#### Need Solution for R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 31 Maths Textbook Solution.

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

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

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

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

Solution: we have,

\begin{aligned} &\left(x^{2}+3 x y+y^{2}\right) d x-x^{2} d y=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{x^{2}+3 x y+y^{2}}{x^{2}} \end{aligned}

It is a homogeneous equation Put    $y=vx$  and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

\begin{aligned} &x \frac{d v}{d x}+v=\frac{x^{2}+3 x^{2}+v^{2} x^{2}}{x^{2}} \\ &\Rightarrow x \frac{d v}{d x}+v=\frac{1+3 v+v^{2}}{1}-v \\ &\Rightarrow x \frac{d v}{d x}+v=1+2 v+v^{2} \\ &\Rightarrow x \frac{d v}{d x}+v=(1+v)^{2}\left[\therefore(v+1)^{2}=1+2 v+v^{2}\right] \end{aligned}

Separating the variable and Integrating bot side we get

\begin{aligned} &\Rightarrow \int \frac{1}{(v+1)^{2}} d v=\int \frac{d x}{x} \\ &-\frac{1}{v+1}= \log x+k\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &\Rightarrow \frac{-1}{y / x+1}=\log x+k \\ &\Rightarrow \frac{-x}{x+y}=\log x+k \\ &\Rightarrow \frac{x}{x+y}=\log x=-k \\ &\Rightarrow \frac{x}{x+y}=\log x=c \quad \quad \text { where } c=-\mathrm{k} \end{aligned}

This is a required solution