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

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

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

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

y = c_{1} e^{- t} + c_{2} t e^{- t} - e^{- t} t + e^{- t} t ln (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} - e^{- t} t + e^{- t} t ln (t)

y = c_{1} e^{- t} + c_{2} t e^{- t} - e^{- t} t + e^{- t} t ln (t)

y^{^{''}} - 4 y^{^{'}} + 13 y = - 40 cos (x) + 13 x

y^{^{''}} + 25 y = cos (5 x) csc^{2} (5 x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 e^{- x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{x} - 6

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{t}, y (0) = 0, y^{^{'}} (0) = 1

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