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

The solution for y^{''}+2y^'+100y=0 is y=e^{-t}(c_{1}cos(3sqrt(11)t)+c_{2}sin(3sqrt(11)t))

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

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

y = e^{- t} (c_{1} cos (311 t) + c_{2} sin (311 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (311 t) + c_{2} sin (311 t))

y = e^{- t} (c_{1} cos (311 t) + c_{2} sin (311 t))

\frac{24}{32} x^{^{''}} + 6 x = 0, x (0) = 0.5, x^{^{'}} (0) = 0

\frac{d ^{4} y}{d x ^{4}} - \frac{d y}{d x} = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

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

(D^{3} - 6 D^{2} + 11 D - 6) y = 0

What is the solution for y^{''}+2y^'+100y=0 ?
The solution for y^{''}+2y^'+100y=0 is y=e^{-t}(c_{1}cos(3sqrt(11)t)+c_{2}sin(3sqrt(11)t))