What is the solution for linear 2y^{''}+16y^'+82y=0 ?

The solution for linear 2y^{''}+16y^'+82y=0 is y=e^{-4t}(c_{1}cos(5t)+c_{2}sin(5t))

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linear 2y^{''}+16y^'+82y=0

linear

2 y^{^{''}} + 16 y^{^{'}} + 82 y = 0

y = e^{- 4 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

2 y^{''} (t) + 16 y^{'} (t) + 82 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{- 4 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

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

121 y^{^{''}} + 198 y^{^{'}} + 80 y = 0

2 y^{^{'''}} + 7 y^{^{''}} + 4 y^{^{'}} - 4 y = 0

y^{^{''}} + 25 y^{^{'}} + 2500 y = 0

3 y^{^{''}} - 2 y + 4 y^{^{'}} = 0

What is the solution for linear 2y^{''}+16y^'+82y=0 ?
The solution for linear 2y^{''}+16y^'+82y=0 is y=e^{-4t}(c_{1}cos(5t)+c_{2}sin(5t))