Question

For the differential equation $x\frac{dy}{dx}-y=\sqrt{(x^2+y^2)}$, show that its solution is $y+\sqrt{(x^2+y^2)}=k{x}^2$

Answer 1

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