What is the solution for x^{''}+4x^'+4x=100sin(4t) ?

The solution for x^{''}+4x^'+4x=100sin(4t) is x=c_{1}e^{-2t}+c_{2}te^{-2t}-3sin(4t)-4cos(4t)

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x^{''}+4x^'+4x=100sin(4t)

x^{^{''}} + 4 x^{^{'}} + 4 x = 100 sin (4 t)

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - 3 sin (4 t) - 4 cos (4 t)

Solution steps

x^{''} (t) + 4 x^{'} (t) + 4 x = 100 sin (4 t)

Solve linear ODE:

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - 3 sin (4 t) - 4 cos (4 t)

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - 3 sin (4 t) - 4 cos (4 t)

x^{^{''}} + 50 x^{^{'}} + 400 x = 100 cos (20 t)

y^{^{''}} + y^{^{'}} - 2 y = e^{t} sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 22 y^{^{'}} + 121 y = 16.3 u

y^{^{''}} + y = 2 e^{- 8 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - 6 y^{^{'}} - 7 = 0

What is the solution for x^{''}+4x^'+4x=100sin(4t) ?
The solution for x^{''}+4x^'+4x=100sin(4t) is x=c_{1}e^{-2t}+c_{2}te^{-2t}-3sin(4t)-4cos(4t)