Differential equations?

Show that the function f(t)=(e^-1 +1)^-1 satisfies y^1+y^2=y and y(0)=1/2

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  • Ron W
    Lv 7
    9 years ago
    Favourite answer

    You need to be more careful with your typing; it's f(t) = (e^(-t) + 1)^(-1), (not e^(-1)), and the equation is not y^1 + y^2 = y, it's y' + y^2 = 1 (that is, (dy/dt) + y^2 = y).

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

    y' = (-1)(e^(-t) + 1)^(-2)(-e^(-t))

      = e^(-t) (e^(-t) + 1)^(-2)

    y^2 = (e^(-t) + 1)^(-2)

    y' + y^2 = e^(-t) (e^(-t) + 1)^(-2) + (e^(-t) + 1)^(-2)

      = (e^(-t) + 1)^(-2) [e^(-t) + 1]

      = (e^(-t) + 1)^(-1)

      = y

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