## Calculus: Early Transcendentals 8th Edition

(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.
(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.