Question

The solution of the differential equation $xdy-ydx=\sqrt{x^2-y^2}$ is?

Answer 1

$xdy-ydx=\sqrt{x^2-y^2}dx$

put $y=vx$ and $dy=vdx+xdv$

$x(vdx+xdv)-vxdx=\sqrt{x^2-(vx{)}^2}dx$

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

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

$xdv=\sqrt{1-v^2}dx$

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

integrate both side

${\sin }^{-1}v=\ln x+C$

put value of v

${\sin }^{-1}\frac{y}{x}=\ln x+C$