What is the solution for y^{'''}+2y^{''}+4y^'+8y=0 ?

The solution for y^{'''}+2y^{''}+4y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}cos(2t)+c_{3}sin(2t)

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y^{'''}+2y^{''}+4y^'+8y=0

y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} + 8 y = 0

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

Solution steps

y^{'''} (t) + 2 y^{''} (t) + 4 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = - 2, y^{^{'}} (0) = 5

y^{^{''}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 8

2 x^{^{''}} + 4 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1

What is the solution for y^{'''}+2y^{''}+4y^'+8y=0 ?
The solution for y^{'''}+2y^{''}+4y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}cos(2t)+c_{3}sin(2t)