# Thread: how to solve this differential equation

1. ## 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?

2. Originally Posted by szpengchao
$F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi)$ 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?

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

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.

3. Originally Posted by szpengchao
$F''(\xi)+\frac{\xi}{2}F'(\xi) = F(\xi)$ 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

4. ## can u make it more clear

Originally Posted by danny arrigo

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

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

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

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

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

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.