What is the solution for 0.5y^{''}+2y^'+6y=0 ?

The solution for 0.5y^{''}+2y^'+6y=0 is y=e^{-2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

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0.5y^{''}+2y^'+6y=0

0.5 y^{^{''}} + 2 y^{^{'}} + 6 y = 0

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

Solution steps

0.5 y^{''} (t) + 2 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 2 y^{^{''}} - 15 y^{^{'}} = 0

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

y^{^{''}} - 5 y^{^{'}} - 24 y = 0, y (0) = 7, y^{^{'}} (0) = - 3

81 y^{^{''}} + 36 y^{^{'}} + 40 y = 0, y (0) = 6, y^{^{'}} (0) = 7

2 y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 0, y (2) = 1

What is the solution for 0.5y^{''}+2y^'+6y=0 ?
The solution for 0.5y^{''}+2y^'+6y=0 is y=e^{-2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))