- #1
ericm1234
- 73
- 2
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)" )
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)" )