Phase line problem, my given solution is wrong I believe

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In summary, the conversation discusses the solution to the problem y'=-y^2+2y-1, including plotting a phase line and determining the behavior of the solution as t goes to infinity. The solution is y= 1+ (y0-1)/(1+(y0-1)t), but there is disagreement about the analysis of the behavior for different values of y0. Ultimately, it is determined that for y<1, the solution approaches 1, and for y>1, the solution approaches infinity.
  • #1
ericm1234
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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)" )
 
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  • #2
hi ericm1234! :smile:

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

(which is exactly what you get if you solve the equation directly)
 
  • #3
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
yes, it's wrong :redface:
 
  • #5
In terms of the phase line, [itex]y'= -(y- 1)^2[/itex] so that y= 1 is an equilibrium point. For y< 1, y- 1< 0 but it is squared so that [itex]y'= -(y-1)^2[/itex] 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 [itex]y'= -(y-1)^2[/itex] positive and y increases away from y= 1 to infinity.
 
  • #6
I think you disregarded the negative sign then. y' is negative for all y, except y=1.
 

1. What is the phase line problem?

The phase line problem is a mathematical concept in differential equations that involves graphing the behaviors of a dynamic system over time.

2. How does the given solution for the phase line problem differ from the correct solution?

The given solution may be wrong due to errors in calculations or assumptions made during the problem-solving process. It is important to carefully check each step of the solution to identify where the mistake may have occurred.

3. What are some common mistakes made when solving the phase line problem?

Some common mistakes include incorrect application of mathematical formulas, incorrect interpretation of the given information, and errors in graphing or plotting the solution.

4. How can I improve my problem-solving skills for the phase line problem?

Practicing and reviewing examples of the phase line problem is the best way to improve your problem-solving skills. Additionally, seeking help from a tutor or classmate can provide valuable insights and tips.

5. Can the phase line problem be applied in real-life situations?

Yes, the phase line problem can be applied to various real-life scenarios, such as predicting population growth, analyzing economic trends, and understanding the behavior of physical systems.

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