What is the solution for y^{''}+4y^'+40y=0 ?

The solution for y^{''}+4y^'+40y=0 is y=e^{-2t}(c_{1}cos(6t)+c_{2}sin(6t))

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y^{''}+4y^'+40y=0

y^{^{''}} + 4 y^{^{'}} + 40 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 40 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 12 y^{^{''}} + 48 y^{^{'}} - 64 y = 0

x^{^{''}} + 4 x^{^{'}} + 20 x = 0

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

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

(D^{4} - 2 D^{3} + 2 D^{2} - 2 D + 1) y = 0

What is the solution for y^{''}+4y^'+40y=0 ?
The solution for y^{''}+4y^'+40y=0 is y=e^{-2t}(c_{1}cos(6t)+c_{2}sin(6t))