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Homework Help: Differential equation

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Make the following change of variables:

    [tex]x = r \cos \theta[/tex]

    [tex]y = r \sin \theta[/tex]

    and integrate the following equation:

    [tex](xy'-y)^2 = a(1+y'^2)\sqrt{x^2+y^2}[/tex]

    3. The attempt at a solution
    First it's worth noting that the equation [tex]x^2+y^2=a^2[/tex] (even without changing variables) is solution for the above differential equation.

    Now, making the substitution of variables, I'm able to reduce the equation down to:

    [tex]\left(\frac{dr}{d\theta}\right)^2 = \frac{r^2(r-a)}{a}[/tex]

    Looking at this equation, we see that if we take either r=0, or else r=a, then we get:

    [tex]\frac{dr}{d\theta} = 0[/tex]

    [tex]r(\theta) = K[/tex]

    which, indeed, fitting w/ the "boundary condition" r=0 or r=a, gives us a consistent solution.

    My question is: is this the "correct" way to solve the equation? Just by looking at the equation and making a "guess" that happens to work? Or is there a more "formal" way to take the equation:

    [tex]\frac{dr}{r\sqrt{r-a}} = \pm\frac{d\theta}{a}[/tex]

    and "derive" the "correct" solution?
  2. jcsd
  3. Sep 22, 2010 #2
    With respect to what do you differentiate? i.e. what does ' mean? Derivative w.r.t x or some external time, or else?
  4. Sep 22, 2010 #3
    In the original equation, differentiate y wrt x.

    That is:

    [tex]y' = \frac{dy}{dx}[/tex]
  5. Sep 24, 2010 #4
    My question is not so much about how to do the change of variables..

    That I'm pretty sure I can do.

    My question is more that, once I do the change of variables, I get an expression like:

    [tex]\left(\frac{dr}{d\theta}\right)^2 = \frac{r^2(r-a)}{a}[/tex]

    or, perhaps "simplifying":

    [tex]\frac{dr}{r\sqrt{r-a}} = \pm \frac{d\theta}{\sqrt{a}}[/tex]

    My point only is that this seems like a very difficult form to integrate (if it's possible to integrate at all... is it??)

    And, going back to the first equation above, I "know" that I'm looking for a solution which is a circle, and indeed I "notice" that if I take r=a, then the right hand side of this equation becomes zero, and the left hand side likewise becomes zero, and so r=a becomes a "solution" of the original equation.

    My question is only whether this kind of "guessing"... i.e., looking at the equation and then noticing which solutions "work" just by inspection, is a valid way to "solve" the equation. After all, r=a does "solve" the equation, and it's the "solution" I'm looking for, but is there a more rigorous way to get there?
  6. Sep 24, 2010 #5
    My answers are "pretty much" and "no". Pretty much that's what you do to solve equations. You try things. Sometimes many things before you get it right. And certainly the more you solve, the better you get at it and can recognize patterns and other means of solutions to help steer you in the right direction. Surely the [itex]x^2+y^2[/itex] lends itself to a possible "try" of polar coordinates. This happens to work in this case.

    And no, it's not too difficult to integrate:
    \frac{dr}{r\sqrt{r-a}} = \pm \frac{d\theta}{\sqrt{a}}
    and solve for r. Just integrate both sides, add const of integration, keep the [itex]\pm[/itex] where it needs to be.

    Dang it! I can't even integrate that left side. Now guess what I'm going to do? Well, I can look in my Calculus text book for similar problems and probably work a few of this or that one. Then I could just ask Mathematica to solve it. The Mathematica solution will have a particular form. Now, why is that? So I could then look in the Calculus text for that particular form and often that is enough to figure out how to do it. But as you see, I'm tryin' things. :)
    Last edited: Sep 24, 2010
  7. Oct 9, 2010 #6
    Yes, so according to Mathematica the form works out to:

    [tex]\int \frac{dx}{x\sqrt{x-a}} = \frac{2}{\sqrt{a}}\tan^{-1}\left(\sqrt{\frac{x-a}{a}}\right)[/tex]

    which is fine. I think that answers my question.

    I have another related question though. This whole question/thread is taken from a relatively old work on Differential Equations (Ince, published in the early 20th century). The full problem statement is as follows (I'm going to highlight the part I don't understand in red):

    The bit about changing the variables and integrating is what we've been discussing, and I'm pretty sure I understand that ... but the "if/then" statement there (highlighted in red) doesn't seem to be complete. It gives a condition for "Q", but then doesn't mention Q further. It's a bit confusing (to me).

    It's easy enough to draw a perpendicular from the origin to the tangent emanating from a point P on the curve, and to call this point H. And it's easy enough to drop another perpendicular from the point H back onto OP, and to call the point of intersection Q. But without stipulating any further conditions on Q, how would you get that P has to lie on a circle?

    As I said in the original post, it's clear that the equation for a circle satisfies this differential equation, but I'm not sure how you can derive the same from the "analytical geometry" question that is posed above in red.
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