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

The solution for y^{'''}-2y^{''}-y^'+2y=0,y(0)=2 is y=(2-c_{2}-c_{3})e^{2t}+c_{2}e^{-t}+c_{3}e^t

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{3}{4}

y^{^{''}} + 4 y^{^{'}} + cy = 0, y (0) = 1, y^{^{'}} (0) = - \frac{10}{3}

(D^{2} + 8 D + 16) y = 0

y^{^{''}} - \frac{1}{9} y = 0

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