Confused about continuity of this function

[wumple]

Homework Statement

For y'=1/(x+y), sketch a direction field and the solution through y(0)=0.

Homework Equations

I'm confused as to why there is a solution through y(0) - I thought that the existence theorem says that if y' is continuous in a box, then there are solutions through all points in the box.

The Attempt at a Solution

y' is not continuous in the box surrounding the y=-x line. So why does a solution exist there? Is y' actually continuous there? It approaches negative infinity from one side and positive infinity from the other.

 

[JThompson]

"If P, then Q" is not the same thing as "If Q, then P".

So yes.. "if y' is continuous in a box, then there are solutions through all points in the box", but this does not mean "if a solution exists at point y(a)=b, then this solution is differentiable at a".

Take $$y=\sqrt{x}$$. y(0)=0 is defined, but y'(0) is undefined.

 

[HallsofIvy]

Another way of saying the same thing- "if P then Q" tells you what happens if P is false. It tells you nothing about what happens if P is false. In particular, it does not tell you that Q is false.