Answer
(a) $k = \frac{5}{2}, -\frac{5}{2}$
(b) Every member of this family of solutions $~~y = A~sin~kt+ B~cos~kt~~$ is also a solution.
Work Step by Step
(a) $y = cos~kt$
$y' = -k~sin~kt$
$y'' = -k^2~cos~kt$
We can find the values of $k$:
$4y'' = -25y$
$4(-k^2~cos~kt) = -25~cos~kt$
$4k^2 = 25$
$k^2 = \frac{25}{4}$
$k = \frac{5}{2}, -\frac{5}{2}$
(b) $y = A~sin~kt+ B~cos~kt$
$y' = k~A~cos~kt-k~B~sin~kt$
$y'' = -k^2~A~sin~kt-k^2~B~cos~kt$
We can evaluate the differential equation:
$4y''$
$= 4(-k^2~A~sin~kt-k^2~B~cos~kt)$
$ = -4k^2(A~sin~kt+~B~cos~kt)$
$ = -4(\frac{25}{4})(A~sin~kt+~B~cos~kt)$
$ = -25(A~sin~kt+~B~cos~kt)$
$ = -25~y$
Then every member of this family of solutions $~~y = A~sin~kt+ B~cos~kt~~$ is also a solution.
