What is the solution for y^{''}-4y^'-y=0 ?

The solution for y^{''}-4y^'-y=0 is y=c_{1}e^{(2+sqrt(5))t}+c_{2}e^{(2-sqrt(5))t}

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

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

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

Solution steps

y^{''} (t) - 4 y^{'} (t) - y = 0

Solve linear ODE:

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

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

What is the solution for y^{''}-4y^'-y=0 ?
The solution for y^{''}-4y^'-y=0 is y=c_{1}e^{(2+sqrt(5))t}+c_{2}e^{(2-sqrt(5))t}