# Thread: Find and sketch region of existence and uniqueness

1. ## Find and sketch region of existence and uniqueness

I have been asked to look at the following question over the weekend.

The problem is I do not really understand how to start to tackle this problem.

Consider the following equation y' = xLN(x+y)

I've had calc 1,2,3 and linear algebra but for some reason I'm not sure how to start this problem. Not sure how to sketch it. It's been a while for me... I had linear in the spring of 2011, Calc 3 in the fall of 2010, Calc 2 in the spring of 2010 and Calc 1 in the fall of 2009.

I appricate any pointers!

2. ## Re: Find and sketch region of existence and uniqueness

Do you mean

$y'=x\ln(x+y)?$

If so, there's a well-known existence and uniqueness theorem that you should apply to this DE to answer your question.

3. ## Re: Find and sketch region of existence and uniqueness

That's exactly what I mean!

I'll try to research for a existence formula for this question.

4. ## Re: Find and sketch region of existence and uniqueness

I have found the theorm though I'm not sure I fully understand it.

I have taken the original formula and found the deriviate with respect to y to be

df/dy = x/(y+x) by using the chain rule.

y' = x(LN(X+Y)) => x * 1/(x+y) which simplifies to x / (y+x)

sorry for not using latex. I will re learn how to use the tags so that it's done properly in the future.

I'm not sure it's unique though? I know it's only zero when both X and Y are 0.

This point is undefined as it would be 0/0 correct?

so my original differential equations derivative is not continuous everywhere?

5. ## Re: Find and sketch region of existence and uniqueness

The region x + y < 0 is not in the domain of your DE, since the logarithm function is not defined there. Rearranging, you get that y must be strictly less than -x. If it is true that y < -x, then your df/dy is differentiable in y, and continuous in x. Thus, by the Picard-Lindelöf Theorem, your solution exists and is unique. So, sketch that region, and you're done. Make sense?