How do I solve for y

I got my differential equation down to this and now only need to solve for y, but how do I do it o.O (e^y)/(e^y+1) = Ce^(0.5x^2), where C is a constant.

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 can use the natural log (ln)...

Take the ln of both sides and use the power rule for natural logs.

For example, ln(e^(y+1))=y+1 since you can rewrite ln(e^(y+1)) as ln(e)^(y+1) and then bring the y+1 in front and cancel ln(e) because ln(e)=1.