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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 5 y^{^{''}} + 4 y = 0

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

y^{^{'''}} + 5 y^{^{''}} - 25 y^{^{'}} - 125 y = 0

35 y^{^{''''}} + 237 y^{^{''}} + 400 y = 0

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 400 y = 0

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