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

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

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

y^{^{''}} + 2 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 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}

\frac{d}{d x} (arctan (x^{2} + y^{2}) + ln (x y) - 1)

8 cos (2 x)

y^{^{''}} + y^{^{'}} + k y = 0

\frac{d L}{d x}

\frac{d}{d x} (ln (x (x^{2} + 3)^{3}))

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