Question

Solve the following differential equation:
$(3xy+y^2)dx+(x^2+xy)dy=0$

Answer 1

$(3xy+y^2)dx+(x^2+xy)dy=0$
$\frac{dy}{dx}=-\left(\frac{3xy+y^2}{x^2+xy}\right)$ (Homogeneous)

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

$\frac{dv}{dx}\cdot x+v=\left(\frac{3x.vx+v^2{x}^2}{x^2+x.vx}\right)$

$\frac{dv}{dx}\cdot x=\frac{-3v-v^2}{1+v}-v$

$x\frac{dv}{dx}=\frac{-3v-v^2-v-v^2}{1+v}$

Solving $x\frac{dv}{dx}=\frac{-2v^2-4v}{1+v}$

$\int \frac{1+v}{2v^2+4v}dv=\int -\frac{1}{x}dx$
or $\frac{1}{4}\int \frac{2+2v}{2v+v^2}dv=-\int \frac{1}{x}dx$

$\frac{1}{4}\log |v^2+2v|=-\log |x|+\log c$

$\frac{1}{4}\log (\frac{y^2}{x^2}+2\frac{y}{x})=\log \frac{c}{x}$

$\log (\frac{y^2}{x^2}+\frac{2y}{x})=4\log \frac{c}{x}$

$\log (\frac{y^2}{x^2}+\frac{2y}{x})=\log \frac{c^4}{x^4}$

$\frac{y^2}{x^2}+\frac{2y}{x}=\frac{c^4}{x^4}$

$x^2{y}^2+2x^{3}y=c^4$

Let $c^4$ be equal to $k$

Then answer is $x^2{y}^2+2x^{3}y=k$