Question

Solve the differential equation: $\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$

Answer 1

$\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$ ...(i)
Putting $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Also $v=\frac{y}{x}$
$\therefore$ Equation (i) becomes
$v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^2+v^2{x}^2}}{x}$
$\therefore v+x\frac{dv}{dx}=v+\sqrt{1+v^2}$
$\Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}$
$\Rightarrow \frac{1}{\sqrt{1+v^2}}dx=\frac{1}{x}dx$
On integration, we get
$\int \frac{1}{\sqrt{1+v^2}}dv=\int \frac{1}{x}dx$
$\Rightarrow \log |v+\sqrt{1+v^2}|=\log \left|x\right|+\log c$
$\Rightarrow \log |\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}|=\log \left|cx\right|$
$\Rightarrow \frac{y+\sqrt{x^2+y^2}}{x}=cx$
$\Rightarrow y+\sqrt{x^2+y^2}=c{x}^2$
This is the general solution.