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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = - sec^{2} (x)

y^{^{''}} + y = x^{2} + 2

D^{2} (D + 3)^{2} y = 9 e^{- 3 x}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{6 t}

2 y^{^{''}} + 3 y^{^{'}} + y = sin (x), y (0) = 1, y^{^{'}} (0) = 2

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