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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 392 y = cos (2 t)

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 3 t^{2} - 18

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x), y (0) = 0

y^{^{''}} + y^{^{'}} + y 2 = sin (x^{2})

10 x^{^{''}} + 45 x^{^{'}} + 2500 x = 180 cos (7 π)

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