#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 6 Maths textbook Solution.

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

Given:

Hint: in homogeneous differential equation put $y=vx$

To solve: we have to solve the given differential Eqn.

Solution: we have

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

Here it is a homogeneous equation.

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

So,$v+x\frac{dv}{dx}=\frac{1+v}{1-v}$

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

\begin{aligned} &\Rightarrow \frac{1-v}{1+v^{2}} d v=\frac{d v}{d x} \\ &\Rightarrow \int \frac{1-v}{1+v^{2}} d v=\int \frac{d x}{x} \end{aligned}                                                                            $[ Integrating \: on \: both \: side]$

\begin{aligned} &\Rightarrow \int \frac{1}{1+v^{2}} d v-\frac{1}{2} \int \frac{2 v}{1+v^{2}} d v=\int \frac{d x}{x} \\ &\Rightarrow \tan ^{-1} v-\frac{1}{2} \log \left(1+v^{2}\right)= \log x+c .\left[\therefore \int \frac{d x}{1+x^{2}}=\tan ^{-1} x\right] \\ &\Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(1+^{y^{2}} /_{x^{2}}\right)=\log x+c .[\therefore v=y / x] \\ &\Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)=\log x+c \\ &\Rightarrow \tan ^{-1} \frac{y}{x}=\frac{1}{2} \log \left(x^{2}+y^{2}\right)+c . \end{aligned}

This is required solution.