What is the solution for y^{''}-10y^'+2y=0,y(0)=0 ?

The solution for y^{''}-10y^'+2y=0,y(0)=0 is y=-c_{2}e^{(5+sqrt(23))t}+c_{2}e^{(5-sqrt(23))t}

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y^{''}-10y^'+2y=0,y(0)=0

y^{^{''}} - 10 y^{^{'}} + 2 y = 0, y (0) = 0

y = - c_{2} 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_{2} e^{(5 + 23) t} + c_{2} e^{(5 - 23) t}

y = - c_{2} e^{(5 + 23) t} + c_{2} e^{(5 - 23) t}

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

u^{^{''}} + H (t) = 0

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 8

What is the solution for y^{''}-10y^'+2y=0,y(0)=0 ?
The solution for y^{''}-10y^'+2y=0,y(0)=0 is y=-c_{2}e^{(5+sqrt(23))t}+c_{2}e^{(5-sqrt(23))t}