What is the solution for -x^{''}-3x^'-2=t ?

The solution for -x^{''}-3x^'-2=t is x=c_{1}e^{-3t}+c_{2}-(t^2)/6-(5t)/9

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-x^{''}-3x^'-2=t

- x^{^{''}} - 3 x^{^{'}} - 2 = t

x = c_{1} e^{- 3 t} + c_{2} - \frac{t ^{2}}{6} - \frac{5 t}{9}

Solution steps

- x^{''} (t) - 3 x^{'} (t) - 2 = t

Solve linear ODE:

x = c_{1} e^{- 3 t} + c_{2} - \frac{t ^{2}}{6} - \frac{5 t}{9}

x = c_{1} e^{- 3 t} + c_{2} - \frac{t ^{2}}{6} - \frac{5 t}{9}

y^{^{''}} - y = - 4 t + 3

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

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

y^{^{''}} + 49 y = cos (7 t), y (0) = - 5, y^{^{'}} (0) = - 56

y^{^{''}} + 5 y + 4 = 0, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for -x^{''}-3x^'-2=t ?
The solution for -x^{''}-3x^'-2=t is x=c_{1}e^{-3t}+c_{2}-(t^2)/6-(5t)/9