What is the solution for x^{''}+8x^'+80x=0 ?

The solution for x^{''}+8x^'+80x=0 is x=e^{-4t}(c_{1}cos(8t)+c_{2}sin(8t))

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x^{''}+8x^'+80x=0

x^{^{''}} + 8 x^{^{'}} + 80 x = 0

x = e^{- 4 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

Solution steps

x^{''} (t) + 8 x^{'} (t) + 80 x = 0

Solve linear ODE:

x = e^{- 4 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

x = e^{- 4 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 5, y^{^{'}} (- 1) = 4

r^{^{''}} - 4 r^{^{'}} - 53 r = 0

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

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

y^{^{''''}} + y^{^{'''}} - 2 y^{^{''}} = 0

What is the solution for x^{''}+8x^'+80x=0 ?
The solution for x^{''}+8x^'+80x=0 is x=e^{-4t}(c_{1}cos(8t)+c_{2}sin(8t))