What is the solution for y^{''}+2y^'+y=4e^{-2t}+t+t^2 ?

The solution for y^{''}+2y^'+y=4e^{-2t}+t+t^2 is y=c_{1}e^{-t}+c_{2}te^{-t}+4e^{-2t}+t^2-3t+4

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y^{''}+2y^'+y=4e^{-2t}+t+t^2

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 2 t} + t + t^{2}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 4 e^{- 2 t} + t^{2} - 3 t + 4

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + 4 e^{- 2 t} + t^{2} - 3 t + 4

y = c_{1} e^{- t} + c_{2} t e^{- t} + 4 e^{- 2 t} + t^{2} - 3 t + 4

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{''}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 32 y = - 80 t + 64 t^{2}

D^{3} y = e^{3 x}

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

y^{^{''}} + 4 y^{^{'}} - 2 y = \frac{1}{x ^{2} e ^{2 x}}

What is the solution for y^{''}+2y^'+y=4e^{-2t}+t+t^2 ?
The solution for y^{''}+2y^'+y=4e^{-2t}+t+t^2 is y=c_{1}e^{-t}+c_{2}te^{-t}+4e^{-2t}+t^2-3t+4