What is the solution for y^{''}-2y^'-3y=3t^2+4t-5 ?

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

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y^{''}-2y^'-3y=3t^2+4t-5

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t^{2} + 4 t - 5

y = c_{1} e^{3 t} + c_{2} e^{- t} - t^{2} + 1

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 3 t^{2} + 4 t - 5

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - t^{2} + 1

y = c_{1} e^{3 t} + c_{2} e^{- t} - t^{2} + 1

y^{^{''}} - y = 7.1 cos (t), y (0) = 0, y^{^{'}} (0) = 7.1

y^{^{''}} - 4 y^{^{'}} + 3 y = 8 e^{x}

y^{^{''}} + 3 y = t^{3}, y (0) = 0, y^{^{'}} (0) = 0

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

(D^{2} + 1) y = sec^{2} (x)

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