Find the particular solution of the differential equation $\frac{dy}{dx}= \frac{xy}{x^{2}+y^{2}}.$ given that $y= 1$ when $x= 0$ .

Answers (1)

As the Differential equation can be written as
$\frac{dy}{dx}= \frac{xy}{x^{2}+y^{2}}= \frac{\frac{y}{x}}{1+\frac{y^{2}}{x^{2}}}$ so its homogeneous
put $y= v{x}\Rightarrow \frac{dy}{dx}= v+\frac{xdv}{dx}$
$\therefore v+x\frac{dv}{dx}= \frac{v}{1+v^{2}}$
$\Rightarrow x\, \frac{dv}{dx}= \frac{v}{1+v^{2}}-v= \frac{v-v-v^{3}}{1+v^{2}}$
$\Rightarrow \int \frac{1+v^{2}}{v^{3}}dv= -\int \frac{dx}{x}$
$\Rightarrow \int\left ( \frac{1}{v^{3}} +\frac{1}{v}\right )dv= -\int \frac{dx}{x}$
$\Rightarrow -\frac{1}{2v^{2}}+\log \left | v \right |= -\log \left | x \right |+c$
$\Rightarrow -\frac{x^{2}}{2y^{2}}+\log \left | \frac{y}{x} \right |= -\log \left | x \right |+c$
$\Rightarrow -\frac{x^{2}}{2y^{2}}+\log \left | y \right |= c$
As $y= 1$ when $x= 0,$ so $\frac{-0^{2}}{2\times 12}+\log \left | 1 \right |= c\Rightarrow c= 0$
Here, the required solution is $\frac{-x^{2}}{2y^{2}}+\log \left | y \right |= 0$
or $\log \left | y \right |= \frac{x^{2}}{2y^{2}}$