Thread: differential question
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Old July 5th, 2009, 01:13 PM
superdude superdude is offline
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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
Last edited by superdude; July 5th, 2009 at 06:54 PM.
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