Question

(a) For what values of k does the function y = cos kt satisfy the differential equation 4y" = -25y?(b) For those values of k, verify that every member of the family of functions y = A sin kt + B cos kt is also a solution.

Answer 1

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 $\cos(-x)=\cos x$.)