Question

The solution of differential equation $x\frac{dy}{dx}=y+\sqrt{x^2+y^2}$ is:
(where $C$ is integration constant)

• A. $|x|-\sqrt{x^2+y^2}=C{y}^2$
• B. $|x+\sqrt{x^2+y^2}|=C{y}^2$
• C. $|y+\sqrt{x^2+y^2}|=C{x}^2$
• D. $|y-\sqrt{x^2+y^2}|=C{x}^2$

Answer 1

The correct option is C $|y+\sqrt{x^2+y^2}|=C{x}^2$

$x\frac{dy}{dx}=y+\sqrt{x^2+y^2}\Rightarrow \frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}\cdots \left(i\right)$
This is an homogeneous D.E.

Put $y=vx,$ so that,
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Then, equation $\left(i\right)$ becomes: $v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^2+v^2{x}^2}}{x}\Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}$
$\Rightarrow \int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}$
$\Rightarrow \log |v+\sqrt{1+v^2}|=\log |x|+\log c\Rightarrow |v+\sqrt{1+v^2}|=|cx|$
$\Rightarrow |y+\sqrt{x^2+y^2}|=C{x}^2;$
where $|c|=C$