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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 8 y = 8 x^{2} + 8 x + 1

x^{^{''}} + 2 x^{^{'}} + 5 x = 1, x (0) = 0, x^{^{'}} (0) = 0

y^{^{''}} + 9 y = e^{- 2 x}

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = x^{2} + 4 x + 8

y^{^{''}} + 4 y^{^{'}} + 4 y = - 2 sin (3 x)

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