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Math Help - Separation of variables problem, please check.

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

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

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

    My answer: 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.
    \frac{du}{dt} = e^{4u+3t} = e^{4u}e^{3t}, u(0) = 10

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

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

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

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

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

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

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

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

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

    or possibly

    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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