What is the solution for x^{''}+4x^'+2x=0 ?

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

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x^{''}+4x^'+2x=0

x^{^{''}} + 4 x^{^{'}} + 2 x = 0

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

Solution steps

x^{''} (t) + 4 x^{'} (t) + 2 x = 0

Solve linear ODE:

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

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

\int (5 e^{- 0.8 x}) d x

y = 2 x e^{2 x} - e^{2 x}

\int 18 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

(x - 8 y) d x + (4 x) d y = 0

\frac{( s + 1 ) ^{3}}{s ^{5}}

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