What is the solution for y^{''}+40y^'+490y=0 ?

The solution for y^{''}+40y^'+490y=0 is y=e^{-20t}(c_{1}cos(3sqrt(10)t)+c_{2}sin(3sqrt(10)t))

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y^{''}+40y^'+490y=0

y^{^{''}} + 40 y^{^{'}} + 490 y = 0

y = e^{- 20 t} (c_{1} cos (310 t) + c_{2} sin (310 t))

Solution steps

y^{''} (t) + 40 y^{'} (t) + 490 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (310 t) + c_{2} sin (310 t))

y = e^{- 20 t} (c_{1} cos (310 t) + c_{2} sin (310 t))

y^{^{''}} - 3 y^{^{'}} + 1 y = 0

y^{^{''}} - 9 y = 0, y (0) = 3, y^{^{'}} (0) = - 9

y^{^{''}} - 25 y^{^{'}} + 144 y = 0, y^{^{'}} (0) = 5

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.455

3 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

What is the solution for y^{''}+40y^'+490y=0 ?
The solution for y^{''}+40y^'+490y=0 is y=e^{-20t}(c_{1}cos(3sqrt(10)t)+c_{2}sin(3sqrt(10)t))