Ncert Solutions For Class 12 Maths Ex 9.4

Ncert Solutions For Class 12 Maths Chapter 9 Ex 9.4

Q.1: \(\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}\)

Solution:

The given differential equation is:

\(\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}\) \(\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}\\ \\ \Rightarrow \frac{dy}{dx}=\frac{2\sin ^{2}\frac{x}{2}}{2\cos ^{2}\frac{x}{2}}=\tan ^{2}\frac{x}{2}\\ \\ \Rightarrow \frac{dy}{dx}=(\sec ^{2}\frac{x}{2}-1)\)

Separate the variables, we get:

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

Now, integrating both sides of this equation, we get:

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

This is the required general solution of the given differential equation.

Q.2: \(\frac{dy}{dx}=\sqrt{4-y^{2}}\: (-2<y<2)\)

Solution:

The given differential equation is:

\(\frac{dy}{dx}=\sqrt{4-y^{2}}\)

Separate the variables, we get:

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

Now, integrating both sides of this equation, we get:

\(\int \frac{dy}{\sqrt{4-y^{2}}}=\int dx\\ \\ \Rightarrow \sin ^{-1}\frac{y}{2}=x+C\\ \\ \Rightarrow \frac{y}{2}=\sin (x+C)\\ \\ \Rightarrow y=2\sin (x+C)\)

This is the required general solution of the given differential equation.

 

 

Q.3: \(\frac{dy}{dx}+y=1\: (y\neq 1)\)

Solution:

The given differential equation is:

\(\frac{dy}{dx}+y=1\\ \\ \Rightarrow dy+y\: dx=dx\\ \\ \Rightarrow dy=(1-y)dx\)

Separate the variables, we get:

\(\Rightarrow\frac{ dy}{1-y}=dx\)

Now, integrating both sides, we get:

\(\int \frac{ dy}{1-y}=\int dx\\ \\ \Rightarrow log(1-y)=x+log\, C\\ \\ \Rightarrow -log\,C-log(1-y)=x\\ \\ \Rightarrow log\,C(1-y)=e^{-x}\\ \\ \Rightarrow 1-y=\frac{1}{C}e^{-x}\\ \\ \Rightarrow y=1-\frac{1}{C}e^{-x}\\ \\ \Rightarrow y=1+Ae^{-x}\: (where\: A=-\frac{1}{C})\)

This is the required general solution of the given differential equation.

 

 

Q.4: \(\sec ^{2}x\tan\: y \: dx+\sec ^{2}\: y\, \tan \: x\, dy=0\)

Solution:

The given differential equation is:

\(\sec ^{2}x\tan\: y \: dx+\sec ^{2}\: y\, \tan \: x\, dy=0\\ \\ \Rightarrow \frac{\sec ^{2}x\tan\: y \: dx+\sec ^{2}\: y\, \tan \: x\, dy}{\tan \, x\, \tan \, y}=0\\ \\ \Rightarrow \frac{\sec ^{2}x}{\tan x}dx+\frac{\sec ^{2}y}{\tan y}dy=0\\ \\ \Rightarrow \frac{\sec ^{2}x}{\tan x}dx=-\frac{\sec ^{2}y}{\tan y}dy\)

Integrating both sides of this equation, we get:

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

Let, tan x = t

Therefore, \(\frac{d}{dx}(\tan x)= \frac{dt}{dx}\\ \\ \Rightarrow \sec ^{2}x=\frac{dt}{dx}\\ \\ \Rightarrow \sec ^{2}xdx=dt\)

Now,

\(\int \frac{\sec ^{2}x}{\tan x}dx=\int \frac{1}{t}dt\;=\;log\, t\;=\;log(\tan x)\\\)

Similarly, \(\int \frac{\sec ^{2}x}{\tan x}dx=log(\tan y)\)

Substituting these values in equation (1), we get:

\(log(\tan x)=-log(\tan y)+log\, C\\ \\ \Rightarrow log(\tan x)=log(\frac{C}{\tan y})\\ \\ \Rightarrow \tan x=\frac{C}{\tan y}\\ \\ \Rightarrow \tan x\, \tan y=C\)

This is the required general solution of the given differential equation.

Q.5: \((e^{x}+e^{-x})dy-(e^{x}-e^{-x})dx=0\)

Solution:

The given differential equation is:

\((e^{x}+e^{-x})dy-(e^{x}-e^{-x})dx=0\\ \\ \Rightarrow (e^{x}+e^{-x})dy=(e^{x}-e^{-x})dx\\ \\ \Rightarrow dy=[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}]dx\)

Integrating both sides of this equation, we get:

\(\int dy=\int [\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}]dx+C\\ \\ \Rightarrow y=\int [\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}]dx+C\: \: \: \: \: ….(1)\)

Let \((e^{x}+e^{-x})\)

\(\frac{d}{dx}(e^{x}+e^{-x})=\frac{dt}{dx}\\ \\ \Rightarrow e^{x}-e^{-x}=\frac{dt}{dx}\\ \\ \Rightarrow (e^{x}-e^{-x})dx=dt\)

Substituting this value in equation (1), we get:

\(y=\int \frac{1}{t}dt+C\\ \\ \Rightarrow y=log(t)+C\\ \\ \Rightarrow y=log(e^{x}+e^{-x})+C\)

This is the required general solution of the given differential equation.

 

 

Q.6: \(\frac{dy}{dx}=(1+x^{2})(1+y^{2})\)

Solution:

The given differential equation is:

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

Integrating both sides of this equation, we get:

\(\int \frac{dy}{1+y^{2}}=\int (1+x^{2})dx\\ \\ \Rightarrow \tan ^{-1}y=\int dx+\int x^{2}dx\\ \\ \Rightarrow \tan ^{-1}y=x+\frac{x^{3}}{3}+C\)

This is the required general solution of the given differential equation.

 

 

Q.7: \(y\, log\, y\, dx-x\, dy=0\)

Solution:

The given differential equation is:

\(y\, log\, y\, dx-x\, dy=0\\ \\ \Rightarrow y\, log\, y\, dx=x\, dy\\ \\ \Rightarrow \frac{dy}{y\, log\, y}=\frac{dx}{x}\)

Integrating both sides, we get:

\(\int \frac{dy}{y\, log\, y}=\int \frac{dx}{x}\) . . . . . . . . . . . . . . (1)

Let, log y = t

Therefore, \(\frac{d}{dy}(log\, y)=\frac{dt}{dy}\\ \\ \Rightarrow \frac{1}{y}=\frac{dt}{dy}\\ \\ \Rightarrow \frac{1}{y}dy=dt\)

Substituting this value in equation (1), we get:

\(\int \frac{dt}{t}=\int \frac{dx}{x}\\ \\ \Rightarrow log \, t=log\, x+log \, C\\ \\ \Rightarrow log(log\, y)=log\, Cx\\ \\ \Rightarrow log\, y=Cx\\ \\ \Rightarrow y=e^{Cx}\)

This is the required general solution of the given differential equation.

 

 

Q.8: \(x^{5}\frac{dy}{dx}=-y^{5}\)

Solution:

The given differential equation is:

\(x^{5}\frac{dy}{dx}=-y^{5}\\ \\ \Rightarrow \frac{dy}{y^{5}}=-\frac{dx}{x^{5}}\\ \\ \Rightarrow \frac{dx}{x^{5}}+\frac{dy}{y^{5}}=0\)

Integrating both sides, we get:

\(\int \frac{dx}{x^{5}}+\int \frac{dy}{y^{5}}=k\) (where k is any constant)

\(\int {x^{-5}}dx+\int {y^{-5}dy}=k\\ \\ \Rightarrow \frac{x^{-4}}{-4}+\frac{y^{-4}}{-4}=k\\ \\ \Rightarrow x^{-4}+y^{-4}=-4k\\ \\ \Rightarrow x^{-4}+y^{-4}=C\) (C=-4k)

This is the required general solution of the given differential equation.

Q.9. \(\frac{dy}{dx}=\sin ^{-1}x\)

Solution:

The given differential equation is:

\(\frac{dy}{dx}=\sin ^{-1}x\\ \\ \Rightarrow dy=\sin ^{-1}x\, dx\)

Integrating both sides, we get:

\(\int dy=\int \sin ^{-1}x\, dx\\ \\ \Rightarrow y=\int (\sin ^{-1}x-1)dx\\ \\ \Rightarrow y=\sin ^{-1}x\cdot \int (1)dx-\int [(\frac{d}{dx}(\sin ^{-1}x)\cdot \int (1)dx)]dx\\ \\ \Rightarrow y=\sin ^{-1}x\cdot x-\int (\frac{1}{\sqrt{1-x^{2}}}\cdot x)dx\\ \\ \Rightarrow y=x\sin ^{-1}x+\int \frac{-x}{\sqrt{1-x^{2}}}dx\: \: \: \: \: ….(1)\)

Let, \(1-x^{2}=t\\ \\ \Rightarrow \frac{d}{dx}(1-x^{2})=\frac{dt}{dx}\\ \\ \Rightarrow -2x=\frac{dt}{dx}\\ \\ \Rightarrow x\, dx=-\frac{1}{2}dt\)

Substituting this value in equation (1), we get:

\(y=x\sin ^{-1}x+\int \frac{1}{2\sqrt{t}}dt\\ \\ \Rightarrow y=x\sin ^{-1}x+\frac{1}{2}\cdot \int (t)^{-\frac{1}{2}}dt\\ \\ \Rightarrow y=x\sin ^{-1}x+\frac{1}{2}\cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C\\ \\ \Rightarrow y=x\sin ^{-1}x+\sqrt{t}+C\\ \\ \Rightarrow y=x\sin ^{-1}x+\sqrt{1-x^{2}}+C\)

This is the required general solution of the given differential equation.

 

 

Q.10. \(e^{x}\tan y\, dx+(1-e^{x})\sec ^{2}y\, dy=0\)

Solution:

\(e^{x}\tan y\, dx+(1-e^{x})\sec ^{2}y\, dy=0\\ \\ (1-e^{x})\sec ^{2}y\, dy=-e^{x}\tan y\, dx\)

Separating the variables, we get:

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

Integrating both sides, we get:

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

Let, tan y = u

\(\frac{d}{dy}{\tan y}=\frac{du}{dy}\\ \\ \Rightarrow \sec ^{2}y=\frac{du}{dy}\\ \\ \Rightarrow \sec ^{2}ydy=du\)

Therefore, \(\\\int \frac{\sec ^{2}y}{\tan y}dy=\int \frac{du}{u}=log\, u=log(\tan y)\)

Now, let 1 – ex = t

Therefore, \(\frac{d}{dx}(1-e^{x})=\frac{dt}{dx}\\ \\ \Rightarrow -e^{x}=\frac{dt}{dx}\\ \\ \Rightarrow -e^{x}dx=dt\\ \\ \Rightarrow \int \frac{-e^{x}}{1-e^{x}}dx=\int \frac{dt}{t}=log\:t=log(1-e^{x})\)

Substituting the values of \(\int \frac{\sec ^{2}y}{\tan y}dy\) and \(\int \frac{-e^{x}}{1-e^{x}}dx\)

\(\Rightarrow log(\tan y)=log(1-e^{x})+log\, C\\ \\ \Rightarrow log(\tan y)=log[C(1-e^{x})]\\ \\ \Rightarrow \tan y=C(1-e^{x})\)

This is the required general solution of the given differential equation.