What is the solution for x^{''}+2x^'+1x=e^{-2t} ?

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

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x^{''}+2x^'+1x=e^{-2t}

x^{^{''}} + 2 x^{^{'}} + 1 x = e^{- 2 t}

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + 1 x = e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 4 sin (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} \cdot e^{- 2 x} + 1

y^{^{''}} + 7 y = 5

x^{^{''}} + x^{^{'}} - 2 x = 8 sin (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = (12 x - 7) e^{- x}

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