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Math Help - differential functions

  1. #1
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    differential functions

    I am trying to do this problem using a somewhat similiar book example...

    dy/dx = (e^y + 1)(x-2)^9
    d/dy (e^y+1) = d/dx (x-2)^9
    integral d/dy (e^y +1) = integral d/dx (x-2)^9
    ln[y+1]^10 = 9(x-2)^8
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    I am trying to do this problem using a somewhat similiar book example...

    dy/dx = (e^y + 1)(x-2)^9
    d/dy (e^y+1) = d/dx (x-2)^9
    integral d/dy (e^y +1) = integral d/dx (x-2)^9
    ln[y+1]^10 = 9(x-2)^8
    y' = (e^y + 1)(x - 2)^9 .............divide both sides by e^y + 1

    \Rightarrow \frac {y'}{e^y + 1} = (x - 2)^9 ........now split up the y' and set up the integration

    \Rightarrow \int \frac {1}{e^y + 1}dy = \int (x - 2)^9 dx ........as it stands, the left hand side is hard to integrate

    \Rightarrow \int \frac {e^y + 1 - e^y}{e^y + 1} dy = \int (x - 2)^9dx ......add e^y - e^y (which is zero so we're not changing anything) to the numerator on the left.

    \Rightarrow \int 1dy - \int \frac {e^y}{e^y + 1}dy = \int (x - 2)^9dx ........split up the integral on the left

    \Rightarrow y - \ln \left|e^y + 1 \right| = \frac {1}{10}(x - 2)^{10} + C .....integrated, using substituion on the second integral on the left

    if you have any questions say so
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  3. #3
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    Quote Originally Posted by startingover View Post
    I am trying to do this problem using a somewhat similiar book example...

    dy/dx = (e^y + 1)(x-2)^9
    We have,
    y' = (e^y+1)(x-2)^9
    Divide,
    \frac{y'}{e^y+1} = (x - 2)^9
    Integrate,
    \int \frac{y'}{e^y+1} dx = \int (x -2)^9 dx
    Substitution rule,
    \int \frac{1}{e^y+1} dy = \int (x-2)^9 dx
    Now you need to find,
    \int \frac{1}{e^y+1} dy = \int \frac{e^y}{e^y(e^y+1)}dy
    Let t=e^y \mbox{ with }t'=e^y thus,
    \int \frac{1}{t(t+1)} dt = \int \frac{1}{t} - \frac{1}{t+1} dt = \ln |t| - \ln |t+1| +C= \ln \left| \frac{t}{t+1} \right| +C = \ln \frac{e^y}{e^y+1} +C =  - \ln (1+e^{-y})+C
    Thus, we find that,
    - \ln (1+e^{-y}) = \frac{1}{10}(x-2)^{10} +C
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