# how to solve this differential equation

• Jan 19th 2009, 05:52 AM
szpengchao
how to solve this differential equation
$\displaystyle F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi)$ with boundary conditions:
$\displaystyle F(0)=1, \ \ \ \ F\rightarrow0 \ \ as \ \ \xi\rightarrow\infty$

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

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

but i just dont know how to get that?
• Jan 19th 2009, 07:34 AM
Jester
Quote:

Originally Posted by szpengchao
$\displaystyle F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi)$ with boundary conditions:
$\displaystyle F(0)=1, \ \ \ \ F\rightarrow0 \ \ as \ \ \xi\rightarrow\infty$

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

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

but i just dont know how to get that?

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

so you can use reduction of order $\displaystyle F(\xi) = \left( \xi^2+2 \right) G(\xi)$ to reduce it to first order (which is separable) giving rise to error functions and exponentals.
• Jan 19th 2009, 07:37 AM
DeMath
Quote:

Originally Posted by szpengchao
$\displaystyle F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi)$ with boundary conditions:
$\displaystyle F(0)=1, \ \ \ \ F\rightarrow0 \ \ as \ \ \xi\rightarrow\infty$

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

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

but i just dont know how to get that?

Use the Laplace transform
• Jan 19th 2009, 01:05 PM
szpengchao
can u make it more clear
Quote:

Originally Posted by danny arrigo

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

so you can use reduction of order $\displaystyle F(\xi) = \left( \xi^2+2 \right) G(\xi)$ 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
• Jan 19th 2009, 01:15 PM
Jester
Quote:

Originally Posted by szpengchao
can u make it more clear? i still cant get the point

Sure, if $\displaystyle F = (\xi^2+2)G$ taking derivatives gives

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

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

Expanding gives

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

If you let $\displaystyle G' = H$ then

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

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