What is the solution for y^{''}+2y^'+4y=0,y(0)=1 ?

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

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y^{''}+2y^'+4y=0,y(0)=1

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 7

y^{^{''''}} - 4 y^{^{'''}} + 8 y^{^{''}} - 16 y^{^{'}} + 16 y = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

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