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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

c y^{^{''}} + ab y^{^{'}} = 0

3 y^{^{''}} + 35 y = 0, y (0) = 2, y^{^{'}} (0) = 9

(D^{3} + D^{2} - 2 D + 12) y = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + y = 0, y (π) = 1, y (π) = - 5

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