Taylor Series Solution for y'' = e^y at x=0, y(0)=0, y'(0)=-1

[dduardo]

This problem has been bugging me and I can't seem to figure it out:

y'' = e^y where y(0)= 0 and y'(0)= -1

I'm supposed to get the first 6 nonzero terms

I know the form is:

y(x) = y(0) + y'(0)x/1! + y''(0)x^2/2! + y'''(0)x^3/3! +...

and I know the first two terms are

y(x) = 0 - x +...

But what's troubling is the e^y. How would I go about getting the y''(0) term. I tried a subsitution of u = y', but the integrals gets very messy. I'm thinking there is either a typo in the book or there is a simpler way to get the answer.

All I need is an example of how to get the next term. I can figure out the rest.

 

[matt grime]

y''(0) = e^{y(0)} = e^0=1

if you don't mind I'll also say that y''' = (y'')' = d/dx(e^y) = y'e^y

 

[dduardo]

Ah, ok, thanks a lot. I understand now.