# Thread: solving diffyQ for calc2

1. ## solving diffyQ for calc2

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

2. Originally Posted by RhotelidlewildR
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

You can't! If you write it as $\displaystyle y^2 e^{-y}$ you can integrate by parts.

3. my only problem with that is that i already integrated. That is how i ended up in that dilemma

4. Originally Posted by RhotelidlewildR
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

$\displaystyle \frac{dy}{dx} = \frac{e^y \cdot \sin^2{x}}{y \cdot \sec{x}}$
$\displaystyle \int y \cdot e^{-y} \, dy = \sin^2{x} \cdot \cos{x} \, dx$
$\displaystyle -(y+1)e^{-y} = \frac{\sin^3{x}}{3} + C$