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

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

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

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

y = c_{1} + c_{2} e^{- 2 t} + c_{3} e^{- t} + c_{4} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} + c_{3} e^{- t} + c_{4} e^{t}

y = c_{1} + c_{2} e^{- 2 t} + c_{3} e^{- t} + c_{4} e^{t}

\frac{1}{4} x^{^{''}} + 10 x^{^{'}} + 1000 x = 0

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

4 y^{^{'''}} - 7 y^{^{''}} - 34 y^{^{'}} - 8 y = 0

y^{^{''}} - 5 y^{^{''}} + 6 y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 3

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