Answer: k = 4, k = -4 and k = 0.
Step-by-step explanation:
If we have y = sin(kt)
then:
y' = k*cos(kt)
y'' = -k^2*son(x).
then, if we have the relation:
y'' - y = 0
we can replace it by the things we derivated previously and get:
-k^2*sin(kt) + 16*sin(kt) = 0
we can divide by sin in both sides (for t ≠0 and k ≠0 because we can not divide by zero)
-k^2 + 16 = 0
the solutions are k = 4 and k = -4.
Now, we have another solution, but it is a trivial one that actually does not give any information, but for the diff equation:
-k^2*sin(kt) + 16*sin(kt) = 0
if we take k = 0, we have:
-0 + 0 = 0.
So the solutions are k = 4, k = -4 and k = 0.
