What is the solution for (y^{''}+2y^'+5y)^3=0 ?

The solution for (y^{''}+2y^'+5y)^3=0 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))

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(y^{''}+2y^'+5y)^3=0

(y^{^{''}} + 2 y^{^{'}} + 5 y)^{3} = 0

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

Solution steps

(y^{''} (t) + 2 y^{'} (t) + 5 y)^{3} = 0

Solve linear ODE:

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

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

5 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

y^{^{''}} - 22 y^{^{'}} + 2 y = 0

1 y^{^{''}} + 10 y^{^{'}} + \frac{32}{10} y = 0

y^{^{''}} - 20 y^{^{'}} - a y = 0, y (0) = 1, y^{^{'}} (0) = 15

y^{^{''}} - y^{^{'}} + \frac{y}{4} = 0

What is the solution for (y^{''}+2y^'+5y)^3=0 ?
The solution for (y^{''}+2y^'+5y)^3=0 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))