What is the solution for 8y^{''}-32y=16cos(x) ?

The solution for 8y^{''}-32y=16cos(x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-2/5 cos(x)

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8y^{''}-32y=16cos(x)

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{2}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{2}{5} cos (x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{2}{5} cos (x)

y^{^{''}} - 6 y^{^{'}} + 5 y = x, y (0) = e^{x}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = 6 e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (2 x)

y^{^{''}} + 9 y = cos (3 t), y (0) = 2, y^{^{'}} (0) = 5

y^{^{''}} - y^{^{'}} - 6 y = (x^{2} + 3 x) e^{2 x} + cos (2 x)

What is the solution for 8y^{''}-32y=16cos(x) ?
The solution for 8y^{''}-32y=16cos(x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-2/5 cos(x)