What is the solution for y^{''}-2y^'-8y=0 ?

The solution for y^{''}-2y^'-8y=0 is y=c_{1}e^{4t}+c_{2}e^{-2t}

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y^{''}-2y^'-8y=0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- 2 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 2 t}

y = c_{1} e^{4 t} + c_{2} e^{- 2 t}

\frac{d}{d x} ((\frac{3}{2})^{x} ln (\frac{3}{2}))

\int x^{3} 2 + x^{4} d x

\frac{\partial}{\partial y} (x^{2} + 3 x - 4 x y)

\int - (\frac{1}{x}) d x

\int - \frac{1}{t} d t

What is the solution for y^{''}-2y^'-8y=0 ?
The solution for y^{''}-2y^'-8y=0 is y=c_{1}e^{4t}+c_{2}e^{-2t}