What is the solution for x^{''}+2x^'+8x=0 ?

The solution for x^{''}+2x^'+8x=0 is x=e^{-t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))

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x^{''}+2x^'+8x=0

x^{^{''}} + 2 x^{^{'}} + 8 x = 0

x = e^{- t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

x^{''} (t) + 2 x^{'} (t) + 8 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (7 t) + c_{2} sin (7 t))

x = e^{- t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y^{^{'''}} - y^{^{''}} + 9 y^{^{'}} - 9 y = 0

y^{^{''}} - 12 y^{^{'}} + 136 y = 0

256 y^{^{''''}} = 16 y

y^{^{''}} - 7 y^{^{'}} - 4 y = 0

y^{^{''}} - 15 y^{^{'}} + 50 y = 0, y (0) = 8, y^{^{'}} (0) = - 55

What is the solution for x^{''}+2x^'+8x=0 ?
The solution for x^{''}+2x^'+8x=0 is x=e^{-t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))