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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 1 - 8 x^{2} + 48 e^{- 4 x}

y^{^{'''}} + 15 y^{^{''}} - 500 y = e^{- 10 x}

(D^{2} - D) y = e^{x} cos (x)

y^{^{''}} + 4 y = 2, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 5 sin (x))

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