Question

Solve the differential equation:
$x^2\frac{dy}{dx}=x^2-2y^2+xy$.

Answer 1

Given,

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

$\frac{dy}{dx}=1-2\frac{y^2}{x^2}+\frac{y}{x}$

substitute $y=vx\to \frac{dy}{dx}=x\frac{dv}{dx}+v$

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

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

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

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

integrating on both sides, we get,

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

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

$\frac{1}{2\sqrt{2}}\log \left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right|=\log x+c$