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Math Help - differential equations

  1. #16
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by emily28 View Post
    i dont that i was meant to look at particular solutions and expenituals, is there another way to do it using perhaps a different method, as i havent covered this in class and dont think the teacher would want me to approach it this way?
    Probably Jhevon's integrating factor approach would be the best then.

    -Dan
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  2. #17
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    done a little on using the intergrating factor, but how would be the best way to get the formula into a form to use the intergrating formula?
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  3. #18
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    does anyone have any suggestions/
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  4. #19
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    Quote Originally Posted by emily28 View Post
    \frac{du}{dx} + 4 u = -4x
    Find an integrating factor defined by

    \mu(x)=\exp\left(4\int~dx\right)
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  5. #20
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by emily28 View Post
    does anyone have any suggestions/
    See here.

    -Dan
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  6. #21
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emily28 View Post
    does anyone have any suggestions/
    I will say what Krizalid said, but in a more general way.

    For a differential equation of the form: y' + p(x)y = f(x)

    we define \mu (x) = e^{\int p(x)~dx} = \mbox { exp} \left( \int p(x)~dx \right) as the integrating factor of the differential equation. We solve the differential equation by multiplying through by the integrating factor. so we get the new equation:

    \mu (x) y' + \mu (x) p(x) y = \mu (x) f(x)

    it will always be the case that the left hand side is the derivative we would get from applying the product rule. so the equation becomes:

    \left( \mu (x) y \right)' = \mu (x) f(x)

    \Rightarrow \mu (x) y = \int \mu (x) f(x)~dx

    \Rightarrow y = \frac { \int \mu (x) f(x) ~dx}{ \mu (x)} is our solution

    remember that you will get an arbitrary constant from the integration, you have to divide this by \mu (x) as well
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  7. #22
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    after doing some reading on intergrating factors, from



    ive used the intergrating factor, whereby f(x) = -4x and a(x) = 4

    so g(x)= exp^(intergral of 4 dx)

    so  g(x) = e^{4x}

    now using the formula y= 1/g(x).[intergral of f(x)g(x) dx]

    so  y=1/e^{4x} [intergral  -4xe^{4x}]<br />

    is this correct and how would i carry on from this?
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  8. #23
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emily28 View Post
    after doing some reading on intergrating factors, from



    ive used the intergrating factor, whereby f(x) = -4x and a(x) = 4

    so g(x)= exp^(intergral of 4 dx)

    so  g(x) = e^{4x}

    now using the formula y= 1/g(x).[intergral of f(x)g(x) dx]

    so  y=1/e^{4x} [intergral  -4xe^{4x}]<br />

    is this correct and how would i carry on from this?
    yes, that is fine, and incidentally, that is exactly the process i described to you above. i tried to save you the trouble of trying to get it from you text book, sometimes those things are ... let's just say, I hate the text that i used for differential equations. the only thing i like about it is that it gives the answers for even as well as odd problems, which is rare for textbooks to do
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