#### Please Solve R.D. Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 8 Maths textbook Solution.

Answer: $\frac{x+\sqrt{2y}}{x-\sqrt{2y}}=\left ( cx^{2} \right )^{\sqrt{2}}$

Given:$x^{2}\frac{dy}{dx}=x^{2}-2y^{2}+xy$

To find: we have to find the solution of given differential equation

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

Solution: we have,

$x^{2}\frac{dy}{dx}=x^{2}-2y^{2}+xy$

$\frac{dy}{dx}=\frac{x^{2}-2y^{2}+xy}{x^{2}}$

This is a homogeneous differential equation.

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

We have,

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

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

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

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

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

$\Rightarrow \frac{\sqrt{2}}{2} \log \left|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\right|=2 \log x+\log c \\$

$\left.\Rightarrow \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right] \\$

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

\begin{aligned} &\Rightarrow \frac{1}{\sqrt{2}} \log \left(\frac{x+y \sqrt{2}}{x-y \sqrt{2}}\right)=\log x^{2}+\log c \\ &\Rightarrow \log \left(\frac{x+y \sqrt{2}}{x-y \sqrt{2}}\right)^{\frac{1}{\sqrt{2}}}=\log c x^{2} \\ &\Rightarrow \frac{x+y \sqrt{2}}{x-y \sqrt{2}}=\left(c x^{2}\right)^{\sqrt{2}} \end{aligned}

Hence this is required solution.