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

The solution for x^{''}+2x^'+x=2te^{-t} is x=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^3)/3

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x^{''}+2x^'+x=2te^{-t}

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

x = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{3}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{3}

x = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{3}

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

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{2} e^{t}

2 y^{^{''}} + 3 y^{^{'}} - 5 y = 2 cos (x)

y^{^{''}} - 4 y = e^{- x}

y^{^{''}} + 4 y = 10 x (4 - x), y (0) = - 5, y (4) = 10

What is the solution for x^{''}+2x^'+x=2te^{-t} ?
The solution for x^{''}+2x^'+x=2te^{-t} is x=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^3)/3