# Phase line problem, my given solution is wrong I believe

1. Jul 13, 2012

### ericm1234

I was given the solution to this problem:
y'=-y^2+2y-1, plot phase line, solve it, in terms of y(0)=y0, what happens as t goes to infinity.
So then the phase line shows x=1 is 'semistable'. Also the solution is y= 1+ (y0-1)/(1+(y0-1)t) but now he claims this, which I believe is wrong:
if y0=1: y(t)=0 for all t>0
if y0>1, y(t) goes to 0 as t goes to infin.
if y0<1, y(t) goes to neg. infin. (also confused here because he now writes, "as t (arrow up) 1/(1-y0)" )

2. Jul 13, 2012

### tiny-tim

hi ericm1234!

it's easier to analyse if you write it as y - 1 = 1/(t - C)

(which is exactly what you get if you solve the equation directly)

3. Jul 13, 2012

### ericm1234

Agreed Tiny-Tim. But now, when t=0, c=-1/(y0-1). Ok, my issue now is with the analysis that I listed above, which I believe is wrong, as I wrote it (from the solutions to the problem given to me by the teacher)

4. Jul 14, 2012

### tiny-tim

yes, it's wrong

5. Jul 14, 2012

### HallsofIvy

Staff Emeritus
In terms of the phase line, $y'= -(y- 1)^2$ so that y= 1 is an equilibrium point. For y< 1, y- 1< 0 but it is squared so that $y'= -(y-1)^2$ is positive and y is increasing toward y= 1 NOT toward negative infinity. For y> 1, y- 1> 0 but with the squaring we still have $y'= -(y-1)^2$ positive and y increases away from y= 1 to infinity.

6. Jul 14, 2012

### ericm1234

I think you disregarded the negative sign then. y' is negative for all y, except y=1.