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

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

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

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

y = c_{1} e^{3 t} + c_{2} e^{- t} - t^{2} + \frac{4 t}{3} - \frac{14}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - t^{2} + \frac{4 t}{3} - \frac{14}{9}

y = c_{1} e^{3 t} + c_{2} e^{- t} - t^{2} + \frac{4 t}{3} - \frac{14}{9}

(D^{2} - 2 D + 3) y = x^{3} + sin (x)

\frac{d ^{2} y}{d x ^{2}} + 9 y = sec (3 x)

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} cos (2 x)

y^{^{''}} + y = 16 cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = 2 x - 5

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