What is the solution for 3y^{''''}+y^{'''}=2 ?

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

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3y^{''''}+y^{'''}=2

3 y^{^{''''}} + y^{^{'''}} = 2

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

Solution steps

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

Solve linear ODE:

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

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

(D + 2) (D - 1)^{3} y = e^{x}

y^{^{''}} + 16 y = 5 sin (4 x)

y^{^{''}} + 9 y = cos (x), y (\frac{π}{2}) = - \frac{5}{3}, y^{^{'}} (0) = 5

(D^{2} - 3 D + 2) y = x + 1

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

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