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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = x + 1

y^{^{''}} + y = e^{- x} (2 sin (x) + 4 cos (x))

y^{^{''}} + y^{^{'}} - 7 y = 3 e^{2 x} - 2 e^{3 x}

y^{^{'''}} + y^{^{'}} - 10 = x e^{2 x}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 (1 + e^{- x})

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