What is the solution for y^{''}-2y^'-3y=9t^2 ?

The solution for y^{''}-2y^'-3y=9t^2 is y=c_{1}e^{3t}+c_{2}e^{-t}-3t^2+4t-14/3

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y^{''}-2y^'-3y=9t^2

y^{^{''}} - 2 y^{^{'}} - 3 y = 9 t^{2}

y = c_{1} e^{3 t} + c_{2} e^{- t} - 3 t^{2} + 4 t - \frac{14}{3}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 9 t^{2}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - 3 t^{2} + 4 t - \frac{14}{3}

y = c_{1} e^{3 t} + c_{2} e^{- t} - 3 t^{2} + 4 t - \frac{14}{3}

y^{^{''}} - 3 y^{^{'}} - 40 y = (t + 1) e^{t}

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} \cdot e^{2 x} - e^{2 x}

y^{^{''}} + 0.7 y + 4 = cos (3 t)

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

What is the solution for y^{''}-2y^'-3y=9t^2 ?
The solution for y^{''}-2y^'-3y=9t^2 is y=c_{1}e^{3t}+c_{2}e^{-t}-3t^2+4t-14/3