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

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

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

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

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - \frac{1}{8} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - \frac{1}{8} cos (2 t)

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - \frac{1}{8} cos (2 t)

y^{^{''}} + 16 y = 17 sec^{2} (4 t)

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

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

2 y^{^{''}} + y^{^{'}} = cos^{2} (x)

y^{^{''}} - 5 y^{^{'}} + 6 y = (2 x - 1) e^{4 x}

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