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Math Help - Integrating factor on Exact, Homogeneous DE

  1. #1
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    Integrating factor on Exact, Homogeneous DE

    Problem: y*(1+x)+2xy'=0

    I have rearranged the equation to put it in exact form, resulting in the equation:
    (y+xy)dx+2xdy=0

    After that I found my \frac{\partial M}{\partial y} and \frac{\partial N}{\partial x}

    This leads to a complicated integrating factor of e^{-\frac{1}{2}(ln|x|-x)}

    Am I on the correct track?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Integrating factor on Exact, Homogeneous DE

    Quote Originally Posted by cheme View Post
    Problem: y*(1+x)+2xy'=0

    I have rearranged the equation to put it in exact form, resulting in the equation:
    (y+xy)dx+2xdy=0

    After that I found my \frac{\partial M}{\partial y} and \frac{\partial N}{\partial x}

    This leads to a complicated integrating factor of e^{-\frac{1}{2}(ln|x|-x)}

    Am I on the correct track?
    Dividing throughout by 2x, we get

    y'+\left(\frac{1+x}{2x}\right)y=0

    The integrating factor is:

    exp\left[\int \left(\frac{1+x}{2x}\right)dx\right]

    =exp\left[\int \left(\frac{1}{2x}+\frac{1}{2}\right) dx\right]

    =exp\left(\frac{1}{2}ln\ x+\frac{x}{2}\right)

    =exp\left(\frac{1}{2}ln\ x\right)*exp\left(\frac{x}{2}\right)

    =\sqrt{x}e^{x/2}

    (Note that this isn't an exact differential equation as \frac{\partial M}{\partial y} \neq\frac{\partial N}{\partial x}.)
    Last edited by alexmahone; September 12th 2011 at 04:54 PM. Reason: Further simplification
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  3. #3
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    Re: Integrating factor on Exact, Homogeneous DE

    I did not see to divide through by 2x. If I do that, isn't it just a separable equation?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Integrating factor on Exact, Homogeneous DE

    Quote Originally Posted by cheme View Post
    I did not see to divide through by 2x. If I do that, isn't it just a separable equation?
    Yes, this is a separable equation.
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