I am having trouble towards the end of this problem:
dy/dx=(e^(y)sin^(2)x)/ysecx
towards the end i get (y^(2)/2e^(y))=sin^(3)x/3
i am not sure really how to isolate y
Can anyone please help?
your solution is incorrect ...
$\displaystyle \frac{dy}{dx} = \frac{e^y \cdot \sin^2{x}}{y \cdot \sec{x}}$
$\displaystyle \int y \cdot e^{-y} \, dy = \sin^2{x} \cdot \cos{x} \, dx$
using parts on the left side ...
$\displaystyle -(y+1)e^{-y} = \frac{\sin^3{x}}{3} + C$
you're stuck here as far as solving for y is concerned ... unless you are skilled in using the Lambert W function.