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Confused about continuity of this function

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    For y'=1/(x+y), sketch a direction field and the solution through y(0)=0.

    2. Relevant 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.

    3. 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.
  2. jcsd
  3. Sep 30, 2010 #2
    "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 [tex]y=\sqrt{x}[/tex]. y(0)=0 is defined, but y'(0) is undefined.
  4. Sep 30, 2010 #3


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    Science Advisor

    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.
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