Question

Find the general solution for the following differential equation:
$x^{2}dy+y(x+y)dx=0$

Answer 1

$x^{2}dy=-y(x+y)dx$

$⟹\frac{dy}{dx}=-\frac{y(x+y)}{x^2}$

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

$\frac{dy}{dx}=-\frac{y}{x}-(\frac{y}{x}{)}^2$

The above differential equation is homogeneous.

$\therefore$ put $vx=y........\left(1\right)$

$\frac{dy}{dx}=-v-v^2.....\left(2\right)$

differentiating equation (1),

$\frac{dy}{dx}=v+x\frac{dv}{dx}......\left(3\right)$

by equation (2) & (3),

$v+x\frac{dv}{dx}=-v^2-v$

$x\frac{dv}{dx}=-v^2-2v$

$⟹\frac{1}{v^2+2v}dv=-\frac{1}{x}dx$

integrating both sides,

$\int \frac{1}{v^2+2v}dv=-\int \frac{1}{x}dx$

$\int \frac{1}{(v^2+2v+1)-1}dv=-ln|x|+ln|c|$

$\int \frac{1}{(v+1{)}^2-1^2}dv=-ln|x|+ln|c|$

using $\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}ln|\frac{x-a}{x+a}|+c$

$\frac{1}{2}ln\left|\frac{v+1-1}{v+1+1}\right|=-ln|x|+ln|c|$

$\frac{1}{2}ln\left|\frac{v}{v+2}\right|=-ln|x|+ln|c|$

$ln\left|\frac{\sqrt{v}}{\sqrt{v+2}}\right|=-ln|x|+ln|c|$

$ln\left|\frac{\sqrt{v}}{\sqrt{v+2}}\right|+ln|x|=ln|c|$

$ln\left|\frac{x\sqrt{v}}{\sqrt{v+2}}\right|=ln|c|$

$⟹\frac{x\sqrt{v}}{\sqrt{v+2}}=c$

from equation (1) $v=\frac{y}{x}$,

$\frac{x\sqrt{\frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c$

squaring both sides,

$\frac{xy}{\left(\frac{y+2x}{x}\right)}=c^2$

Therefore,

general solution$:x^{2}y=c^2(y+2x)$