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)

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)