What is the solution for y^{''}-4y^'-y-2=0 ?

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

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y^{''}-4y^'-y-2=0

y^{^{''}} - 4 y^{^{'}} - y - 2 = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) - y - 2 = 0

Solve linear ODE:

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

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

x^{^{''}} + 4 x = 10 cos (3 t)

y^{^{''}} + 9 y = 12 x e^{2 x}

y^{^{''}} - y^{^{'}} - 2 y = 7 e^{2 x} - 3 e^{x}

y^{^{''}} + y^{^{'}} = 4 x + 10 sin (x)

y^{^{''}} - 4 y^{^{'}} + 3 y = e t - 2 t + 3

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