What is the solution for y^{''}+8y^'+16y=2sin(4x) ?

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

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y^{''}+8y^'+16y=2sin(4x)

y^{^{''}} + 8 y^{^{'}} + 16 y = 2 sin (4 x)

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} - \frac{1}{16} cos (4 x)

Solution steps

y^{''} (x) + 8 y^{'} (x) + 16 y = 2 sin (4 x)

Solve linear ODE:

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} - \frac{1}{16} cos (4 x)

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} - \frac{1}{16} cos (4 x)

y^{^{''}} + 4 y^{^{'}} + 4 = \frac{1}{x ^{2} e ^{2 x}}

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

(D^{4} - 1) y = x

y^{^{''}} + 4 y = tan^{2} (2 x)

y^{^{''}} - 4 y = e - 2 x

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