What is the solution for y^{'''}+4y^{''}+13y^'=0 ?

The solution for y^{'''}+4y^{''}+13y^'=0 is y=c_{1}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))

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y^{'''}+4y^{''}+13y^'=0

y^{^{'''}} + 4 y^{^{''}} + 13 y^{^{'}} = 0

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

Solution steps

y^{'''} (t) + 4 y^{''} (t) + 13 y^{'} (t) = 0

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 3, y^{^{'}} (0) = 2

y^{^{''''}} + 3 y^{^{'''}} + 4 y^{^{''}} + 12 y^{^{'}} = 0

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

81 y^{^{''}} = - 16 y, y (0) = cos (k t)

3 y^{^{''}} - 15 y^{^{'}} + 18 y = 0

What is the solution for y^{'''}+4y^{''}+13y^'=0 ?
The solution for y^{'''}+4y^{''}+13y^'=0 is y=c_{1}+e^{-2t}(c_{2}cos(3t)+c_{3}sin(3t))