Find the solution to the given differential equation.

$x^2{y}^{\text{'}}-2x^{2}y=0$

$($a$)y=x^2$

$($b$)y=x^2+C$

$($c$)y=e^{x^2}+C$

$($d$)y=C{e}^{x^2}$

$($e$)$ None of these