What is the solution for y^{''}+18y^'-y=t^{-1}e^t ?

The solution for y^{''}+18y^'-y=t^{-1}e^t is

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y^{''}+18y^'-y=t^{-1}e^t

y^{^{''}} + 18 y^{^{'}} - y = t^{- 1} e^{t}

y = c_{1} e^{(- 9 + 82) t} + c_{2} e^{(- 9 - 82) t} + e^{- 9 t + 82 t} \cdot \int \frac{82 e ^{10 t - 82 t}}{164 t} d t + e^{- 9 t - 82 t} \cdot \int - \frac{82 e ^{10 t + 82 t}}{164 t} d t

Solution steps

y^{''} (t) + 18 y^{'} (t) - y = t^{- 1} e^{t}

Solve linear ODE:

y = c_{1} e^{(- 9 + 82) t} + c_{2} e^{(- 9 - 82) t} + e^{- 9 t + 82 t} \cdot \int \frac{82 e ^{10 t - 82 t}}{164 t} d t + e^{- 9 t - 82 t} \cdot \int - \frac{82 e ^{10 t + 82 t}}{164 t} d t

y = c_{1} e^{(- 9 + 82) t} + c_{2} e^{(- 9 - 82) t} + e^{- 9 t + 82 t} \cdot \int \frac{82 e ^{10 t - 82 t}}{164 t} d t + e^{- 9 t - 82 t} \cdot \int - \frac{82 e ^{10 t + 82 t}}{164 t} d t

D B = 10 lo g_{10} (\frac{0.1}{1 \cdot 1 0 ^{- 12}})

y^{^{''}} - 8 y^{^{'}} + 16 y = 32 t + 64, y (0) = 3, y^{^{'}} (0) = - 3

y^{^{''}} + 4 y^{^{'}} + 40 y = 12.5 sin (6 t)

y^{^{''}} + 3 y^{^{'}} - 28 y = - 96 t + 56 t^{2}

θ^{^{''}} (t) - θ (t) = t

What is the solution for y^{''}+18y^'-y=t^{-1}e^t ?
The solution for y^{''}+18y^'-y=t^{-1}e^t is