Initial conditions: y(0) = 0, y'(0) = 1
I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).
I have come to the end of my problem set and the only problem remaining is the one in the thread title.
I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.
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?
Thanks to both of you. For some reason I couldn't see that I should have found my c2 PRIOR to taking ln of both sides. If I take my solution now and stuff it into the ode it yields the desired result, with some very light algebraic manipulation. Solved.