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Math Help - Differntial Equations Problem

  1. #1
    Member eXist's Avatar
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    Differntial Equations Problem

    A problem that my friend gave me. I never took diff-e-q but my guess would be to integrate with respect to x then substitue x(0) to find the constant, but not sure if that's right, doesn't seem like diff-e-q material that way.

    Separate variables and use partial fractions to solve the initial value problem \frac{dt}{dx}=17x(x-15);x(0)=17
    Last edited by eXist; September 24th 2009 at 04:41 PM.
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by eXist View Post
    A problem that my friend gave me. I never took diff-e-q but my guess would to integrate with respect to x then substitue x(0) to find the constant, but not sure if that's right.

    Separate variables and use partial fractions to solve the initial value problem \frac{dt}{dx}=17x(x-15);x(0)=17
    I think you mean ...

    \frac{dx}{dt}=17x(x-15);x(0)=17

    separate variables ...

    \frac{dx}{x(x-15)} = 17 \, dt


    partial fraction decomposition ...

    \frac{1}{x(x-15)} =\frac{A}{x} + \frac{B}{x-15}

    1 = A(x-15) + Bx

    let x = 15 ... B = \frac{1}{15}

    let x = 0 ... A = -\frac{1}{15}

    \frac{1}{x(x-15)} = \frac{1}{15}\left(\frac{1}{x-15} - \frac{1}{x}\right)


    \frac{1}{15}\int \left(\frac{1}{x-15} - \frac{1}{x}\right) \, dx = \int 17 \, dt

    finish from here?
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  3. #3
    Member eXist's Avatar
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    Ya I think I have the rest . Thanks a lot, you're the man skeets.
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  4. #4
    Member garymarkhov's Avatar
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    Quote Originally Posted by skeeter View Post
    partial fraction decomposition ...

    \frac{1}{x(x-15)} =\frac{A}{x} + \frac{B}{x-15}

    1 = A(x-15) + Bx

    let x = 15 ... B = \frac{1}{15}

    let x = 0 ... A = -\frac{1}{15}

    \frac{1}{x(x-15)} = \frac{1}{15}\left(\frac{1}{x-15} - \frac{1}{x}\right)


    \frac{1}{15}\int \left(\frac{1}{x-15} - \frac{1}{x}\right) \, dx = \int 17 \, dt
    Can you explain the partial fraction decomposition steps you've taken here?
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