What is the solution for y^{''}+9y^'+18y=0 ?

The solution for y^{''}+9y^'+18y=0 is y=c_{1}e^{-3t}+c_{2}e^{-6t}

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y^{''}+9y^'+18y=0

y^{^{''}} + 9 y^{^{'}} + 18 y = 0

y = c_{1} e^{- 3 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) + 9 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = c_{1} e^{- 3 t} + c_{2} e^{- 6 t}

y = c_{1} e^{- 3 t} + c_{2} e^{- 6 t}

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x}

\frac{\partial}{\partial x} ((1 + x y) e^{x y})

h \to 0 lim (2 x^{2} + 3 x h + 2)

y^{^{'}} = \frac{( 3 e ^{x} )}{e ^{y}}

x tan (\frac{y}{x}) + y - x y^{^{'}} = 0

What is the solution for y^{''}+9y^'+18y=0 ?
The solution for y^{''}+9y^'+18y=0 is y=c_{1}e^{-3t}+c_{2}e^{-6t}