# Diff Eq Region of Unique Solution

1. Jun 4, 2006

### DieCommie

I have just started my first diff eq class, and I am struggling with the concepts. In the section 'Preliminary Theory' we are given Picard's Theorem and a few odd examples. Then this is the question from the homework:

"Determine a region of the xy-plane for which the given differential equation would have a unique solution through a point (x_0, y_0) in the region. dy/dx = y^(2/3) "

The wording has got me confused, and Im not really sure what they want. The solution is a family of curves, so they want the particular curve at the point (x_0, y_0)? Or they want a region of something?

2. Jun 5, 2006

### HallsofIvy

Staff Emeritus
The "wording" is "Determine a region" and doesn't say anything about solving the equation. I don't see anything confusing about that. Picard's theorem says that the d.e. dy/dx= f(x,y), with boundary condition y(x0)= y0 has a unique solution in a region containing (x0, y0) if f(x,y) is continuous in that region and f(x,y) is "Lipschitz" in y in that region (beginning texts use the simpler but less general condition that fy(x,y) be continuous in the region). So: where are those not true? As long as your region avoids places where that is not true, you are okay.

In this particular case, it is easy to find the general solution and to see if it is possible to find another solution for some points (x0,y0) but that is not required.

3. Jun 5, 2006

### DieCommie

Ok, so the only place place these conditions are not true is at y=0, so my region is everywhere but y=0. And this region is the region where the family of solutions occupy?

THx for the help, I think I can do the problems though I dont fully grasp what I am doing.
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