What is the solution for x^{''}+6x^'+8x=0 ?

The solution for x^{''}+6x^'+8x=0 is x=c_{1}e^{-2t}+c_{2}e^{-4t}

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x^{''}+6x^'+8x=0

x^{^{''}} + 6 x^{^{'}} + 8 x = 0

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 6

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for x^{''}+6x^'+8x=0 ?
The solution for x^{''}+6x^'+8x=0 is x=c_{1}e^{-2t}+c_{2}e^{-4t}