_1} \times I.F.} \right)dx + C} \\ &\Rightarrow \; x{e^{ - y}} = \int {\left( {y.{e^{ - y}}} \right)dy} + C\\ &\Rightarrow \; x{e^{ - y}} = y.\int {{e^{ - y}}dy - \int {\left[ {\frac{d}{{dy}}\left( y \right).\int {{e^{ - y}}dy} } \right]dy} } + C\\ &\Rightarrow \; x{e^{ - y}} = y\left( {\frac{{{e^{ - y}}}}{{ - 1}}} \right) - \int {\left( {\frac{{{e^{ - y}}}}{{ - 1}}} \right)dy + C} \\& \Rightarrow \; x{e^{ - y}} = - y{e^{ - y}} + \int {{e^{ - y}}dy + C} \\ &\Rightarrow \; x{e^{ - y}} = - y{e^{ - y}} - {e^{ - y}} + C\\& \Rightarrow \; x = - y - 1 + C{e^y}\\ &\Rightarrow \; x + y + 1 = C{e^y}\end{align}

## Chapter 9 Ex.9.6 Question 11

For the below differential equations find the particular solution satisfying the given condition:

$$ydx + \left( {x - {y^2}} \right)dy = 0$$

### Solution

\begin{align}&ydx + \left( {x - {y^2}} \right)dy = 0\\& \Rightarrow \; ydx - \left( {{y^2} - x} \right)dy = 0\\ &\Rightarrow \; \frac{{dx}}{{dy}} = \frac{{{y^2} - x}}{y} = y - \frac{x}{y}\\ &\Rightarrow \; \frac{{dx}}{{dy}} + \frac{x}{y} = y\end{align}

Put in form $$\frac{{dx}}{{dy}} + {p_1}x = {Q_1}$$

$$\left( {{\text{where, }}{p_1} = \frac{1}{y}{\text{ and }}{Q_1} = y} \right)$$

\begin{align}I.F. &= {e^{\int {{p_1}dy} }} = \int {{e^{\frac{1}{y}dy}}} = {e^{\log y}} = y\\y\left( {I.F} \right) &= \int {\left( {{Q_1} \times I.F.} \right)dx + C}\\ &\Rightarrow \; xy = \int {\left( {y.y} \right)dy} + C\\& \Rightarrow \; xy = \int {{y^2}dy} + C\\&\Rightarrow \; xy = \frac{{{y^3}}}{3} + C\\& \Rightarrow \; x = \frac{{{y^3}}}{3} + \frac{C}{y}\end{align}

## Chapter 9 Ex.9.6 Question 12

For the below differential equations find the particular solution satisfying the given condition:

$$\left( {x + 3{y^3}} \right)\frac{{dy}}{{dx}} = y\left( {y > 0} \right)$$

### Solution

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

\begin{align} &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{y}{{x + 3{y^2}}}\\& \Rightarrow \; \frac{{dx}}{{dy}} = \frac{{x + 3{y^2}}}{y} = \frac{x}{y} + 3y\\& \Rightarrow \; \frac{{dx}}{{dy}} - \frac{x}{y} = 3y\end{align}

$$\frac{{dx}}{{dy}} + {p_1}x = {Q_1}$$

$$\left( {{\text{where, }}{p_1} = - \frac{1}{y}{\text{ and }}{Q_1} = 3y} \right)$$

\begin{align}I.F. &= {e^{\int {{p_1}dy} }} = {e^{ - \int {\frac{{dy}}{y}} }} = {e^{ - \log y}} = {e^{\log {y^{ - 1}}}} = \frac{1}{y}\\y\left( {I.F} \right) &= \int {\left( {{Q_1} \times I.F.} \right)dx + C} \\& \Rightarrow \; x \times \frac{1}{y} = \int {\left( {3y \times \frac{1}{y}} \right)dy} + C\\& \Rightarrow \; \frac{x}{y} = 3y + C\\ &\Rightarrow \; x = 3{y^2} + Cy\end{align}

## Chapter 9 Ex.9.6 Question 13

For the below differential equations find the particular solution satisfying the given condition:

$$\frac{{dy}}{{dx}} + 2y\tan x = \sin x;{\text{ }}y = 0{\text{ when }}x = \frac{\pi }{3}$$

### Solution

\begin{align}\frac{{dy}}{{dx}} + 2y\tan x &= \sin x\\\frac{{dy}}{{dx}} + py &= Q\end{align}

$$\left( {{\text{where, }}p = 2\tan x{\text{ and }}Q = \sin x} \right)$$

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{\int {2\tan xdx} }} = {e^{2\log \left| {\sec x} \right|}} = {e^{\log \left( {{{\sec }^2}x} \right)}} = {\sec ^2}x\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; y({\sec ^2}x) = \int {\left( {\sin x.{{\sec }^2}x} \right)dx} + C\\ &\Rightarrow \; y{\sec ^2}x = \int {\left( {\sec x.\tan x} \right)dx} + C\\ &\Rightarrow \; y{\sec ^2}x = \sec x + C\\&y = 0{\text{ at }}x = \frac{\pi }{3}\\&0 \times {\sec ^2}\frac{\pi }{3} = \sec \frac{\pi }{3} + C\\& \Rightarrow \; 0 = 2 + C\\& \Rightarrow \; C = - 2\\&y{\sec ^2}x = \sec x - 2\\ &\Rightarrow \; y = \cos x - 2{\cos ^2}x\end{align}

## Chapter 9 Ex.9.6 Question 14

For the below differential equations find the particular solution satisfying the given condition:

$$\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = \frac{1}{{1 + {x^2}}};{\text{ }}y = 0{\text{ when }}x = 1$$

### Solution

\begin{align}&\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = \frac{1}{{1 + {x^2}}}\\& \Rightarrow \; \frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}\end{align}

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = \frac{{2x}}{{1 + {x^2}}}{\text{ and }}Q = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right)$$

\begin{align}&I.F. = {e^{\int {pdx} }} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}dx} }} = {e^{\log \left( {1 + {x^2}} \right)}} = 1 + {x^2}\\&y\left( {I.F} \right) = \int {\left( {Q \times I.F.} \right)dx + C} \\&\Rightarrow \; y\left( {1 + {x^2}} \right) = \int {\left[ {\frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}.\left( {1 + {x^2}} \right)} \right]dx} + C\\ &\Rightarrow \; y\left( {1 + {x^2}} \right) = \int {\frac{1}{{1 + {x^2}}}dx} + C\\ &\Rightarrow \; y\left( {1 + {x^2}} \right) = {\tan ^{ - 1}}x + C{\text{ }}....{\text{(1)}}\\\\y &= 0{\text{ at }}x = 1\\& \Rightarrow \; C = - \frac{\pi }{4}\\y\left( {1 + {x^2}} \right) &= {\tan ^{ - 1}}x - \frac{\pi }{4}\end{align}

## Chapter 9 Ex.9.6 Question 15

For the below differential equations find the particular solution satisfying the given condition:

$$\frac{{dy}}{{dx}} - 3y\cot x = \sin 2x;{\text{ }}y = 2{\text{ when }}x = \frac{\pi }{2}$$

### Solution

$$\frac{{dy}}{{dx}} - 3y\cot x = \sin 2x$$

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = - 3\cot x{\text{ and }}Q = \sin 2x} \right)$$

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{ - 3\int {\cot xdx} }} = {e^{ - 3\log \left| {\sin x} \right|}} = {e^{\log {{\left| {\sin x} \right|}^{ - 3}}}} = {e^{\log \left| {\frac{1}{{{{\sin }^3}x}}} \right|}} = \frac{1}{{{{\sin }^3}x}}\\y\left( {I.F} \right)& = \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; y.\frac{1}{{{{\sin }^3}x}} = \int {\left[ {\sin 2x.\frac{1}{{{{\sin }^3}x}}} \right]dx} + C\\& \Rightarrow \; y\cos e{c^3}x = 2\int {\left( {\cot x\cos ecx} \right)dx} + C\\& \Rightarrow \; y\cos e{c^3}x = - 2\cos ecx + C\\& \Rightarrow \; y = - \frac{2}{{\cos e{c^2}x}} + \frac{C}{{\cos e{c^3}x}}\\& \Rightarrow \; y = - 2{\sin ^2}x + C{\sin ^3}x\end{align}

\begin{align}y &= 2{\text{ at }}x = \frac{\pi }{2}\\2 &= - 2 + C\\& \Rightarrow \; C = 4\\y& = - 2{\sin ^2}x + 4{\sin ^3}x\\& \Rightarrow \; y = 4{\sin ^3}x - 2{\sin ^2}x\end{align}

## Chapter 9 Ex.9.6 Question 16

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $$\left( {x,y} \right)$$ is equal to the sum of the coordinates of the point.

### Solution

$${\text{Let }}F\left( {x,y} \right)$$be the curve passing through origin.

At $$\left( {x,y} \right)$$, slope of curve will be$$\frac{{dy}}{{dx}}$$

\begin{align}\frac{{dy}}{{dx}}& = x + y\\& \Rightarrow \; \frac{{dy}}{{dx}} - y = x\\&\frac{{dy}}{{dx}} + py = Q\end{align}

$$\left( {{\text{where, }}p = - 1{\text{ and }}Q = x} \right)$$

\begin{align}I.F. &= {e^{\int {pdx} }} = {e^{\int {\left( { - 1} \right)dx} }} = {e^{ - x}}\\y\left( {I.F} \right) &= \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; y{e^{ - x}} = \int {x{e^{ - x}}} dx + C\\ &\Rightarrow \; y{e^{ - x}} = x\int {{e^{ - x}}dx} - \int {\left[ {\frac{d}{{dx}}\left( x \right).\int {{e^{ - x}}dx} } \right]} dx + C\\ &\Rightarrow \; y{e^{ - x}} = - x{e^{ - x}} + \int {{e^{ - x}}dx} + C\\ &\Rightarrow \; y{e^{ - x}} = - x{e^{ - x}} + \left( { - {e^{ - x}}} \right) + C\\& \Rightarrow \; y{e^{ - x}} = - {e^{ - x}}\left( {x + 1} \right) + C\\ &\Rightarrow \; y = - \left( {x + 1} \right) + C{e^x}\\ &\Rightarrow \; x + y + 1 = C{e^x}\end{align}

Curve passes through origin.

\begin{align}1 &= C\\ &\Rightarrow \; x + y + 1 = {e^x}\end{align}

## Chapter 9 Ex.9.6 Question 17

Find the equation of a curve passing through the point $$\left( {0,2} \right)$$ given that the sum of the coordinates of any point in the curve exceeds the magnitude of the slope of the tangent to the curve at any point by $$5.$$

### Solution

$$F\left( {x,y} \right)$$ be curve and let $$\left( {x,y} \right)$$ be a point on the curve. Slope of the tangent to curve at $$\left( {x,y} \right)$$ is $$\frac{{dy}}{{dx}}$$

\begin{align}&\frac{{dy}}{{dx}} + 5 = x + y\\& \Rightarrow \; \frac{{dy}}{{dx}} - y = x - 5\\&\frac{{dy}}{{dx}} + py = Q\end{align}

$$\left( {{\text{where, }}p = - 1{\text{ and }}Q = x - 5} \right)$$

\begin{align}&I.F. = {e^{\int {pdx} }} = {e^{\int {\left( { - 1} \right)dx} }} = {e^{ - x}}\\&y\left( {I.F} \right) = \int {\left( {Q \times I.F.} \right)dx + C} \\ &\Rightarrow \; y{e^{ - x}} = \int {\left( {x - 5} \right){e^{ - x}}} dx + C\\&\int {\left( {x - 5} \right){e^{ - x}}} dx = \left( {x - 5} \right)\int {{e^{ - x}}dx} - \int {\left[ {\frac{d}{{dx}}\left( {x - 5} \right).\int {{e^{ - x}}} dx} \right]} dx\\& = \left( {x - 5} \right)\left( { - {e^{ - x}}} \right) - \int {\left( { - {e^{ - x}}} \right)dx} \\ &= \left( {5 - x} \right){e^{ - x}} - \left( { - {e^{ - x}}} \right)\\& = \left( {4 - x} \right){e^{ - x}}\\& \Rightarrow \; y{e^{ - x}} = \left( {4 - x} \right){e^{ - x}} + C\end{align}

Curve passes through $$\left( {0,2} \right)$$

\begin{align}&0 + 2 - 4 = C.0\\& \Rightarrow \; - 2 = C\\ &\Rightarrow \; C = - 2\\&x + y - 4 = - 2{e^x}\\ &\Rightarrow \; y = 4 - x - 2{e^x}\end{align}

## Chapter 9 Ex.9.6 Question 18

The integrating factor of the differential equation $$x\frac{{dy}}{{dx}} - y = 2{x^2}$$

(A) $${e^{ - x}}$$

(B) $${e^{ - y}}$$

(C) $$\frac{1}{x}$$

(D) $$x$$

### Solution

\begin{align}&x\frac{{dy}}{{dx}} - y = 2{x^2}\\& \Rightarrow \; \frac{{dy}}{{dx}} - \frac{y}{x} = 2x\end{align}

$$\frac{{dy}}{{dx}} + py = Q$$

$$\left( {{\text{where, }}p = - \frac{1}{x}{\text{ and }}Q = 2x} \right)$$

$$\therefore I.F. = {e^{\int { - \frac{1}{x}dx} }} = {e^{ - \log x}} = {e^{\log \left( {{x^{ - 1}}} \right)}} = {x^{ - 1}} = \frac{1}{x}$$

Thus, the correct option is C.

## Chapter 9 Ex.9.6 Question 19

The integrating factor of the differential equation $$\left( {1 - {y^2}} \right)\frac{{dx}}{{dy}} + yx = ay\left( { - 1 > y < 1} \right)$$

(A) $$\frac{1}{{{y^2} - 1}}$$

(B) $$\frac{1}{{\sqrt {{y^2} - 1} }}$$

(C) $$\frac{1}{{1 - {y^2}}}$$

(D) $$\frac{1}{{\sqrt {1 - {y^2}} }}$$

### Solution

\begin{align}&\left( {1 - {y^2}} \right)\frac{{dx}}{{dy}} + yx = ay\\ &\Rightarrow \; \frac{{dx}}{{dy}} + \frac{{yx}}{{1 - {y^2}}} = \frac{{ay}}{{1 - {y^2}}}\\&\frac{{dx}}{{dy}} + {p_1}x = {Q_1}\end{align}

$$\left( {{\text{where, }}{p_1} = \frac{y}{{1 - {y^2}}}{\text{ and }}{Q_1} = \frac{{ay}}{{1 - {y^2}}}} \right)$$

$$\therefore I.F. = {e^{\int {{p_1}dy} }} = {e^{\int {\frac{y}{{1 - {y^2}}}} dy}} = {e^{ - \frac{1}{2}\log \left( {1 - {y^2}} \right)}} = {e^{\log \left[ {\frac{1}{{\sqrt {1 - {y^2}} }}} \right]}} = \frac{1}{{\sqrt {1 - {y^2}} }}$$

Thus, the correct option is D.

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