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  • Need Solution for R.D .Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 32 Maths Textbook Solution.

Need Solution for R.D .Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 32 Maths Textbook Solution.

Answers (1)

Answer: \log \left|x^{2}+x y+y^{2}\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3 x}}\right)+c

Given:\left ( x-y \right )\frac{dy}{dx}=x+2y

To solve: have to solve the given differential equation.

Hint: Put y=vx and \frac{dy}{dx}=v+x\frac{dv}{dx}

Solution: Here,

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

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

It is homogeneous equation

Put y=vx\Rightarrow\frac{dy}{dx}=v+x\frac{dv}{dx}

So, v+x \frac{d y}{d x}=\frac{x+2 v x}{x-v x} \\

\Rightarrow x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v \\

\Rightarrow x \frac{d v}{d x}=\frac{1+2 v-v+v^{2}}{1-v} \\

\Rightarrow x \frac{d v}{d x}=\frac{1+v+v^{2}}{1-v}

Separating the variable and Integrating both side we get

\int \frac{1-v}{1+v+v^{2}} d v=\int \frac{d x}{x} \\

\Rightarrow \frac{1}{2} \int \frac{2 v-2}{1+v+v^{2}}=-\int \frac{1}{x} d x \\

\Rightarrow \int \frac{(2 v+1)-3}{1+v+v^{2}} d v=-2 \int \frac{1}{x} d x \\

\Rightarrow \int \frac{(2 v+1)-3}{1+v+v^{2}} d v=-\int \frac{3}{v^{2}+2 v\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x=-2 \int \frac{1}{x} d x \\

\Rightarrow \log \left|1+v+v^{2}\right|-3 x \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{v+\frac{1}{2}}{\sqrt{3 / 2}}\right)=-2 \log x+c \\

\Rightarrow \log \left|1+v+v^{2}\right|+\log x^{2}-2 \sqrt{3} \tan ^{-1}\left(\frac{v+\frac{1}{2}}{\sqrt{3 / 2}}\right)=c . \\

\Rightarrow \log \left[x^{2}\left(1+\frac{y}{x}+{ }^{y^{2}} /_{x^{2}}\right]-2 \sqrt{3} \tan ^{-1}\left(\frac{\frac{2 y}{x}+1}{\sqrt{3}}\right)=c . \quad\left[\therefore v=\frac{y}{x}\right]\right.

\Rightarrow \log \left(y^{2}+x y+x^{2}\right)=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3 x}}\right)+c

This is required solution.

 

Posted by

infoexpert21

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