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Intergrating a Differential Equation using the intergrating factor method

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{dy}{dx} + \frac{y}{2x} = -x^{\frac{1}{2}}[/tex]

    2. Relevant equations

    [tex]yT =\int{QT}dx + C[/tex]
    where T is the intergrating factor

    T = [tex]e^{\int{P}dx[/tex]
    and P is the co-efficient of y from the differential equations

    3. The attempt at a solution

    well to find T we need to do:

    [tex] = x^{\frac{1}{2}}[/tex]

    so using [tex]yT =\int{QT}dx + C[/tex]

    you get

    [tex] yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}x^-{\frac{1}{2}}}dx[/tex]
    [tex] yx^{\frac{1}{2}}= \int{1}dx[/tex]
    [tex] yx^{\frac{1}{2}}= x + c[/tex]

    the answer in the back of the book says [tex]yx^{\frac{1}{2}} = \frac{1}{2}x^{2}[/tex]
    Where have I gone wrong?

    Thanks :)
  2. jcsd
  3. Sep 30, 2008 #2

    Which makes your integrating factor [tex]x^\frac 1 2[/tex]

    [tex]yx^\frac 1 2=-\int x^\frac 1 2 x^\frac 1 2dx=-\int xdx[/tex]

    Watch your inputting.
  4. Sep 30, 2008 #3
    yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}}x^-{\frac{1}{2}}}dx

    woops. above is what is ment to say
    simplifies to 1 right?

    Thanks :)
    Last edited: Sep 30, 2008
  5. Sep 30, 2008 #4
    No!!! Is the negative with x^1/2 or with just the 1/2?
  6. Sep 30, 2008 #5

    is what i mean

    thanks :)
  7. Sep 30, 2008 #6
    I knew what you meant, but why would the negative be with the 1/2 making it [tex]x^{-\frac 1 2}[/tex]???

    Isn't the right side [tex]-\int x^\frac 1 2dx[/tex] then multiplied by your integrating factor, [tex]-\int x^\frac 1 2x^\frac 1 2dx[/tex].
  8. Sep 30, 2008 #7
    your a genius :)
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