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

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

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

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 0.1 x^{^{'}} + 1.0025 x = 0

y^{^{'''}} + 6 y = 0

\frac{1}{2} y^{^{''}} + 20 y = 0

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

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