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Thread: Differential Equations Intergating Factors

  1. #1
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    Differential Equations Intergating Factors

    Hi

    Having trouble with the following problem:

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

    $\displaystyle e^{\int p(x) dx}$

    $\displaystyle e^{\int-4y dx}$

    $\displaystyle e^{-2y^2}$

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

    $\displaystyle e^{-2y^2}y=e^{-2y^2}x^5$

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

    $\displaystyle udv = uv - \int vdu$

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

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

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

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

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

    $\displaystyle e^{\int p(x) dx}$

    $\displaystyle e^{\int-4y dx}$

    $\displaystyle e^{-2y^2}$

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

    $\displaystyle e^{-2y^2}y=e^{-2y^2}x^5$

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

    $\displaystyle udv = uv - \int vdu$

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

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

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

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

    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 $\displaystyle x$ without an $\displaystyle x$ ending up attached to $\displaystyle \displaystyle \frac{dy}{dx}$.
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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.

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

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

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

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

    $\displaystyle x^4y = \int x^8$

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

    $\displaystyle y = \frac{x^5}{9} + \frac{C}{x^4}$
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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.

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

    $\displaystyle e^{\int \ln(x^4) \, dx} = x^4$ 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.

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

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

    $\displaystyle x^4y = \int x^8$

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

    $\displaystyle y = \frac{x^5}{9} + \frac{C}{x^4}$
    ..
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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 \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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