What is the solution for x^{''}+8x^'+4x=0 ?

The solution for x^{''}+8x^'+4x=0 is x=c_{1}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}

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x^{''}+8x^'+4x=0

x^{^{''}} + 8 x^{^{'}} + 4 x = 0

x = c_{1} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t}

Solution steps

x^{''} (t) + 8 x^{'} (t) + 4 x = 0

Solve linear ODE:

x = c_{1} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t}

x = c_{1} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t}

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = 0

4 y + 9 y^{^{''}} = - 12 y^{^{'}}

9 y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d ^{2}}{d x ^{2}} ((u (x)) e^{2 x}) = 0

What is the solution for x^{''}+8x^'+4x=0 ?
The solution for x^{''}+8x^'+4x=0 is x=c_{1}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}