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

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

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y^{''}+2y^'+4y=0

y^{^{''}} + 2 y^{^{'}} + 4 y = 0

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

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