Explain Solution R.D. Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 39 Maths Textbook Solution.

Answer: $\frac{x^{2}}{2y^{2}}=log\mid y\mid$

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

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,

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

$\frac{dy}{dx}=\left ( \frac{1}{\frac{x}{y}+\frac{y}{x}} \right )$            ...(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}=\left(\frac{1}{\frac{1}{v}+v}\right) \\ &\Rightarrow x \frac{d v}{d x}=\frac{x}{1+v^{2}}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{v-v-v^{3}}{1+v^{2}} \\ &\Rightarrow x \frac{d v}{d x}=\frac{-v^{3}}{1+v^{2}} \end{aligned}

Separating the variables and integrating both side we get

\begin{aligned} &-\left(\frac{1+v^{2}}{v^{3}}\right) d v=\frac{d x}{x} \\ &\Rightarrow \int\left(-\frac{1}{v^{3}}-\frac{1}{v}\right) d v=\int \frac{d x}{x} \\ &\Rightarrow \frac{1}{2 v^{2}}-\log |v|=\log |x|+c \end{aligned}

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

\begin{aligned} &\Rightarrow \frac{x^{2}}{2 y^{2}}-\log \left|\frac{y}{x}\right|=\log |x|+c\\ &\Rightarrow \frac{x^{2}}{2 y^{2}}=\log |x|+\log \left|\frac{y}{x}\right|+c\\ &\Rightarrow \frac{x^{2}}{2 y^{2}}=\log |y|+c \end{aligned}            ...(ii)

It is given that y=1 when x=0

Putting y=1, x=0 in equation (ii) we get

\begin{aligned} &\Rightarrow 0=\log (1)+c \\ &\Rightarrow c=0 \\ \end{aligned}

Putting value of c in equation (ii) we get

\begin{aligned} &\Rightarrow \frac{x^{2}}{2 y^{2}}=\log |y| \end{aligned}

This is required solution.