# Calculus

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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 -

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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