What is the solution for 0.5y^{''}+6y^'+50y=0 ?

The solution for 0.5y^{''}+6y^'+50y=0 is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))

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0.5y^{''}+6y^'+50y=0

0.5 y^{^{''}} + 6 y^{^{'}} + 50 y = 0

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

Solution steps

0.5 y^{''} (t) + 6 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t))

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

\frac{d ^{3} y}{d x ^{3}} = \frac{d y}{d x}

16 y^{^{''''}} = 625 y

y^{^{''}} + 64 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - \frac{4}{3}

(4 D^{3} + 4 D^{2} + D) y = 0

What is the solution for 0.5y^{''}+6y^'+50y=0 ?
The solution for 0.5y^{''}+6y^'+50y=0 is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))