What is the solution for y^{''}-32y=16cos(4x) ?

The solution for y^{''}-32y=16cos(4x) is y=c_{1}e^{4sqrt(2)x}+c_{2}e^{-4sqrt(2)x}-1/3 cos(4x)

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y^{''}-32y=16cos(4x)

y^{^{''}} - 32 y = 16 cos (4 x)

y = c_{1} e^{42 x} + c_{2} e^{- 42 x} - \frac{1}{3} cos (4 x)

Solution steps

y^{''} (x) - 32 y = 16 cos (4 x)

Solve linear ODE:

y = c_{1} e^{42 x} + c_{2} e^{- 42 x} - \frac{1}{3} cos (4 x)

y = c_{1} e^{42 x} + c_{2} e^{- 42 x} - \frac{1}{3} cos (4 x)

What is the solution for y^{''}-32y=16cos(4x) ?
The solution for y^{''}-32y=16cos(4x) is y=c_{1}e^{4sqrt(2)x}+c_{2}e^{-4sqrt(2)x}-1/3 cos(4x)