What is the solution for y^{''}-10y^'-98y=0 ?

The solution for y^{''}-10y^'-98y=0 is y=c_{1}e^{(5+sqrt(123))t}+c_{2}e^{(5-sqrt(123))t}

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y^{''}-10y^'-98y=0

y^{^{''}} - 10 y^{^{'}} - 98 y = 0

y = c_{1} e^{(5 + 123) t} + c_{2} e^{(5 - 123) t}

Solution steps

y^{''} (t) - 10 y^{'} (t) - 98 y = 0

Solve linear ODE:

y = c_{1} e^{(5 + 123) t} + c_{2} e^{(5 - 123) t}

y = c_{1} e^{(5 + 123) t} + c_{2} e^{(5 - 123) t}

y^{^{''}} - 4 y^{^{'}} + y 5 = 0

y^{^{''}} - 9 y^{^{'}} + 7 y = 0

3 y^{^{'''}} + 5 y^{^{''}} = 0

5 y^{^{''}} + 5 y^{^{'}} + 10 y = 0

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

What is the solution for y^{''}-10y^'-98y=0 ?
The solution for y^{''}-10y^'-98y=0 is y=c_{1}e^{(5+sqrt(123))t}+c_{2}e^{(5-sqrt(123))t}