What is the solution for 20y^{''}+240y^'+800y=0 ?

The solution for 20y^{''}+240y^'+800y=0 is y=e^{-6t}(c_{1}cos(2t)+c_{2}sin(2t))

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20y^{''}+240y^'+800y=0

20 y^{^{''}} + 240 y^{^{'}} + 800 y = 0

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

Solution steps

20 y^{''} (t) + 240 y^{'} (t) + 800 y = 0

Solve linear ODE:

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

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

a y^{^{''}} + b y^{^{'}} - cy = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 3, y (0) = 5

y^{^{''}} + 1 y = 0, y (\frac{π}{2}) = 2, y^{^{'}} (\frac{π}{2}) = 3

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{5}{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 15, y (1) = 0

What is the solution for 20y^{''}+240y^'+800y=0 ?
The solution for 20y^{''}+240y^'+800y=0 is y=e^{-6t}(c_{1}cos(2t)+c_{2}sin(2t))