What is the solution for y^{'''}+6y^{''}+13y^'=0 ?

The solution for y^{'''}+6y^{''}+13y^'=0 is y=c_{1}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+6y^{''}+13y^'=0

y^{^{'''}} + 6 y^{^{''}} + 13 y^{^{'}} = 0

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

Solution steps

y^{'''} (t) + 6 y^{''} (t) + 13 y^{'} (t) = 0

Solve linear ODE:

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

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

0.2 y^{^{''}} + 5 y^{^{'}} + \frac{1}{0.002} y = 0

y^{^{''}} + 3 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 8

y^{^{'''}} + 6 y^{^{''}} - 15 y^{^{'}} - 100 y = 0

y^{^{''''}} + 4 y^{^{'''}} + 10 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

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

What is the solution for y^{'''}+6y^{''}+13y^'=0 ?
The solution for y^{'''}+6y^{''}+13y^'=0 is y=c_{1}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))