Question

Solve the homogeneous differential equation: $(x^2+y^2)dx+2xydy=0$.

Answer 1

Differential equation $(x^2+y^2)dx+2xydy=0$.

$2xydy=-(x^2+y^2)dx$

$\frac{dy}{dx}=-\frac{x^2+y^2}{2xy}$ it is homogeneous equation

Let $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=-\frac{(x^2+v^2{x}^2)}{2x\cdot vx}$

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

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

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

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

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

$\frac{2}{6}\int \frac{dt}{t}=-\int \frac{dx}{x}$

$\frac{2}{6}\int \log tdt=-\log x+\log c$ Let $I+3v^2=t$

$\frac{1}{3}\log (1+3v^2)=-\log x+\log c$ $6vdv=dt$

$\frac{1}{3}\log (1+\frac{3y^2}{x^2})=-\log x+\log c$ $vdv=\frac{1}{6}dt$

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

$\log {\left(\frac{x^2+3y^2}{x^2}\right)}^{1/3}=-\log x+\log c$

$\log x+\log {\left(\frac{x^2+3y^2}{x^2}\right)}^{1/3}=\log c$

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

$x{\left(\frac{x^2+3y^2}{x^2}\right)}^{1/3}=c$

Where $c$ is an integration constant.