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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?

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

    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

    $\displaystyle \int(y+1)e^{-y} ~dy = e^t+C$
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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?

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

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

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

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

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

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


    Now use integration by parts:

    Let $\displaystyle u = y + 1$ so that $\displaystyle du = 1$

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


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

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

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

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

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

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

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

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