Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+16y=0. Separate your answers by commas. isn't the answer just ±4?

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.

21

step-by-step explanation:

11279

a

explanation: