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Math Help - Problem with Integrating Factor

  1. #1
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    Problem with Integrating Factor

    I'm trying to solve the differential equation dy/dx + (2-3x^2)(x^-3)y = 1. I've integrated (2-3x^2)(x^-3) and taken the exponential of this to give an integrating factor of (x^-3)e^(x^2). However, when I differentiate the integrating factor multiplied by y, I don't end up with the left hand side of the equation. I must have gone wrong somewhere in my working, however, I've checked it several times!
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  2. #2
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    Quote Originally Posted by StaryNight View Post
    I'm trying to solve the differential equation dy/dx + (2-3x^2)(x^-3)y = 1. I've integrated (2-3x^2)(x^-3) and taken the exponential of this to give an integrating factor of (x^-3)e^(x^2). However, when I differentiate the integrating factor multiplied by y, I don't end up with the left hand side of the equation. I must have gone wrong somewhere in my working, however, I've checked it several times!
    I think you'll find that in the integrating factor it's e^{-\frac{1}{x^2} not e^{x^2}. Note: -\frac{1}{x^2} \neq x^2 .... it seems you're confusing -\frac{1}{x^2} with \frac{1}{x^{-2}} ....
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by StaryNight View Post
    I'm trying to solve the differential equation dy/dx + (2-3x^2)(x^-3)y = 1. I've integrated (2-3x^2)(x^-3) and taken the exponential of this to give an integrating factor of (x^-3)e^(x^2). However, when I differentiate the integrating factor multiplied by y, I don't end up with the left hand side of the equation. I must have gone wrong somewhere in my working, however, I've checked it several times!
    \displaystyle \int \frac{2-3x^2}{x^3}=-\frac{1}{x^2}-3lnx+C

    Integrating Factor = \displaystyle e^{-\frac{1}{x^2}-3lnx}=e^{-\frac{1}{x^2}}e^{-3lnx}=x^{-3}e^{-\frac{1}{x^2}}
    Last edited by alexmahone; January 24th 2011 at 03:26 AM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    We can rewrite the equation in the form:

    Pdx+Qdy=0\;,\quad P=(1-3x^2)y\;,\;Q=x^3

    and:


    \dfrac{1}{Q}\left(\dfrac{{\partial P}}{{\partial y}}-\dfrac{{\partial Q}}{{\partial x}}\right)=\ldots=\dfrac{1}{x^3}-\dfrac{6}{x}=F(x)

    so, an integrating factor is:


    \mu(x)=e^{\int F(x)\;dx}



    Fernando Revilla

    Edited: Sorry, I misread the OP's equation.
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