What is the solution for y^{''}+8y^'+16y=8sin(4x) ?

The solution for y^{''}+8y^'+16y=8sin(4x) is y=c_{1}e^{-4x}+c_{2}xe^{-4x}-1/4 cos(4x)

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y^{''}+8y^'+16y=8sin(4x)

y^{^{''}} + 8 y^{^{'}} + 16 y = 8 sin (4 x)

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

Solution steps

y^{''} (x) + 8 y^{'} (x) + 16 y = 8 sin (4 x)

Solve linear ODE:

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

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

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

y^{^{''}} + y^{^{'}} - 2 y = t e^{3 t}

y^{^{''}} + 24 y^{^{'}} + 144 y = t^{2} e^{- 12 t} + e^{- 12 t}

y^{^{''}} - 81 y = 3

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{''}+8y^'+16y=8sin(4x) ?
The solution for y^{''}+8y^'+16y=8sin(4x) is y=c_{1}e^{-4x}+c_{2}xe^{-4x}-1/4 cos(4x)