A theorem we will discuss later says that for a “reasonable” initial value problem, there will be one and only one solution, which will be defined on some interval about the initial point. That there is a solution should seem reasonable at this point, just start at the initial value and follow the slope field and you will get the graph of the solution. The last example should illustrate why you might get a solution defined only in an interval.
4. In solving the first two problems, you almost certainly observed and used (if only implicitly) the fact that solution curves didn’t cross to narrow the range of possible initial values to try. Explain how this fact implies that a “reasonable” initial value problem can’t have two solutions.
That's pasted from my class's lab directions page online... it's using a slope field line java program. "the last example" refers to y'=-(x+2)/y which you can view by plugging it into our program at http://www.math.ksu.edu/math240/java/lab1/lab1.html
the best answer I can come up with is that there's only a single solution that doesn't immediately tend toward infinity, or (in reference to the example if you put it into that program) that there's only a limited part of the graph where a solution curve can exist because of the infinite switching of slope.
Any help? I'm not great at this stuff as it is and I just have to get through it, so trying to figure this out after just a couple class periods isn't going to well.
thanks a lot