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}

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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}