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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

q^{^{''}} + 0.1 q^{^{'}} + q = v sin (0.5 t), q (0) = 0

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 30 e^{2 t}

y^{^{''}} + 25 y = 14 cos (5 x) + 21 sin (5 x)

y^{^{''}} - y = t, y (0) = 0, y^{^{'}} (0) = 1

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