What is the solution for y^{''}-2y^'-3y=3t^2-5 ?

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

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y^{''}-2y^'-3y=3t^2-5

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t^{2} - 5

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 3 t^{2} - 5

Solve linear ODE:

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

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

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

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

y^{^{''}} + y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

3 y^{^{''}} + 5 y^{^{'}} + 2 y = t + e^{t}

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

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