What is the solution for y^{''}-10y^'-16y=0 ?

The solution for y^{''}-10y^'-16y=0 is y=c_{1}e^{(5+sqrt(41))t}+c_{2}e^{(5-sqrt(41))t}

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y^{''}-10y^'-16y=0

y^{^{''}} - 10 y^{^{'}} - 16 y = 0

y = c_{1} e^{(5 + 41) t} + c_{2} e^{(5 - 41) t}

Solution steps

y^{''} (t) - 10 y^{'} (t) - 16 y = 0

Solve linear ODE:

y = c_{1} e^{(5 + 41) t} + c_{2} e^{(5 - 41) t}

y = c_{1} e^{(5 + 41) t} + c_{2} e^{(5 - 41) t}

y^{^{'}} + y = y^{^{''}}

\frac{5}{6} y^{^{''}} + 10 y^{^{'}} + 60 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + (4 - 1) y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 6 y = 0

What is the solution for y^{''}-10y^'-16y=0 ?
The solution for y^{''}-10y^'-16y=0 is y=c_{1}e^{(5+sqrt(41))t}+c_{2}e^{(5-sqrt(41))t}