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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = 6 e^{2 x}, y (0) = 4, y^{^{'}} (0) = 4

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{2 t}

(D^{4} - 1) \cdot y = sin (2 x)

y^{^{''}} + y^{^{'}} - 7 y = 42 t

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