Calculus
posted by Tania on .
Solve the following initial value problem explicitly:
y'= ye^x , y(0)=1
How do you solve this problem? I'm really confused!

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.)