What is the solution for x^{''}+12x^'+40x=0 ?

The solution for x^{''}+12x^'+40x=0 is x=e^{-6t}(c_{1}cos(2t)+c_{2}sin(2t))

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x^{''}+12x^'+40x=0

x^{^{''}} + 12 x^{^{'}} + 40 x = 0

x = e^{- 6 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

x^{''} (t) + 12 x^{'} (t) + 40 x = 0

Solve linear ODE:

x = e^{- 6 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

x = e^{- 6 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y^{^{'''}} - 3 y^{^{''}} - 13 y^{^{'}} + 15 y = 0

y^{^{''}} - 3 y = 0, y (0) = 13, y^{^{'}} (0) = 0

27 y^{^{''}} + (15 - 95) y^{^{'}} - 55 y = 0

324 y^{^{''}} - 612 y^{^{'}} + 289 y = 0, y (0) = 8, y^{^{'}} (0) = 5

y^{^{''}} - 13 y - 40 y = 0

What is the solution for x^{''}+12x^'+40x=0 ?
The solution for x^{''}+12x^'+40x=0 is x=e^{-6t}(c_{1}cos(2t)+c_{2}sin(2t))