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Uniqueness - exact differential equation

  1. May 21, 2013 #1
    Hi folks! This one got me in doubts...

    1. The problem statement, all variables and given/known data

    Solve IVP (Initial Value Problem): [itex](2xy+sin(x))dx+(x^{2}+1)dy=0, y(0)=2[/itex]

    Is the solution unique? Motivate why!

    2. Relevant equations

    Relevant equations for solving the exact equation...

    3. The attempt at a solution

    I can solve this without any trouble. Since the answer is an implicit solution, I get:

    [itex]F(x,y)=x^{2}y-cos(x)+y= C [/itex]

    Putting in the IVP value I get for y:

    [itex]y(x)=\frac{1+cos(x)}{x^{2}+1}[/itex]

    Now, to the question, is this solution unique? Why? Why not? And how does it relate to implicit solutions (here, exact differential equation)?

    What about if this IVP was a separable, or a linear differential equation?

    I am thankful for any hints, because this one got me really thinking...
     
  2. jcsd
  3. May 21, 2013 #2

    LCKurtz

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  4. May 21, 2013 #3
    Thank you LCKurtz for the fast reply! I have studied the sheet, as I have with my coursebook, but didn't get any smarter. Here's where I don't get it:

    Yes, the paper says and explains clearly why there is y(0)≠0 gets no solution. But how does this applies to this particular equation, meaning, which first gives an implicit solution? Or does explicit/implicit not matter here?

    I get a solution to this exact differential equation, even though the paper says I shouldn't get one. Or am I misreading this entirely?

    Thanks for your help!
     
  5. May 21, 2013 #4

    LCKurtz

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    Have you figured out what ##f(x,y)## is in your equation and checked the hypotheses of the uniqueness theorem against it?
     
  6. May 21, 2013 #5
    [itex]dF/dy=\frac{-2x}{x^{2}+1}[/itex]

    So you are saying that the "test" against the hypotheses of uniqueness states the fact, independent whether the differential equation (IVP) has any solution - as it does in this case?
     
    Last edited: May 21, 2013
  7. May 21, 2013 #6

    LCKurtz

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    That link I gave you has both an existence and uniqueness theorem. If a function satisfies the hypotheses of the uniqueness theorem, doesn't it automatically satisfy those for the existence theorem? So if your ##f(x,y)## satisfies the hypotheses of the uniqueness theorem, then it has a solution and it is unique. Does your ##f(x,y)## satisfy the hypotheses? That is your only question. You hardly need the existence theorem given that you already have a solution.
     
  8. May 22, 2013 #7
    Thanks for your effort!

    However I am still confused with this, especially with implicit solutions.

    As a last try, let me rephrase my question:

    The paper says: "no solution if x0=0 and y0≠0." . But here, I do have a solution for the differential equation. Maybe that last summary is irrelevant if I have this single solution.
     
  9. May 22, 2013 #8

    LCKurtz

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    They are talking about their equation. You need to check the theorems for your equation. And whether or not a solution exists has nothing to do with whether you have an explicit or implicit formula.
     
  10. May 22, 2013 #9
    Whoops!
    So let me try it out in this case, and please correct me if I am wrong:

    [itex]dy/dx=-\frac{(2xy+sin(x))}{x^{2}+1}[/itex]

    is 0 for x=0, y=2.

    [itex]dF/dy=-\frac{(2x)}{x^{2}+1}[/itex]

    is 0 for x=0, y=2. However, this still means that the solution is unique, because both functions are continuous and defined near x=0, y=2. Meaning, the solution exists, and it is unique.
     
    Last edited: May 22, 2013
  11. May 22, 2013 #10

    HallsofIvy

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    Yes, the first thing you should have done, after writing the equation as
    [tex]\frac{dy}{dx}= \frac{2xy+ sin(x)}{x^2+ 1}[/tex]
    is recognize that the denominator is never 0 and so the function is continuous for all x. Since the derivative, using the quotient rule will just have the square of [itex]x^2+ 1[/itex] in the denominator, it is still never 0 and so the derivative is continuous for all x.
     
  12. May 22, 2013 #11
    Thanks to both of you for your hints and guidance!
     
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