What is the solution for y^{'''}-2y^{''}-4y^'+8=0 ?

The solution for y^{'''}-2y^{''}-4y^'+8=0 is y=c_{1}+c_{2}e^{(1+sqrt(5))t}+c_{3}e^{(1-sqrt(5))t}+2t

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y^{'''}-2y^{''}-4y^'+8=0

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 = 0

y = c_{1} + c_{2} e^{(1 + 5) t} + c_{3} e^{(1 - 5) t} + 2 t

Solution steps

y^{'''} (t) - 2 y^{''} (t) - 4 y^{'} (t) + 8 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{(1 + 5) t} + c_{3} e^{(1 - 5) t} + 2 t

y = c_{1} + c_{2} e^{(1 + 5) t} + c_{3} e^{(1 - 5) t} + 2 t

y^{^{''}} - 8 y^{^{'}} - 9 y = 2 e^{- x} - 9 x

y^{^{''}} - 5 y^{^{'}} + 4 y = x, y^{1} (0) = e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{- x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 5 y = e^{t}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = - 2 sin (x) + 3 e^{x}

What is the solution for y^{'''}-2y^{''}-4y^'+8=0 ?
The solution for y^{'''}-2y^{''}-4y^'+8=0 is y=c_{1}+c_{2}e^{(1+sqrt(5))t}+c_{3}e^{(1-sqrt(5))t}+2t