Question

For what values of k

Answer 1

If y = cos(kt), then its first two derivatives are

y' = -k sin(kt)

y'' = -k² cos(kt)

Substituting y and y'' into 49y'' = -16y gives

-49k² cos(kt) = -15 cos(kt)

⇒   49k² = 15

⇒   k² = 15/49

⇒   k = ±√15/7

Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.