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Integrating factor strategy

  1. Apr 11, 2005 #1
    Q. Motivate the Integrating factor strategy for U ( "Mew" ) of y


    I know how to prove it for "Mew" of x but how to do for "mew" of y


    Maybe something like this.

    Mdx (x.y) + Ndy ( x, y ) = 0

    Assume this is differentiable so let us multiply by "mew" of x on both sides to make it exact.

    Then M ( tilda ) the left term and N ( tilda ) equal to the right term

    Then may be, Find partial with respect to x in the M terms. and partial with respect to y in the N terms. Is this idea/approach correct.

    Thanks, for your help
     
  2. jcsd
  3. Apr 12, 2005 #2

    saltydog

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    Naeem, no offense but this is not clear and the notations is awkward. Perhaps if you specify a specific problem we can help you.
     
  4. Apr 12, 2005 #3
    Well, I need to come up with the following final formula used for finding the Integrating factor, for a linear differential equation.


    e ^ Integral Nx-My / M = Greek Letter U ( Mew) (y)


    This is the formula used to find the Integrating factor, with respect to y in Linear differential equation.
     
  5. Apr 12, 2005 #4

    Hurkyl

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    Just using 'u' is acceptable. However, you can create a Greek mu (that's how it's spelled) with the following incantation, if you remove the spaces:

    & m u ;

    That will be turned into the symbol μ. (Yes, I know the default font doesn't render it very well. :frown: If you want, you might write your post in the Times New Roman font -- it does Greek characters well)
     
  6. Apr 13, 2005 #5

    Andrew Mason

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    I am not sure what you are asking. Perhaps this will help:

    For a first order differential equation put into the form:

    [tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]

    you want to find a function [itex]\mu(x)[/itex] such that:

    [tex]\mu\frac{dy}{dx} + \mu P(x)y = \mu Q(x)[/tex] where:

    [tex]\mu\frac{dy}{dx} + \mu P(x)y = \frac{d}{dx}(\mu y)[/tex]

    This reduces to:

    [tex]\mu P(x)y = y\frac{d\mu}{dx}[/tex]

    [tex]\frac{d\mu}{dx} = \mu P(x)[/tex]

    Dividing by [itex]\mu[/itex] and integrating both sides:

    [tex]\int \frac{1}{\mu}d\mu = \int P(x) dx[/tex]

    [tex]ln\mu = \int P(x) dx[/tex]

    So the general solution for [itex]\mu[/itex] is:

    [tex]\mu = \pm e^{\int P(x) dx}[/tex]

    AM
     
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