Question

Solve the differential equation $x\frac{dy}{dx}=y+2\sqrt{(y^2-x^2)}$. If its solution is $y+\sqrt{(y^2-x^2)}=k{x}^C$, here k is the constant of integration, find value of C

Answer 1

$x\frac{dy}{dx}=y+2\sqrt{y^2-x^2}$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore x(x\frac{dv}{dx}+v)=2\sqrt{-x^2+x^2{v}^2}+xv$
$\Rightarrow \frac{dv}{dx}=\frac{2\sqrt{v^2-1}}{x}\Rightarrow \frac{1}{\sqrt{v^2-1}}\frac{dv}{dx}=\frac{2}{x}$
Integrating both sides w.r.t $x$ we get
$\int \frac{1}{\sqrt{v^2-1}}\frac{dv}{dx}dx=\int \frac{2}{x}dx\Rightarrow \log (\sqrt{v^2-1}+v)+2\log x+c\Rightarrow y+\sqrt{y^2-x^2}=k{x}^3$

$\therefore c=3$