Solve by variable separation methods.

$1\mathrm{.}1\frac{dy}{dx}=(x1{)}^2$

$1\mathrm{.}2x\frac{dy}{dx}=4y$

$1\mathrm{.}3\frac{dy}{dx}=-2x{y}^2$

$1\mathrm{.}4\frac{dy}{dx}=(\frac{2y3}{4x5}{)}^2$

$1\mathrm{.}5,\frac{dy}{dx}2y=0$

$(35$ points$)$