Question

Solve: $2xydx+(x^2+2y^2)dy=0$.

Answer 1

Given,

$2xydx+(x^2+2y^2)dy=0$

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

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

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

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

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

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

integrating on both sides, we get,

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

substitute $3v+2v^3=t\to (3+6v^2)dv=dt$

$\frac{1}{3}\log (3v+2v^3)=-\log x+\log c$

$\log (3v+2v^3{)}^{\frac{1}{3}}=-\log x+\log c$

$\log (3v+2v^3{)}^{\frac{1}{3}}+\log x=\log c$

$\log \left[x\right(3v+2v^3{)}^{\frac{1}{3}}]=\log c$

$x(3v+2v^3{)}^{\frac{1}{3}}=c$

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