What is the solution for y^{''''}+2y^{'''}+2y^{''}=0 ?

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

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y^{''''}+2y^{'''}+2y^{''}=0

y^{^{''''}} + 2 y^{^{'''}} + 2 y^{^{''}} = 0

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

Solution steps

y^{(4)} (t) + 2 y^{'''} (t) + 2 y^{''} (t) = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 25 y = 0

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0, y (0) = 2

y^{^{''}} - 9 y = 0, y (0) = 2, y^{^{'}} (0) = 4

(4 D^{2} - 3 D - 10) y = 0

x^{^{'''}} - 4 x^{^{''}} - 3 x^{^{'}} + 18 x = 0

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