# differential equation proving

• Mar 27th 2008, 08:16 PM
chris_panda85
differential equation proving
Given a differential equation
y"= -(2g'/g)y' + [V-E-(g"/g)]y

V(x)= x^(2k) & g(x) = exp(-βx^2/2)

subtitute V(x) & g(x), found

y" = 2βxy' + [x^(2k) - (βx)^2 + β - E]y

the problem i encounter is the subtitution I done is missing β

y" = 2βxy' + [x^(2k) - (βx)^2 - E]y

P/s: ^ represent power of
• Mar 28th 2008, 02:13 PM
Aryth
Well, it's actually just an error in differentiating:

$g(x) = e^{-\frac{\beta x^2}{2}}$

$g'(x) = -\beta xe^{-\frac{\beta x^2}{2}}$

The first one was an exponential, so we used the chain rule, but notice that this is a product, so we must use the product rule:

$g''(x) = -\beta e^{-\frac{\beta x^2}{2}} + (\beta x)^2e^{-\frac{\beta x^2}{2}}$

Now we do the division:

$-\frac{g''(x)}{g(x)} = \frac{\beta e^{-\frac{\beta x^2}{2}} - (\beta x)^2e^{-\frac{\beta x^2}{2}}}{e^{-\frac{\beta x^2}{2}}}$

Which gives:

$-\frac{g''(x)}{g(x)} = \beta - (\beta x)^2$

$= -(\beta x)^2 + \beta$

And there is the right answer.
• Mar 31st 2008, 06:01 AM
chris_panda85
Thanks ...
Thanks for helping me out. Guess I was kinda careless as well. Anyway, thanks again ...