What is the solution for y^{''}+6y^'+8y=0,y(0)=9 ?

The solution for y^{''}+6y^'+8y=0,y(0)=9 is y=(9-c_{2})e^{-2t}+c_{2}e^{-4t}

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y^{''}+6y^'+8y=0,y(0)=9

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 9

y = (9 - c_{2}) e^{- 2 t} + c_{2} e^{- 4 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = (9 - c_{2}) e^{- 2 t} + c_{2} e^{- 4 t}

y = (9 - c_{2}) e^{- 2 t} + c_{2} e^{- 4 t}

cos (e^{- 4 θ^{5}})

\frac{d}{d x} (\frac{9 x ^{2} - 6}{x ^{2}})

\frac{\partial}{\partial x} (\frac{4}{1 + x ^{2}})

\frac{d}{d x} (1 - x y cos (π y))

f (x) = 5^{x} ln (5)

What is the solution for y^{''}+6y^'+8y=0,y(0)=9 ?
The solution for y^{''}+6y^'+8y=0,y(0)=9 is y=(9-c_{2})e^{-2t}+c_{2}e^{-4t}