for what values of k does the function y=costkt satisfy the differential equation 4y''=-25y?

Printable View

- Jul 3rd 2009, 02:53 PMsuperdudedifferential question
for what values of k does the function y=costkt satisfy the differential equation 4y''=-25y?

- Jul 3rd 2009, 02:54 PMmr fantastic
- Jul 3rd 2009, 06:55 PMcalc101
$\displaystyle y = \cos{(kt)}$

$\displaystyle y' = -k\sin{(kt)}$

$\displaystyle y''= -k^2\cos{(kt)}$

$\displaystyle 4(-k^2\cos{(kt)}) = -25(\cos{(kt)})$

__Spoiler__: - Jul 3rd 2009, 07:06 PMmr fantastic
I was rather hoping that the OP would make an attempt and say where s/he was stuck if help was still required. Short of using my editing powers, this is now impossible.

As it is, you've left only one trivial (but important) thing for the OP to do. I hope that this one small thing remains left for the OP. - Jul 3rd 2009, 11:05 PMCaptainBlack
- Jul 5th 2009, 11:13 AMsuperdude
yes I did get that answer, plus or minus 5/2

part b) I'm stuck on again. "For those values of k, verify that every member of the family of functions y=Asinkt+Bcoskt is also a solution." (where A,B, and of course k, are constants). This is what I did:

$\displaystyle

y=A\sin(kt)+B\cos(kt)\\

y'=Ak\cos(kt)-bk\sin(kt)\\

y''=-Bk^2\cos(kt)-Ak^2\sin(kt)\\

$

then I plug in the value of k and get

$\displaystyle

4(\frac{-25b\cos(5t/2)}{4}-\frac{25a\sin(5t/2)}{4})=-25b\cos(5k/2)-25a\sin(5k/2)

$

I'm uncertain what to do next or if the question is completed - Jul 5th 2009, 11:41 AMChris L T521
Note my correction in red first (you had k, when it should have been t).

Now, simplify the LHS to see that LHS = RHS which implies that the family of functions satisfies the differential equation.

To answer you're last question on LaTeX, click on the edited LaTeX image to see the code. - Jul 5th 2009, 04:57 PMsuperdudeAnother question
Thanks, what I'm doing now makes sense.