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kramer
4 years ago
8

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

Mathematics
1 answer:
emmainna [20.7K]4 years ago
4 0

Answer:

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

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