What is the solution for y^{''}-20y^'+12y=0 ?

The solution for y^{''}-20y^'+12y=0 is y=c_{1}e^{2(5+sqrt(22))t}+c_{2}e^{2(5-sqrt(22))t}

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y^{''}-20y^'+12y=0

y^{^{''}} - 20 y^{^{'}} + 12 y = 0

y = c_{1} e^{2 (5 + 22) t} + c_{2} e^{2 (5 - 22) t}

Solution steps

y^{''} (t) - 20 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{2 (5 + 22) t} + c_{2} e^{2 (5 - 22) t}

y = c_{1} e^{2 (5 + 22) t} + c_{2} e^{2 (5 - 22) t}

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

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

y^{^{'''}} - 3 y^{^{''}} - 10 y^{^{'}} = 0

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 9

y^{^{'''}} + 11 y^{^{''}} + 38 y^{^{'}} + 40 y = 0

What is the solution for y^{''}-20y^'+12y=0 ?
The solution for y^{''}-20y^'+12y=0 is y=c_{1}e^{2(5+sqrt(22))t}+c_{2}e^{2(5-sqrt(22))t}