What is the solution for y^{''}-8y^'+5y=0 ?

The solution for y^{''}-8y^'+5y=0 is y=c_{1}e^{(4+sqrt(11))t}+c_{2}e^{(4-sqrt(11))t}

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y^{''}-8y^'+5y=0

y^{^{''}} - 8 y^{^{'}} + 5 y = 0

y = c_{1} e^{(4 + 11) t} + c_{2} e^{(4 - 11) t}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = c_{1} e^{(4 + 11) t} + c_{2} e^{(4 - 11) t}

y = c_{1} e^{(4 + 11) t} + c_{2} e^{(4 - 11) t}

1 x^{^{''}} + 6 x^{^{'}} - 1 x = 0

6 y^{^{''''}} + 77 y^{^{''}} + 25 y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = - 6

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1

y^{^{''}} + 15 y^{^{'}} + 44 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-8y^'+5y=0 ?
The solution for y^{''}-8y^'+5y=0 is y=c_{1}e^{(4+sqrt(11))t}+c_{2}e^{(4-sqrt(11))t}