NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.6

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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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