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

The solution for 1/4 y^{''}+y^'+4y=0 is y=e^{-2t}(c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t))

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

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

y = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

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

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

15 y^{^{''''}} + 128 y^{^{''}} + 25 y = 0

y^{^{''}} + 8 y^{^{'}} + 7 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

What is the solution for 1/4 y^{''}+y^'+4y=0 ?
The solution for 1/4 y^{''}+y^'+4y=0 is y=e^{-2t}(c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t))