What is the solution for (y^{''}+4y^'+20y)^2=0 ?

The solution for (y^{''}+4y^'+20y)^2=0 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))

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(y^{''}+4y^'+20y)^2=0

(y^{^{''}} + 4 y^{^{'}} + 20 y)^{2} = 0

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

Solution steps

(y^{''} (t) + 4 y^{'} (t) + 20 y)^{2} = 0

Solve linear ODE:

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

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

What is the solution for (y^{''}+4y^'+20y)^2=0 ?
The solution for (y^{''}+4y^'+20y)^2=0 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))