#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 15 Maths Textbook Solution.

Answer: $\left ( x+y \right )\left ( 2y-x \right )^{2}=C.$

Given: Here,$\frac{dv}{dx}=\frac{x}{2y+x}$

To solve: we have to solve given differential equation

Hint: put   $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$in homogeneous differential equation

Solution: we have,

$\frac{dv}{dx}=\frac{x}{2y+x}$

It is homogeneous equation.

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

So,$v+x\frac{dv}{dx}=\frac{x}{2vx+x}$

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{1}{2 v+1}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{1-2 v^{2}-v}{2 v+1} \\ &\Rightarrow \int \frac{2 v+1 }{2 v^{2}+v-1}dv=-\int \frac{d x}{ x} \\ &\int \frac{2 v+1}{(2 v-1)+(v+1)} d v=-\log x+\log c \text { (1) } \end{aligned}

Using partial fraction

$\frac{2 v+1}{(2 v-1)+(v+1)}=\frac{A}{(2 v-1)}+\frac{B}{v+1} \\$

$2 v+1=A(v+1)+B(2 v-1) \\$

$\text { Put } v=\frac{1}{2} \\$

$\Rightarrow 2 \times \frac{1}{2}+1=A\left|\frac{1}{2}+1\right|+0 \\$

$\Rightarrow A=4/3$

Again Put $v=-1$

$2\left ( -1 \right )+1=0+3\left ( 2x-1-1 \right )$

$-2+1=B\left ( -3 \right )\Rightarrow B=\frac{1}{3}$

Putting the value of A and B in equation

$\Rightarrow \frac{4}{3}\int \frac{d v}{2 v-1}+\frac{1}{3} \int \frac{d v}{v+1}=-\log x+\log c \\$

$\Rightarrow \frac{4 \log |2 v-1|}{2}+\log |v+1|=\log c^{3}-\log x^{3}$

$\Rightarrow \log \left|(2 v-1)^{2}\right|+\log |v+1|=\log C-\log x^{3} \\$

$\Rightarrow \log \left|(2 v-1)^{2}(v+1)\right|=\left.\log \right|\frac{C}{x^{3}} \mid \\$

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

$\Rightarrow\left(\frac{2 y}{x}-1\right)^{2}\left(\frac{2 y}{x}+1\right)=\frac{C}{x^{3}}\left[\therefore v={ }^{y} / x\right] \\$

$\Rightarrow\left(\frac{2 y-x}{x^{2}}\right)^{2}\left(\frac{y+x}{x}\right)=\frac{C}{x^{3}} \\$

$\Rightarrow(2 y-x)^{2}(x+y)=C$

Hence this is required solution.