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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 5 y = 0

81 y^{^{''}} + 72 y^{^{'}} + 16 y = 0

y^{^{''}} + y^{^{'}} + 1.25 y = 0

121 y^{^{''}} + 88 y^{^{'}} + 16 y = 0

y^{^{'''}} - 4 y^{^{''}} + 3 y^{^{'}} = 0

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