Question

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

Answer 1

$2xydy=(x^2+y^2)dx$
$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$
$\frac{2.dy}{dx}=\frac{x}{y}+\frac{y}{x}$.
Let
$y=vx$ then $\frac{dy}{dx}=v+\frac{xdv}{dx}$
Hence
$2(v+\frac{xdv}{dx})=\frac{1}{v}+v$
$2v+\frac{2xdv}{dx}=\frac{1}{v}+v$
$\frac{2xdv}{dx}=\frac{1}{v}-v$
$\frac{2xdv}{dx}=\frac{1-v^2}{v}$
$\frac{2v}{1-v^2}dv=\frac{dx}{x}$
$\int \frac{2v}{1-v^2}dv=\int \frac{dx}{x}$
$\int \frac{2v}{1-v^2}dv=ln\left(x\right)+c$
Let
$1-v^2=t$
$-2v.dv=dt$
Hence
$I=\int \frac{2v}{1-v^2}dv=-\int \frac{dt}{t}$
$=-lnt$
$=-ln(1-v^2)$.
Hence the differential equation becomes
$-ln(1-v^2)=lnx+C$
Or
$lnx+ln(1-v^2)+C=0$
$ln(x-x{v}^2)+C=0$
$ln(x-x(\frac{y^2}{x^2})+C=0$
$ln(x-\frac{y^2}{x})+C=0$
$ln\left(\frac{x^2-y^2}{x}\right)+C=0$