Thread: differential question
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Old July 5th, 2009, 01:41 PM
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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.
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