What is the solution for y^{''}+10y^'+2y=0 ?

The solution for y^{''}+10y^'+2y=0 is y=c_{1}e^{(-5+sqrt(23))t}+c_{2}e^{(-5-sqrt(23))t}

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y^{''}+10y^'+2y=0

y^{^{''}} + 10 y^{^{'}} + 2 y = 0

y = c_{1} e^{(- 5 + 23) t} + c_{2} e^{(- 5 - 23) t}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = c_{1} e^{(- 5 + 23) t} + c_{2} e^{(- 5 - 23) t}

y = c_{1} e^{(- 5 + 23) t} + c_{2} e^{(- 5 - 23) t}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - \frac{5}{3}

x^{^{''}} + 5 x^{^{'}} + 3 x = 0, x (0) = 5, x^{^{'}} (0) = 7

y^{^{''}} - 22 y^{^{'}} + 121 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + 9 y^{^{'}} + 14 y = 0, y (0) = 0, y^{^{'}} (0) = - 1

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 9

What is the solution for y^{''}+10y^'+2y=0 ?
The solution for y^{''}+10y^'+2y=0 is y=c_{1}e^{(-5+sqrt(23))t}+c_{2}e^{(-5-sqrt(23))t}