What $is$ the solution for the following initial value problem?

$x^3\frac{dy}{dx}+3y{x}^2=cos(x),y()=0$

$y=\frac{sin(x)}{x^3}+C$

$y=sin(x)+C$

$y=\frac{sin(x)}{x}$

$y=\frac{cos(x)}{x^3}$

$y=\frac{sin(x)}{x^3}$