What is the solution for 10y^{''}+80y^'+200y=0 ?

The solution for 10y^{''}+80y^'+200y=0 is y=e^{-4t}(c_{1}cos(2t)+c_{2}sin(2t))

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10y^{''}+80y^'+200y=0

10 y^{^{''}} + 80 y^{^{'}} + 200 y = 0

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

Solution steps

10 y^{''} (t) + 80 y^{'} (t) + 200 y = 0

Solve linear ODE:

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

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

f (x) = \frac{x}{x ^{2} + 2}

\frac{d}{d r} (6 ((6 r - 1) \cdot k + r) - 1)

\frac{\partial}{\partial x} (ln (\frac{x}{x + y}))

n = 1 \sum \infty \frac{1}{( n + 2 ) ^{2}}

y = \frac{3}{16 - x}, at x = 0

What is the solution for 10y^{''}+80y^'+200y=0 ?
The solution for 10y^{''}+80y^'+200y=0 is y=e^{-4t}(c_{1}cos(2t)+c_{2}sin(2t))