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

The solution for y^{''}-8y^'+2y=0 is y=c_{1}e^{(4+sqrt(14))t}+c_{2}e^{(4-sqrt(14))t}

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

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

y = c_{1} e^{(4 + 14) t} + c_{2} e^{(4 - 14) t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(4 + 14) t} + c_{2} e^{(4 - 14) t}

y = c_{1} e^{(4 + 14) t} + c_{2} e^{(4 - 14) t}

y^{^{''}} + 11 y^{^{'}} + 12 y = 0

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

y^{^{''}} + 4000 y^{^{'}} + 4000000 y = 0

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

x^{^{''}} + x^{^{'}} + \frac{1}{3} x = 0, x (0) = 4, x^{^{'}} (0) = 0

What is the solution for y^{''}-8y^'+2y=0 ?
The solution for y^{''}-8y^'+2y=0 is y=c_{1}e^{(4+sqrt(14))t}+c_{2}e^{(4-sqrt(14))t}