Inactive Tutor answered 07/06/13
Tutor
New to Wyzant
dx/dy + 2x - (1/y)e^(-2y) = 0
dx/dy + 2x = (1/y)e^(-2y)
x' + 2x = (1/y)e^(-2y)
I.F. = e^2y
x = (e^-2y) ∫ dy/y
x = (e^-2y) [ln y + c]
The trick is to integrate wrt y rather than x