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

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.