What is the solution for x^{'''}+x^{''}+4x^'+4x=0 ?

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

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

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

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

Solution steps

x^{'''} (t) + x^{''} (t) + 4 x^{'} (t) + 4 x = 0

Solve linear ODE:

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

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

y^{^{'''}} + 3 y^{^{''}} - 10 y^{^{'}} = 0

2 y^{^{''}} - 9 y^{^{'}} + 9 y = 0

81 y^{^{''}} - 18 y^{^{'}} + 5 y = 0, y (1) = 1, y^{^{'}} (1) = 7

9 x^{^{''}} + 9 x^{^{'}} - 4 x = 0

x^{^{''}} = a^{2} x

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