What is the solution for y^{''}+9y^'+18y=0,y(0)=1,y^'(0)=0 ?

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

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y^{''}+9y^'+18y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 9 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = 2 e^{- 3 t} - e^{- 6 t}

Solution steps

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

Solve linear ODE:

y = 2 e^{- 3 t} - e^{- 6 t}

y = 2 e^{- 3 t} - e^{- 6 t}

\frac{\partial}{\partial x} (x^{3} + x e^{y} - 2 y)

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

\int \frac{x ^{2}}{( x + 1 )} d x

\frac{d}{d x} (- e^{x} cos (x))

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