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


Submit Your Answer