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Thread: Integrating factor on Exact, Homogeneous DE

  1. #1
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    Integrating factor on Exact, Homogeneous DE

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

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

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

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

    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: $\displaystyle y*(1+x)+2xy'=0$

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

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

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

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

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

    The integrating factor is:

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

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

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

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

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

    (Note that this isn't an exact differential equation as $\displaystyle \frac{\partial M}{\partial y} \neq\frac{\partial N}{\partial x}$.)
    Last edited by alexmahone; Sep 12th 2011 at 03:54 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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