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STatiana [176]
4 years ago
15

(a) For what values of k does the function y = cos kt satisfy the differential equation 4y" = -25y?(b) For those values of k, ve

rify that every member of the family of functions y = A sin kt + B cos kt is also a solution.
Mathematics
1 answer:
zalisa [80]4 years ago
6 0

a. y=\cos kt\implies y''=-k^2\cos kt

Substituting these into the ODE gives

-4k^2\cos kt=-25\cos kt\impliesk^2=\dfrac{25}4\implies k=\pm\dfrac52

b. y=A\sin kt+B\cos kt\implies y''=-k^2A\sin kt-k^2B\cos kt

If k=\dfrac52, then

4\left(-\dfrac{25}4A\sin\dfrac52t-\dfrac{25}4B\cos\dfrac52t\right)=-25\left(\sin\dfrac52t+\cos\dfrac52t\right)

as required, and the same can be said for k=-\dfrac52. (Note that \sin(-x)=-\sin x and \cos(-x)=\cos x.)

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