Differential Equations — Maths STD 12 Science — Question

$\frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}$
Put y = vx
$\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1)becomes, $v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^2+v^2 x^2}}{x}$
$\therefore v+x \frac{d v}{d x}=v+\sqrt{1+v^2}$
$\therefore \frac{1}{\sqrt{1+v^2}} d v=\frac{1}{x} d x$
Integrating, we get,
$\begin{aligned} & \int \frac{1}{\sqrt{1+v^2}} d v=\int \frac{1}{x} d x+c_1 \\ & \therefore \log \left|v+\sqrt{1+v^2}\right|=\log |x|+\log c, \text { where }_1=\log c \\ & \therefore \frac{y+\sqrt{x^2+y^2}}{x}=c x \\ & \left(y+\sqrt{x^2+y^2}\right)=c x^2 \text { is the general solution }\end{aligned}$