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Math Help - Differential Equations Intergating Factors

  1. #1
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    Differential Equations Intergating Factors

    Hi

    Having trouble with the following problem:

    Solve: \frac{dy}{dx} - \frac{4y}{x} = x^4
    \frac{dy}{dx} - 4y = x^5

    e^{\int p(x) dx}

    e^{\int-4y dx}

    e^{-2y^2}

    e^{-2y^2}(\frac{dy}{dx}-4y)=e^{-2y^2}x^5

    e^{-2y^2}y=e^{-2y^2}x^5

    e^{-2y^2}y=\int e^{-2y^2}x^5

    udv = uv - \int vdu

    u=e^{-2y^2} du=-4ye^{-2y^2}

    dv=x^5 v=\frac{x^6}{6}

    e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}

    e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} *  -4ye^{-2y^2}}

    i continued doing the integration and i got nowhere, where is my mistake??

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi

    Having trouble with the following problem:

    Solve: \frac{dy}{dx} - \frac{4y}{x} = x^4
    \frac{dy}{dx} - 4y = x^5

    e^{\int p(x) dx}

    e^{\int-4y dx}

    e^{-2y^2}

    e^{-2y^2}(\frac{dy}{dx}-4y)=e^{-2y^2}x^5

    e^{-2y^2}y=e^{-2y^2}x^5

    e^{-2y^2}y=\int e^{-2y^2}x^5

    udv = uv - \int vdu

    u=e^{-2y^2} du=-4ye^{-2y^2}

    dv=x^5 v=\frac{x^6}{6}

    e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}

    e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}

    i continued doing the integration and i got nowhere, where is my mistake??

    P.S
    p(x) = -4/x NOT -4y.
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  3. #3
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    Also, you can't multiply both sides by x without an x ending up attached to \displaystyle \frac{dy}{dx}.
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  4. #4
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    ok i have tried it again my answer does not match the book's answers.

    \frac{dy}{dx} - \frac{4y}{x} = x^4

    e^\int{ln(x^4)} = x^4

    x^4(\frac{dy}{dx} - \frac{4y}{x}) = x^8

    (\frac{d(x^4y)}{dx} = x^8

    x^4y = \int x^8

    x^4y = \frac{x^9}{9} + C

    y = \frac{x^5}{9} + \frac{C}{x^4}
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    ok i have tried it again my answer does not match the book's answers.

    \frac{dy}{dx} - \frac{4y}{x} = x^4

    e^{\int \ln(x^4) \, dx} = x^4 Mr F says: I have made some formatting and notational changes to this line without changing its content. My request to you is this: Please explain where it has come from, noting that, as has previously been pointed out to you, p(x) = -4/x.

    x^4(\frac{dy}{dx} - \frac{4y}{x}) = x^8

    (\frac{d(x^4y)}{dx} = x^8

    x^4y = \int x^8

    x^4y = \frac{x^9}{9} + C

    y = \frac{x^5}{9} + \frac{C}{x^4}
    ..
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  6. #6
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    yep i know where i mistake, was i got the right answer now
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  7. #7
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    As already pointed out you need

    \displaystyle e^{\int \frac{-4}{x}~dx} = e^{-4\ln x} = e^{\ln x^{-4}} = x^{-4}

    Now multiply this guy through the entire equation.
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