#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 11 Maths textbook Solution.

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

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

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

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

Solution: we have,

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

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

It is a homogeneous equation of degree 2.

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

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

\begin{aligned} &\Rightarrow x \frac{d v}{d x}={1+v+v^{2}}-v \\ &\Rightarrow x \frac{d v}{d x}=1+v+v^{2}-v \\ &\Rightarrow x \frac{d v}{d x}=1+v^{2} \\ &\Rightarrow \int \frac{d v}{1+v^{2}}=\int \frac{d x}{x} \end{aligned}

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

This is required solution.