Post a New Question


posted by .

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
    ln(y) = -e^(-x) + C'
    ln(y) = -e^(-x) + C'
    raise to power to base e
    y = Ce^(e^(-x))
    (the negative sign and C' are absorbed in the new constant C)

    1 = Ce^(-e^0) = Ce
    y=e^(-e^(-x) +1)

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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question