What is the solution for y^{''}-6y^'+y=0 ?

The solution for y^{''}-6y^'+y=0 is y=c_{1}e^{(3+2^{1+1/2})t}+c_{2}e^{(3-2^{1+1/2})t}

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y^{''}-6y^'+y=0

y^{^{''}} - 6 y^{^{'}} + y = 0

y = c_{1} e^{(3 + 2^{1 + \frac{1}{2}}) t} + c_{2} e^{(3 - 2^{1 + \frac{1}{2}}) t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{(3 + 2^{1 + \frac{1}{2}}) t} + c_{2} e^{(3 - 2^{1 + \frac{1}{2}}) t}

y = c_{1} e^{(3 + 2^{1 + \frac{1}{2}}) t} + c_{2} e^{(3 - 2^{1 + \frac{1}{2}}) t}

y^{^{''''}} + 5 y^{^{'''}} - 4 y^{^{''}} - 2 y^{^{'}} + \frac{1}{2} y = 0

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

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 1

y^{^{''}} - 8 y^{^{'}} + 14 y = 0, y (0) = 0, y^{^{'}} (0) = 5

What is the solution for y^{''}-6y^'+y=0 ?
The solution for y^{''}-6y^'+y=0 is y=c_{1}e^{(3+2^{1+1/2})t}+c_{2}e^{(3-2^{1+1/2})t}