#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 27 Maths Textbook Solution.

Answer: $\frac{y}{x}+log x =c$

Given:$\left ( x^{2}-2y^{2} \right )dy+\left ( x^{2}-3xy+2y^{2} \right )dx=0$

To Find:  We have to find the solution of given differential equation.

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

Solution: We have,

$\left ( x^{2}-2y^{2} \right )dy+\left ( x^{2}-3xy+2y^{2} \right )dx=0$

$\Rightarrow \frac{dy}{dx}=\frac{x^{2}-3xy+2y^{2}}{2xy-x^{2}}$

It is homogeneous equation.

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

So,$v+x\frac{dv}{dx}=\frac{x^{2}-3xvx+2x^{2}v^{2}}{2vx^{2}-x^{2}}$

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{8-3 v+2 v^{2}}{1+v^{2}}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{1-3 v+2 v^{2}}{2 v-1}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{1-2 v}{2 v-1} \\ &\Rightarrow x \frac{d v}{d x}=-1 \end{aligned}

Separating the variable and integrating both sides, we get,

\begin{aligned} \int d v=-\int \frac{1}{x} d x \\ \end{aligned}

\begin{aligned} &\Rightarrow v=-\log x+c \\ \end{aligned}

\begin{aligned} &\Rightarrow \frac{y}{x}+\log x=c \end{aligned}                                                                                                                                $\left [ \therefore v=\frac{y}{x} \right ]$

Hence this is required solution.