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Y'=(y"+1)/(y-1) is stable at: y= infinity y= 0 y= 1 y= -1 Sorry, I meant: y'=(y"+y)/(y-1)

y'=(y"+1)/(y-1) is stable at:

y*= infinity

y*= 0

y*= 1

y*= -1

Sorry, I meant: y'=(y"+y)/(y-1)

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