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}

y^{^{''}} - 2 y^{^{'}} - 3 y = cos (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{2 x}

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

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0 + 6 sin (x)

y^{^{''}} + y = 2 sin (2 t)

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