What is the solution for 8y^{''}+0.25y^'+512y=2cos(2t) ?

The solution for 8y^{''}+0.25y^'+512y=2cos(2t) is

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8y^{''}+0.25y^'+512y=2cos(2t)

8 y^{^{''}} + 0.25 y^{^{'}} + 512 y = 2 cos (2 t)

y = e^{- \frac{t}{64}} (c_{1} cos (\frac{163839375 t}{1600}) + c_{2} sin (\frac{163839375 t}{1600})) + 4.34027 E - 6 sin (2 t) + \frac{3840}{921601} cos (2 t)

Solution steps

8 y^{''} (t) + 0.25 y^{'} (t) + 512 y = 2 cos (2 t)

Solve linear ODE:

y = e^{- \frac{t}{64}} (c_{1} cos (\frac{163839375 t}{1600}) + c_{2} sin (\frac{163839375 t}{1600})) + 4.34027 E - 6 sin (2 t) + \frac{3840}{921601} cos (2 t)

y = e^{- \frac{t}{64}} (c_{1} cos (\frac{163839375 t}{1600}) + c_{2} sin (\frac{163839375 t}{1600})) + 4.34027 E - 6 sin (2 t) + \frac{3840}{921601} cos (2 t)

y^{^{''}} + y = 1, 6 cos (t) y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = 15 t, y (0) = - 15 t

x^{^{''}} + 6 x^{^{'}} + 5 x = sin (t)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 15 x + 17

What is the solution for 8y^{''}+0.25y^'+512y=2cos(2t) ?
The solution for 8y^{''}+0.25y^'+512y=2cos(2t) is