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))

y^{^{'''}} + 36 y^{^{'}} = 0, y^{^{'}} (0) = 6

y^{^{'''}} - 5 y^{^{''}} - 2 y^{^{'}} + 24 y = 0

f^{^{''}} (x) + f (x) = 0, f (0) = 1

f^{^{''}} (x) + f (x) = 0, f (0) = 0

\frac{d ^{2} x}{d t ^{2}} - 5 \frac{d x}{d t} + 4 x = 0

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))