#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 4 Maths Textbook Solution.

Answer: $y=\frac{x}{1+log|x|}$

Given:$\left ( xy-y^{2} \right )dx=x^{2}dy,y=\left ( 1 \right )=!$

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

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

Solution: we have,

$\left ( xy-y^{2} \right )dx=x^{2}dy,y=\left ( 1 \right )=!$

$\frac{dy}{dx}=\frac{\left ( xy-y^{2} \right )}{x^{2}}$                    ....(i)

It is homogeneous equation.

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

So,

\begin{aligned} &v+x \frac{d v}{d x}=\frac{v x}{x}+\cos e c \frac{v x}{x} \\ &\Rightarrow v+x \frac{d v}{d x}=\frac{x^{2}\left(v-v^{2}\right)}{x^{2}} \\ &\Rightarrow x \frac{d v}{d x}=v-v^{2}-v \\ &\Rightarrow x \frac{d v}{d x}=-v^{2} \end{aligned}

Separating the variables and integrating both side we get

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

Putting $v=\frac{y}{x}$

$\Rightarrow \frac{x}{y}=log|x|+c$            ...(ii)

It is given that y=1 when x=!

Putting y=1, x=!  in equation (ii) we get

$\Rightarrow 1=log|x|+c$

$\Rightarrow c+!$

Putting value of c in equation (ii) we get

\begin{aligned} &\Rightarrow \frac{x}{y}=\log |x|+c \\ &y=\frac{x}{1+\log |x|} \end{aligned}

This is required solution.