What is the solution for 2*y^{''}+4*y^'+8y=0 ?

The solution for 2*y^{''}+4*y^'+8y=0 is y=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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2y^{''}+4y^'+8y=0

2 \cdot y^{^{''}} + 4 \cdot y^{^{'}} + 8 y = 0

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

Solution steps

2 \cdot y^{''} (t) + 4 \cdot y^{'} (t) + 8 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))

5 y^{^{''}} + 18 y^{^{'}} + 9 y = 0, y (0) = 17.4

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (1) = 3

y^{^{''}} + 96 y = 0

2 y^{^{''}} + 3 y^{^{'}} + 5 y = 0

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 2*y^{''}+4*y^'+8y=0 ?
The solution for 2*y^{''}+4*y^'+8y=0 is y=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))