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  • Please Solve R.D.Sharma class 12 Chapter 21 Differential Equations Exercise 21.9 Question 2 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 2 Maths textbook Solution.

Answers (1)

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

Given:\frac{dy}{dx}=\frac{y-x}{y+x}

To solve: we have to solve the given differential equation

Hint: In homogeneous differential equation put

y=vx and \frac{dy}{dx}=v+\frac{xdv}{dx}

Solution:    we have

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

Cleary since each of the functions y-x andy+x  is homogeneous function of degree \partial the given equation is homogenous equation.

Putting y=vx and \frac{dy}{dx}=v+\frac{xdv}{dx}

The given equation becomes

\begin{aligned} &v+x \frac{d v}{d x}=\frac{v x-x}{v x+x} \\ &\Rightarrow v+x \frac{ d v}{d x}=\frac{v-1}{v+1} \\ &\Rightarrow v+x \frac{ d v}{d x}=\frac{v-1}{v+1}-v \\ &\Rightarrow x \frac{ d v}{d x}=\frac{v-1-v^{2}-v}{v+1} \\ &\Rightarrow x \frac{ d v}{d x}=\frac{\left(1+v^{2}\right)}{v+1} \end{aligned}

Separating variables, we have

\Rightarrow \frac{v+1}{v^{2}+1}dv=\frac{dx}{x}

Integrating on both side, we get.

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

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

                                        \Rightarrow \frac{1}{2} \log \left|v^{2}+1\right|+\tan ^{-1} v-\log x+\log c\left[\therefore \int \frac{1 d x}{1+x^{2}}\right]=\tan ^{-1} x \\

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

putting v=y/x

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

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

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

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

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

\Rightarrow \log \left(y^{2}+x^{2}\right)+2 \tan ^{-1}\left(\frac{y}{x}\right)=K,            where K=2logc

It is required solution

Posted by

infoexpert21

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