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

The solution for y^{''}+4y^'+4y=2cos(t) is y=c_{1}e^{-2t}+c_{2}te^{-2t}+8/25 sin(t)+6/25 cos(t)

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

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

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{8}{25} sin (t) + \frac{6}{25} cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{8}{25} sin (t) + \frac{6}{25} cos (t)

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{8}{25} sin (t) + \frac{6}{25} cos (t)

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

y^{^{''}} - 9 y = (x^{2} - 9) sin (3 x)

y^{^{''}} - 36 y = 8, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} - 36 y^{^{'}} = t e^{- 6 t} + 7 cos (6 t)

2 y^{^{''}} + 3 y^{^{'}} + y = g (t)

What is the solution for y^{''}+4y^'+4y=2cos(t) ?
The solution for y^{''}+4y^'+4y=2cos(t) is y=c_{1}e^{-2t}+c_{2}te^{-2t}+8/25 sin(t)+6/25 cos(t)