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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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