What is the solution for 2x^{''}+8x^'+52x=0 ?

The solution for 2x^{''}+8x^'+52x=0 is x=e^{-2t}(c_{1}cos(sqrt(22)t)+c_{2}sin(sqrt(22)t))

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2x^{''}+8x^'+52x=0

2 x^{^{''}} + 8 x^{^{'}} + 52 x = 0

x = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

Solution steps

2 x^{''} (t) + 8 x^{'} (t) + 52 x = 0

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

x = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

y^{^{'''}} + 4 y^{^{''}} - 7 y^{^{'}} - 10 y = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 9, y (1) = 2

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

(y^{^{''''}} + 8 y^{^{''}} + 16 y)^{2} = 0

What is the solution for 2x^{''}+8x^'+52x=0 ?
The solution for 2x^{''}+8x^'+52x=0 is x=e^{-2t}(c_{1}cos(sqrt(22)t)+c_{2}sin(sqrt(22)t))