What is the solution for y^{''}+8y^'+160y=0 ?

The solution for y^{''}+8y^'+160y=0 is y=e^{-4t}(c_{1}cos(12t)+c_{2}sin(12t))

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y^{''}+8y^'+160y=0

y^{^{''}} + 8 y^{^{'}} + 160 y = 0

y = e^{- 4 t} (c_{1} cos (12 t) + c_{2} sin (12 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 160 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (12 t) + c_{2} sin (12 t))

y = e^{- 4 t} (c_{1} cos (12 t) + c_{2} sin (12 t))

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

y^{^{''}} + 0.2 y^{^{'}} + 0.01 y = 0

y^{^{''}} + 9 y^{^{'}} - 20 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

4 y^{^{''}} - 68 y^{^{'}} + 289 y = 0

What is the solution for y^{''}+8y^'+160y=0 ?
The solution for y^{''}+8y^'+160y=0 is y=e^{-4t}(c_{1}cos(12t)+c_{2}sin(12t))