What is the solution for y^{''}+8y^'+16y=2sin(4t) ?

The solution for y^{''}+8y^'+16y=2sin(4t) is y=c_{1}e^{-4t}+c_{2}te^{-4t}-1/16 cos(4t)

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y^{''}+8y^'+16y=2sin(4t)

y^{^{''}} + 8 y^{^{'}} + 16 y = 2 sin (4 t)

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 2 sin (4 t)

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 14 cos (2 x) + 21 sin (2 x)

y^{^{''}} - y^{^{'}} = e^{x} cos (x), y (0) = 0, y^{^{'}} (0) = 0

(D^{2} + 5) y = 3 x e^{x}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 t} sin (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{2} - x + 3

What is the solution for y^{''}+8y^'+16y=2sin(4t) ?
The solution for y^{''}+8y^'+16y=2sin(4t) is y=c_{1}e^{-4t}+c_{2}te^{-4t}-1/16 cos(4t)