What is the solution for x^{''}+4x^'+29x=0 ?

The solution for x^{''}+4x^'+29x=0 is x=e^{-2t}(c_{1}cos(5t)+c_{2}sin(5t))

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x^{''}+4x^'+29x=0

x^{^{''}} + 4 x^{^{'}} + 29 x = 0

x = e^{- 2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

x^{''} (t) + 4 x^{'} (t) + 29 x = 0

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

x = e^{- 2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y^{^{'''}} + 2 y^{^{''}} - 6 y^{^{'}} - 72 y = 0, y (0) = e^{4 x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 2

y^{^{''''}} + 4 y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{33}{4}

81 y^{^{''}} = - 16 y

What is the solution for x^{''}+4x^'+29x=0 ?
The solution for x^{''}+4x^'+29x=0 is x=e^{-2t}(c_{1}cos(5t)+c_{2}sin(5t))