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

The solution for y^{''}+2*y^'+4*y=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 \cdot y^{^{'}} + 4 \cdot y = 0

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

Solution steps

y^{''} (t) + 2 \cdot y^{'} (t) + 4 \cdot 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))

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

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (1) = 2, y^{^{'}} (1) = \frac{1}{6}

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

x^{^{''}} - 0.1 x^{^{'}} + 1.0025 x = 0

y^{^{'''}} + 6 y = 0

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