What is the solution for y^{''}-4y^'-12y=2t^3-t+3 ?

The solution for y^{''}-4y^'-12y=2t^3-t+3 is y=c_{1}e^{6t}+c_{2}e^{-2t}-(t^3)/6+(t^2)/6-t/9-5/27

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y^{''}-4y^'-12y=2t^3-t+3

y^{^{''}} - 4 y^{^{'}} - 12 y = 2 t^{3} - t + 3

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{t ^{3}}{6} + \frac{t ^{2}}{6} - \frac{t}{9} - \frac{5}{27}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = 2 t^{3} - t + 3

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{t ^{3}}{6} + \frac{t ^{2}}{6} - \frac{t}{9} - \frac{5}{27}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{t ^{3}}{6} + \frac{t ^{2}}{6} - \frac{t}{9} - \frac{5}{27}

y^{^{''}} + y^{^{'}} + y = x \cdot e^{x}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = 4 x^{2}

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

y^{^{''}} + y = 3 cos (2 x)

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (t)

What is the solution for y^{''}-4y^'-12y=2t^3-t+3 ?
The solution for y^{''}-4y^'-12y=2t^3-t+3 is y=c_{1}e^{6t}+c_{2}e^{-2t}-(t^3)/6+(t^2)/6-t/9-5/27