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))

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))