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  1. #1
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    Constant



    I need to solve for all constant solutions for the equation above. How do I go about doing this?

    \int(y+1)/e^y dy = \int(e^t)dt

    Am I even going in the right direction? Thanks
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    Quote Originally Posted by cdlegendary View Post

    Am I even going in the right direction? Thanks



    Now integrate LHS by parts

    \int(y+1)e^{-y} ~dy = e^t+C
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  3. #3
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    Quote Originally Posted by cdlegendary View Post


    I need to solve for all constant solutions for the equation above. How do I go about doing this?

    \int(y+1)/e^y dy = \int(e^t)dt

    Am I even going in the right direction? Thanks
    \frac{dy}{dt} = \frac{e^{t + y}}{y + 1}

    (y + 1)\,\frac{dy}{dt} = e^te^y

    e^{-y}(y + 1)\,\frac{dy}{dt} = e^t

    \int{e^{-y}(y + 1)\,\frac{dy}{dt}\,dt} = \int{e^{-t}\,dt}

    \int{e^{-y}(y + 1)\,dy} = -e^{-t} + C_1.


    Now use integration by parts:

    Let u = y + 1 so that du = 1

    Let dv = e^{-y} so that v = -e^{-y}.


    -e^{-y}(y + 1) - \int{-1e^{-y}\,dy} = -e^{-t} + C_1

    -e^{-y}(y + 1) + \int{e^{-y}\,dy} = -e^{-t} + C_1

    -e^{-y}(y + 1) - e^{-y} + C_2 = -e^{-t} + C_1

    -e^{-y}(y + 2) + C_2 = -e^{-t} + C_1

    e^{-y}(y + 2) - C_2 = e^{-t} - C_1

    e^{-y}(y + 2) + C = e^{-t} where C = C_1 - C_2

    -t = \ln{\left[e^{-y}(y + 2) + C\right]}

    t = -\ln{\left[e^{-y}(y + 2) + C\right]}.
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