#### Need Solution for R.D .Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 35 Maths Textbook Solution.

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

Given:$ydx+\left \{ xlog\frac{y}{x} \right \}dy-2xdy=0$

To solve: we have to solve differential equation

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

Solution: we have,

$ydx+\left \{ xlog\frac{y}{x} \right \}dy-2xdy=0$

Divided by dx both side we get

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

It is homogeneous equation

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

So,  $x \frac{d y}{d x}+v=\frac{v x}{2 x-x \log \left(\frac{y}{x}\right)} \\$

$\Rightarrow x \frac{d y}{d x}=\frac{v x}{2 x-\log v}-v \\$

$\Rightarrow x \frac{d y}{d x}=\frac{v x}{2 x-\log v}-v \\$

$\Rightarrow x \frac{d y}{d x}=\frac{v-2 v+2 v x \log v}{2-\log v}$

Separating the variables and integrating both side we get

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

Let $t=logv-1$

$\Rightarrow dt=\frac{1}{v}-1$

Ow equation becomes

$\Rightarrow \frac{d t}{t}-\log v=\log x+\log c\\$

$\Rightarrow \log t-\log v=\log x+\log c\\$

$\Rightarrow \log |\log v-1|-\log v=\log x+\log c \quad[\therefore t=\log v-1]\\$

$\Rightarrow \log (\log v-1)=\log x+\log v+\log c\\$

$\Rightarrow \log (\log v-1)=\log c v x \quad \text { (using logrithm property) }\\$

$\Rightarrow \log \left(\log \frac{y}{x}-1\right)=\log c x x y^{y} / x\left [ \therefore v=\frac{y}{x} \right ]$

$\Rightarrow \log \frac{y}{x}-1=c y\\$

$\Rightarrow c y=\log \left|\frac{y}{x}\right|-1$

This is required solution.