differential question

superdude · #1 $Old$ July 3rd, 2009, 04:53 PM

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

mr fantastic · #2 $Old$ July 3rd, 2009, 04:54 PM

Quote:

Originally Posted by superdude $View Post$

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

Start by differentiating twice the trial solution. Then substitute y and y'' into the DE. Compare the coefficient of cos(kt) on each side.

calc101 · #3 $Old$ July 3rd, 2009, 08:55 PM

$y = \cos{(kt)}$
$y' = -k\sin{(kt)}$
$y''= -k^2\cos{(kt)}$
$4(-k^2\cos{(kt)}) = -25(\cos{(kt)})$

Spoiler:

mr fantastic · #4 $Old$ July 3rd, 2009, 09:06 PM

Quote:

Originally Posted by calc101 $View Post$

$y = cost kt$
$y' = -ksinkt$
$y''= -k^2coskt$
$4(-k^2coskt) = -25(coskt)$

The coskt and the negatives cancel, therefore:
$4k^2 = 25$
$k = \frac{5}{2}$

Therefore, the equation is satisfied when $k=\frac{5}{2}$

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.

CaptainBlack · #5 $Old$ July 4th, 2009, 01:05 AM

Quote:

Originally Posted by calc101 $View Post$

$y = cost kt$
$y' = -ksinkt$
$y''= -k^2coskt$
$4(-k^2coskt) = -25(coskt)$

Please consider improving your use of notation, in particular put brackets around the arguments of functions.

Also \sin and \cos will improve the readability as well

CB

superdude · #6 $Old$ July 5th, 2009, 01:13 PM

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:

$y=A\sin(kt)+B\cos(kt)\\y'=Ak\cos(kt)-bk\sin(kt)\\y''=-Bk^2\cos(kt)-Ak^2\sin(kt)\\$
then I plug in the value of k and get
$4(\frac{-25b\cos(5t/2)}{4}-\frac{25a\sin(5t/2)}{4})=-25b\cos(5k/2)-25a\sin(5k/2)$
I'm uncertain what to do next or if the question is completed

Chris L T521 · #7 $Old$ July 5th, 2009, 01:41 PM

Quote:

Originally Posted by superdude $View Post$

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:

$\begin{aligned}y & = A\sin(kt)+B\cos(kt)\\y' & =Ak\cos(kt)-bk\sin(kt)\\y'' & =-Bk^2\cos(kt)-Ak^2\sin(kt)\end{aligned}$
then I plug in the value of k and get
$4(\frac{-25b\cos(5t/2)}{4}-\frac{25a\sin(5t/2)}{4})=-25b\cos(5{\color{red}t}/2)-25a\sin(5{\color{red}t}/2)$
I'm uncertain what to do next or if the question is completed

I'm sorry, how do I cause a line break with latex?

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.

superdude · #8 $Old$ July 5th, 2009, 06:57 PM

Thanks, what I'm doing now makes sense.