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Thread: Separation of variables problem, please check.

  1. #1
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    Separation of variables problem, please check.

    Question: $\displaystyle \frac{du}{dt} = e^{4u+3t}$, $\displaystyle u(0) = 10$

    My answer: $\displaystyle u = \frac{ln(-\frac{4e^{3t}}{3}+e^{-40}+\frac{4}{3})}{-4}$

    I want to believe this is right, but when I submit the answer it turns out wrong...

    Did I do something wrong here?

    Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Question: $\displaystyle \frac{du}{dt} = e^{4u+3t}$, $\displaystyle u(0) = 10$

    My answer: $\displaystyle u = \frac{ln(-\frac{4e^{3t}}{3}+e^{-40}+\frac{4}{3})}{-4}$

    I want to believe this is right, but when I submit the answer it turns out wrong...

    Did I do something wrong here?

    Thanks.
    $\displaystyle \frac{du}{dt} = e^{4u+3t} = e^{4u}e^{3t}$, $\displaystyle u(0) = 10$

    $\displaystyle e^{-4u}du = e^{3t}dt$

    $\displaystyle \int e^{-4u}du = \int e^{3t}dt$

    $\displaystyle -\frac{1}{4}e^{-4u} = \frac{1}{3}e^{3t} + C$

    We have $\displaystyle u(0) = 10$, so
    $\displaystyle -\frac{1}{4}e^{-4 \cdot 10} = \frac{1}{3}e^{3 \cdot 0} + C$

    $\displaystyle C = -\frac{1}{4}e^{-40} - \frac{1}{3}$.

    Thus
    $\displaystyle -\frac{1}{4}e^{-4u} = \frac{1}{3}e^{3t} - \frac{1}{4}e^{-40} - \frac{1}{3}$

    $\displaystyle e^{-4u} = -\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3}$

    $\displaystyle -4u = ln \left ( -\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3} \right )$

    $\displaystyle u = -\frac{1}{4} \cdot ln \left ( -\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3} \right )$

    It looks good to me.

    Perhaps it's looking for a factored version of the argument of the logarithm?
    $\displaystyle u = -\frac{1}{4} \cdot ln \left ( \frac{4}{3} \left [1 - e^{3t} \right ] + e^{-40} \right )$

    or possibly

    $\displaystyle u = ln \left ( \frac{1}{\sqrt[4]{-\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3}}} \right )$

    or something.

    -Dan
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