What is the solution for 1/4 y^{''}+2y^'+400y=0 ?

The solution for 1/4 y^{''}+2y^'+400y=0 is y=e^{-4t}(c_{1}cos(12sqrt(11)t)+c_{2}sin(12sqrt(11)t))

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1/4 y^{''}+2y^'+400y=0

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 400 y = 0

y = e^{- 4 t} (c_{1} cos (1211 t) + c_{2} sin (1211 t))

Solution steps

\frac{1}{4} y^{''} (t) + 2 y^{'} (t) + 400 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (1211 t) + c_{2} sin (1211 t))

y = e^{- 4 t} (c_{1} cos (1211 t) + c_{2} sin (1211 t))

y^{^{'''}} - y^{^{''}} - 3 y^{^{'}} - y = 0

9 y^{^{''}} - 10 y^{^{'}} + y = 0, y (0) = 9, y^{^{'}} (0) = \frac{1}{9}

3 y^{^{''}} + 12 y^{^{'}} + 24 y = 0, y (0) = 1, y^{^{'}} (0) = 4

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

y^{^{'''}} + 3 y^{^{''}} - 9 y^{^{'}} + 5 y = 0

What is the solution for 1/4 y^{''}+2y^'+400y=0 ?
The solution for 1/4 y^{''}+2y^'+400y=0 is y=e^{-4t}(c_{1}cos(12sqrt(11)t)+c_{2}sin(12sqrt(11)t))