Question

Solve the differential equation $xdy=ydx+\sqrt{x^2+y^2}dx$

Answer 1

Re-arranging the differential equation gives us
$\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt{y^2+x^2}}{x}$
$\frac{dy}{dx}=\frac{y}{x}+\sqrt{\frac{y^2+x^2}{x^2}}$
Let $y=vx$, then $\frac{dy}{dx}=v+\frac{xdv}{dx}$
Hence
$v+\frac{xdv}{dx}=v+\sqrt{1+v^2}$
$\frac{xdv}{dx}=\sqrt{1+v^2}$
$\int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}$
$ln(v+\sqrt{1+v^2})=lnx+C$
Or
$ln(y+\sqrt{x^2+y^2})-lnx=lnx+c$
$ln(y+\sqrt{x^2+y^2})=2lnx+c$