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

The solution for 3y^{''}-2y-y=2t^2-t-3e^{-3t} is y=c_{1}e^t+c_{2}e^{-t}+(-2t^2+t-4)/3-1/8 e^{-3t}

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3y^{''}-2y-y=2t^2-t-3e^{-3t}

3 y^{^{''}} - 2 y - y = 2 t^{2} - t - 3 e^{- 3 t}

y = c_{1} e^{t} + c_{2} e^{- t} + \frac{- 2 t ^{2} + t - 4}{3} - \frac{1}{8} e^{- 3 t}

Solution steps

3 y^{''} (t) - 2 y - y = 2 t^{2} - t - 3 e^{- 3 t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + \frac{- 2 t ^{2} + t - 4}{3} - \frac{1}{8} e^{- 3 t}

y = c_{1} e^{t} + c_{2} e^{- t} + \frac{- 2 t ^{2} + t - 4}{3} - \frac{1}{8} e^{- 3 t}

What is the solution for 3y^{''}-2y-y=2t^2-t-3e^{-3t} ?
The solution for 3y^{''}-2y-y=2t^2-t-3e^{-3t} is y=c_{1}e^t+c_{2}e^{-t}+(-2t^2+t-4)/3-1/8 e^{-3t}