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Math Help - First order differential equation

  1. #1
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    First order differential equation

    Having trouble solving this differential equation:

    <br /> <br />
\frac{\partial \Phi}{\partial x} - K\Phi(x) = -g(x) <br /> <br />

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

    However it seems strange that I get an expression for \Phi in terms of an integral of \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; April 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
    <br />
\frac{\partial \Psi}{\partial t}<br />
    I assume the rest of the question is in Neverland ....?
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  3. #3
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    Hello, peterpan!

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

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

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

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


    Integrate . 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: . I \;=\;e^{\int(-K)dx} \;=\;e^{-Kx}

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

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


    Integrate . 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 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 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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