What is the solution for y^{''}+40y^'+4000y=0 ?

The solution for y^{''}+40y^'+4000y=0 is y=e^{-20t}(c_{1}cos(60t)+c_{2}sin(60t))

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y^{''}+40y^'+4000y=0

y^{^{''}} + 40 y^{^{'}} + 4000 y = 0

y = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

Solution steps

y^{''} (t) + 40 y^{'} (t) + 4000 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

y = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

y^{^{''}} + 20 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

u^{^{''}} + a^{2} H (t) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.459

25 x^{^{''}} + 10 x^{^{'}} + 226 x = 0

What is the solution for y^{''}+40y^'+4000y=0 ?
The solution for y^{''}+40y^'+4000y=0 is y=e^{-20t}(c_{1}cos(60t)+c_{2}sin(60t))