Thread: differential question
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Old July 3rd, 2009, 08:55 PM
calc101 calc101 is offline
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y = \cos{(kt)}
y' = -k\sin{(kt)}
y''= -k^2\cos{(kt)}
4(-k^2\cos{(kt)}) = -25(\cos{(kt)})

Spoiler:

The \cos{(kt)} and the negatives cancel, therefore:
4k^2 = 25
k = \frac{5}{2}

Therefore, the equation is satisfied when k=\frac{5}{2}
Last edited by calc101; July 4th, 2009 at 07:52 AM. Reason: hide explicit solution
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