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Thread: First order differential equation

  1. #1
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    First order differential equation

    Having trouble solving this differential equation:

    $\displaystyle

    \frac{\partial \Phi}{\partial x} - K\Phi(x) = -g(x)

    $

    The answer I have so far is:
    $\displaystyle
    \int_0^{\tau} \frac{\partial \Phi}{\partial x} dx = K\int_0^{\tau}\Phi(x)dx - \int_0^{\tau}g(x) dx
    $
    $\displaystyle
    \Phi = K\int_0^{\tau} \Phi (x)dx - \int_0^{\tau} g(x) dx

    $

    However it seems strange that I get an expression for $\displaystyle \Phi$ in terms of an integral of $\displaystyle \Phi$.

    Any help would be much appreciated.

    P.S. mr fantastic, seeing as this message had to come all the way from Neverland it took a little time to arrive fully! lol!
    Last edited by peterpan; Apr 5th 2008 at 02:20 PM. Reason: Submitted too quickly
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  2. #2
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    Quote Originally Posted by peterpan View Post
    $\displaystyle
    \frac{\partial \Psi}{\partial t}
    $
    I assume the rest of the question is in Neverland ....?
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  3. #3
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    Hello, peterpan!

    $\displaystyle \frac{d\Phi}{dx} - K\Phi \:=\: -g(x) $

    You need an integrating factor: .$\displaystyle I \;=\;e^{\int(-K)dx} \;=\;e^{-Kx}$

    Then we have: . $\displaystyle e^{-Kx}\frac{d\Phi}{dx} - Ke^{-Kx}\Phi \;=\;-e^{-Kx}g(x) $

    . . . $\displaystyle \frac{d}{dx}\left(e^{-Kx}\Phi\right) \;=\;-e^{-Kx}g(x) $


    Integrate . $\displaystyle e^{-Kx}\Phi \;=\;-\int e^{-Kt}g(x)\,dx \quad\hdots\quad \text{etc.}$

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  4. #4
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    I thought that you only normally use an intergrating factor method when you have a differential equation of the form:

    y' + P(x)y = Q(x)

    Which is different to what we have here isnt it ?
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  5. #5
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    Hello,

    P(x) is only a polynomial. And a constant is a polynomial (degree : 0) !
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  6. #6
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    Quote Originally Posted by peterpan View Post
    I thought that you only normally use an intergrating factor method when you have a differential equation of the form:

    y' + P(x)y = Q(x)

    Which is different to what we have here isnt it ?
    In your case, P(x) = -K and Q(x) = -g(x).
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  7. #7
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    I see, thanks for the help. I think the reason I went wrong was because I believed that P(x) _had_ to be a function of x. In general terms is it valid to say P(x) = K?
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  8. #8
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    Quote Originally Posted by peterpan View Post
    I see, thanks for the help. I think the reason I went wrong was because I believed that P(x) _had_ to be a function of x. In general terms is it valid to say P(x) = K?
    Moo has already correctly answered this question:

    Quote Originally Posted by Moo View Post
    [snip]

    P(x) is only a polynomial. And a constant is a polynomial (degree : 0) !
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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, peterpan!


    You need an integrating factor: .$\displaystyle I \;=\;e^{\int(-K)dx} \;=\;e^{-Kx}$

    Then we have: . $\displaystyle e^{-Kx}\frac{d\Phi}{dx} - Ke^{-Kx}\Phi \;=\;-e^{-Kx}g(x) $

    . . . $\displaystyle \frac{d}{dx}\left(e^{-Kx}\Phi\right) \;=\;-e^{-Kx}g(x) $


    Integrate . $\displaystyle e^{-Kx}\Phi \;=\;-\int e^{-Kt}g(x)\,dx \quad\hdots\quad \text{etc.}$


    Is there a way that you can integrate the right had side such that the $\displaystyle e^{(-Kx)}$ dissapears?
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  10. #10
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    Quote Originally Posted by peterpan View Post
    Is there a way that you can integrate the right had side such that the $\displaystyle e^{(-Kx)}$ dissapears?
    No. You'll need to have a specific function for g(x) before you can do anything useful with the right hand side.
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