What is the solution for y^{''}+6y^'+18y=0 ?

The solution for y^{''}+6y^'+18y=0 is y=e^{-3t}(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}+6y^'+18y=0

y^{^{''}} + 6 y^{^{'}} + 18 y = 0

y = e^{- 3 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- 3 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y \to 0 lim (cot (5 y))

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{2}})

x \to 2 lim (\frac{4}{x - 2})

t \to 0 lim (e^{- 2 t})

\frac{d}{d x} (\frac{2 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}})

What is the solution for y^{''}+6y^'+18y=0 ?
The solution for y^{''}+6y^'+18y=0 is y=e^{-3t}(c_{1}cos(3t)+c_{2}sin(3t))