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