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Math Help - how to solve this differential equation

  1. #1
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    how to solve this differential equation

     F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi) with boundary conditions:
     F(0)=1, \ \ \ \ F\rightarrow0 \ \ as \ \ \xi\rightarrow\infty

    i got this problem when doing question about diffusion equation. they give the answer :

     F(\xi)=\frac{1}{\sqrt{\pi}}[(1+\frac{1}{2}\xi^{2})\int_{\xi}^{\infty}{e^{-\frac{x^{2}}{4}} dx} -\xi e^{-\frac{\xi^{2}}{4}}]

    but i just dont know how to get that?
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  2. #2
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    Quote Originally Posted by szpengchao View Post
     F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi) with boundary conditions:
     F(0)=1, \ \ \ \ F\rightarrow0 \ \ as \ \ \xi\rightarrow\infty

    i got this problem when doing question about diffusion equation. they give the answer :

     F(\xi)=\frac{1}{\sqrt{\pi}}[(1+\frac{1}{2}\xi^{2})\int_{\xi}^{\infty}{e^{-\frac{x^{2}}{4}} dx} -\xi e^{-\frac{\xi^{2}}{4}}]

    but i just dont know how to get that?
    Your differential equation admits the following exact solution

    F (\xi)= \xi^2+2

    so you can use reduction of order F(\xi) = \left( \xi^2+2 \right) G(\xi) to reduce it to first order (which is separable) giving rise to error functions and exponentals.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by szpengchao View Post
     F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi) with boundary conditions:
     F(0)=1, \ \ \ \ F\rightarrow0 \ \ as \ \ \xi\rightarrow\infty

    i got this problem when doing question about diffusion equation. they give the answer :

     F(\xi)=\frac{1}{\sqrt{\pi}}[(1+\frac{1}{2}\xi^{2})\int_{\xi}^{\infty}{e^{-\frac{x^{2}}{4}} dx} -\xi e^{-\frac{\xi^{2}}{4}}]

    but i just dont know how to get that?
    Use the Laplace transform
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  4. #4
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    can u make it more clear

    Quote Originally Posted by danny arrigo View Post
    Your differential equation admits the following exact solution

    F (\xi)= \xi^2+2

    so you can use reduction of order F(\xi) = \left( \xi^2+2 \right) G(\xi) to reduce it to first order (which is separable) giving rise to error functions and exponentals.

    can u make it more clear? i still cant get the point
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  5. #5
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    Quote Originally Posted by szpengchao View Post
    can u make it more clear? i still cant get the point
    Sure, if F = (\xi^2+2)G taking derivatives gives

    F' = (\xi^2 + 2) G' + 2 \xi G
    F'' = (\xi^2 + 2) G'' + 4 \xi G' + 2 G .

    Substituting into your equation gives

    (\xi^2 + 2) G'' + 4 \xi G' + 2 G + \frac{\xi}{2} \left( (\xi^2 + 2) G' + 2 \xi G \right) - (\xi^2+2)G = 0

    Expanding gives

    (\xi^2+2)G'' + \left(\frac{\xi^3}{2} + 5 \xi \right) G' = 0

    If you let G' = H then

    (\xi^2+2)H' + \left(\frac{\xi^3}{2} + 5 \xi \right) H = 0

    a separable equation for H. Separate and integrate, replace H with G' and integrate again.
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