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

The solution for y^{''}+2y^'+2y=2t^2e^{-t}sin(t) is

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

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

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) - 2 e^{- t} cos (t) (\frac{1}{3} t^{3} sin^{2} (t) + \frac{1}{6} t^{3} cos (2 t) - \frac{1}{4} t^{2} sin (2 t) - \frac{1}{4} t cos (2 t) + \frac{1}{8} sin (2 t)) + e^{- t} sin (t) (- \frac{1}{2} t^{2} cos (2 t) + \frac{1}{4} (2 t sin (2 t) + cos (2 t)))

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) - 2 e^{- t} cos (t) (\frac{1}{3} t^{3} sin^{2} (t) + \frac{1}{6} t^{3} cos (2 t) - \frac{1}{4} t^{2} sin (2 t) - \frac{1}{4} t cos (2 t) + \frac{1}{8} sin (2 t)) + e^{- t} sin (t) (- \frac{1}{2} t^{2} cos (2 t) + \frac{1}{4} (2 t sin (2 t) + cos (2 t)))

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) - 2 e^{- t} cos (t) (\frac{1}{3} t^{3} sin^{2} (t) + \frac{1}{6} t^{3} cos (2 t) - \frac{1}{4} t^{2} sin (2 t) - \frac{1}{4} t cos (2 t) + \frac{1}{8} sin (2 t)) + e^{- t} sin (t) (- \frac{1}{2} t^{2} cos (2 t) + \frac{1}{4} (2 t sin (2 t) + cos (2 t)))

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

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{4 x}

y^{^{''}} + 2 y^{^{'}} = 1 + (t - 2), y (0) = 0, y^{^{'}} (0) = 1

4 x^{^{''}} + 85 x = cos (6 t)

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