What is the solution for 0.05y^{''}+2y^'+100y=0 ?

The solution for 0.05y^{''}+2y^'+100y=0 is y=e^{-20t}(c_{1}cos(40t)+c_{2}sin(40t))

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0.05y^{''}+2y^'+100y=0

0.05 y^{^{''}} + 2 y^{^{'}} + 100 y = 0

y = e^{- 20 t} (c_{1} cos (40 t) + c_{2} sin (40 t))

Solution steps

0.05 y^{''} (t) + 2 y^{'} (t) + 100 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (40 t) + c_{2} sin (40 t))

y = e^{- 20 t} (c_{1} cos (40 t) + c_{2} sin (40 t))

x^{^{''}} + \frac{0}{\frac{1}{32}} x^{^{'}} + \frac{13}{\frac{1}{32}} x = 0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (π) = 5, y (0) = 6

(D^{3} - 4 D^{2} + 3 D) y = 0

1.5 y^{^{''}} + 12 y^{^{'}} + 24 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y^{0} = 0, y^{^{'}} = 3

What is the solution for 0.05y^{''}+2y^'+100y=0 ?
The solution for 0.05y^{''}+2y^'+100y=0 is y=e^{-20t}(c_{1}cos(40t)+c_{2}sin(40t))