Originally Posted by

**LucasV** Alright, so I let u = y', which makes y" = u * du/dy....

which changes the eq to:

u * du/dy = u * e^y (u's cancel)

I apply the method of separable variables and this gives:

u = e^y + c

when x=0, y(0)=0, y'(0)=1

therefore 1 = e^0 + c which yields c = 0

I then change u back to dy/dx and the eq becomes:

dy/dx = e^y

by performing separable variables again, this yields

-1/(e^y) = x, solving for y gives -> **y = ln(-1/x) + c**

this seems a bit off to me because solving for c requires taking the natural log of infinity which doesn't help.

have I made a mistake somewhere?