Posted by **Tania** on Wednesday, February 2, 2011 at 7:47pm.

Solve the following initial value problem explicitly:

y'= ye^-x , y(0)=1

How do you solve this problem? I'm really confused!

- Calculus - IVP -
**MathMate**, Wednesday, February 2, 2011 at 11:32pm
y' = y e^(-x)

Separate variables to get

dy/y = e^(-x)dx

Integrate

ln(y) = -e^(-x) + C'

ln(y) = -e^(-x) + C'

raise to power to base e

e^(ln(y))=e^(-e(-x)+C')

y = Ce^(e^(-x))

(the negative sign and C' are absorbed in the new constant C)

y(0)=1

=>

1 = Ce^(-e^0) = Ce

C=e

=>

y=e^(-e^(-x) +1)

(Note: it's easier if you always post with the same name, Leah, Grace, and Tania is too confusing.)

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