# Uniqueness Theorem: Finding region

1. Jun 11, 2012

### shaqywacky

Hello everyone!

So today is was my first day of differential equations and I understood most of it until the very end. My professor started talking about partial derivatives which is Calc 3 at my university. He said Calc 3 wasn't required but was recommend for differential equations. He then went on to act like we all knew Calc 3.

So he gave us this homework question:

Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through point (x_0, y_0) in the region(x_0 is x sub 0):

x(dy/dx) = y

I pretty much have no idea how to solve this or really what it's asking. The rest of the homework was really easy so this caught me off guard.

The only information we learned about it is the attachment I uploaded.

Do I need to define a region(IE a box) or is it just an interval. If it's an interval, I think I can answer it. But I seriously don't even understand what the theorem is saying. Is it saying that given a family of differential equations (for example x + C ) and given a point, there is only one C that will cause the function to intersect that point? I don't think that's it because it doesn't involve a region. I think I don't understand it because I have no idea what a partial derivative is.

Any help would be appreciated.

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2. Jun 11, 2012

### Vargo

You need to use the theorem in that attachment. In the theorem, the differential equation looks like $y' = f(x,y)$.

Can you write your differential equation ($xy'=y$) so that it fits that form? If so, then what is f(x,y)? What conditions on f(x,y) does the theorem require to guarantee a unique solution in some interval? Where are those conditions true for your particular function f(x,y)?

Notice the example that is after the theorem. In that example, $f(x,y) = x\sqrt{y}$. At the point (0,0), there are two solutions. So apparently the function $x\sqrt{y}$ does not meet all the criteria given in the theorem.

I hope that helps.

3. Jun 11, 2012

### Vargo

Let me clarify a bit. It is more or less "generic" that if you have a first order differential equation, then through each point, there will be just one solution. That is what the theorem is about in your attachment--conditions on the equation to guarantee that you will not get more than one solution through a point. That is also the relevance of the example after the theorem. It shows that when the conditions on f(x,y) are not met, you can get more than one solution through a point. So for your problem, the question is asking which points can have more than one solution through them and which points can only have one solution through them.

Also, you don't have to worry about the exact box or interval on which the solution exists for this problem.

Oh, and regarding Calc III, all you have to know is how to calculate a partial derivative. Google that because it is really easy. If you can calculate Calc I derivatives, then you can learn to calculate partials after doing a couple examples.

4. Jun 11, 2012

### shaqywacky

Ah, thank you. That makes much more sense. I forgot to mention that I'm taking Calc 3 next semester so I didn't know how to take partial derivatives. I assumed they were really difficult to take but they're really easy and that makes this make much more sense.

Thank you again.