Solve this differential equation. +10BA!!!

I have this question, y' - x = xe^y

So far I've integrated both sides and got it down to the form ln|e^y| - ln|e^y+1| = 0.5x^2 + C

Which I simplified to

(e^y)/(e^y+1) = Ce^0.5x^2

I think there must be something wrong.. Can someone show me the steps on how to get there? Thanks!

e^y / (1 + e^y) = Ce^((1/2) * x^2) <==== long division


. . . . . 1
. . . . . _____
e^y + 1I e^y
. . . . .-
. . . . . .e^y + 1
. . . . . _______
. . . . . 0 - 1

1 - 1/e^y

1 - 1/e^y = Ce^((1/2) * x^2)

1 - Ce^((1/2) * x^2) = 1/e^y

e^y = 1/(1 - Ce^((1/2) * x^2))

ln( e^y ) = ln( 1 / (Ce^((1/2) * x^2) )

y = ln( 1 ) - ln(Ce^((1/2) * x^2))

y = 0 - ln(Ce^((1/2) * x^2))

y = - ln(Ce^((1/2) * x^2))

you are fine....need to solve for e^y...let w = e^y and do the algebra , then y = ln w

{ ln | Ce^(x² / 2 ) / [ 1 - C e^(x² / 2 ) ] | }