If $\frac{dy}{dx}=4xy$ and if $y=1$ when $x=2$ then $y=$

A$.e^{2x^2-8}$

B$.2c^{2x^2-8}-1$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{f}df=$

C$.4e^{2x^2-8}-3$

D$.8e^{2x^2-8}-7$

$lnf=2x$