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}

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