What is the solution for {(d^2y)/(dx^2)+4(dy)/(dx)+20y}^2=0 ?

The solution for {(d^2y)/(dx^2)+4(dy)/(dx)+20y}^2=0 is y=e^{-2x}(c_{1}cos(4x)+c_{2}sin(4x))

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{(d^2y)/(dx^2)+4(dy)/(dx)+20y}^2=0

{\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 20 y}^{2} = 0

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

Solution steps

{\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 20 y}^{2} = 0

Solve linear ODE:

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

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

3 y^{^{'''}} + 17 y^{^{''}} + 31 y^{^{'}} + 14 y = 0

2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

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

\frac{d ^{2} f}{d x ^{2}} = a^{2} f (x)

y^{^{'}} = y^{^{''}}

What is the solution for {(d^2y)/(dx^2)+4(dy)/(dx)+20y}^2=0 ?
The solution for {(d^2y)/(dx^2)+4(dy)/(dx)+20y}^2=0 is y=e^{-2x}(c_{1}cos(4x)+c_{2}sin(4x))