Question
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)
Solutions
Expert Solution
닝t dk dlJ dt dt d匕 -O. 933 d t dt d Y d七 d- 3 2 I dt 3
Po dt da Unstable : unstable cao