What is the solution for y^{''''}+3y^{'''}+4y^{''}+12y^'=0 ?

The solution for y^{''''}+3y^{'''}+4y^{''}+12y^'=0 is y=c_{1}+c_{2}e^{-3t}+c_{3}cos(2t)+c_{4}sin(2t)

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y^{''''}+3y^{'''}+4y^{''}+12y^'=0

y^{^{''''}} + 3 y^{^{'''}} + 4 y^{^{''}} + 12 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- 3 t} + c_{3} cos (2 t) + c_{4} sin (2 t)

Solution steps

y^{(4)} (t) + 3 y^{'''} (t) + 4 y^{''} (t) + 12 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} + c_{3} cos (2 t) + c_{4} sin (2 t)

y = c_{1} + c_{2} e^{- 3 t} + c_{3} cos (2 t) + c_{4} sin (2 t)

(D^{4} - D^{3} - 3 D + D + 2) y = 0

81 y^{^{''}} = - 16 y, y (0) = cos (k t)

3 y^{^{''}} - 15 y^{^{'}} + 18 y = 0

y^{^{'''}} + 5 y^{^{''}} - 4 y^{^{'}} - 60 y = 0

25 y^{^{''}} + 10 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 1

What is the solution for y^{''''}+3y^{'''}+4y^{''}+12y^'=0 ?
The solution for y^{''''}+3y^{'''}+4y^{''}+12y^'=0 is y=c_{1}+c_{2}e^{-3t}+c_{3}cos(2t)+c_{4}sin(2t)