For what value of k, if any, will y = ksin (5x) + 2cos (4x) be a solution to the differential equation y'' +16y =-27sin(5x)?

a. -27
b. -9/5
c. 3
d. There is no such value of k.

c. 3

Step-by-step explanation:

y = k sin(5x) + 2 cos(4x)

y' = 5k cos(5x) − 8 sin(4x)

y" = -25k sin(5x) − 32 cos(4x)

y'' + 16y = -27 sin(5x)

(-25k sin(5x) − 32 cos(4x)) + 16(k sin(5x) + 2 cos(4x)) = -27 sin(5x)

-25k sin(5x) − 32 cos(4x) + 16k sin(5x) + 32 cos(4x) = -27 sin(5x)

-9k sin(5x) = -27 sin(5x)

k = 3

The difference with any previous decade is much greater than the margin of error

Step-by-step explanation: