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Homework Help: D/dx(dy/dx) + d/dy(dx/dy) = 1

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve for either x or y:

    [tex]\frac{d^2y}{dx^2} + \frac{d^2x}{dy^2} = 1 [/tex]

    2. Relevant equations
    I don't know any.

    3. The attempt at a solution
    A little bit of simplification first:

    [tex]\frac{d}{dx}\frac{dy}{dx} + \frac{d}{dy}\frac{dx}{dy} = 1 [/tex]

    does not seem to help.

    Re-writing it as y'' + x'' = 1 does not inspire any further direction either. Actually, it's also ambiguous, because it could mean y''(t) + x''(t) = 1 which is not what I wanted.

    I tried letting u = dy/dx and v = dx/dy which also equals 1/u, like this:

    [tex]\frac{du}{dx} + \frac{dv}{dy} = 1 [/tex]

    Then, reworking v = dx/dy into 1/v = dy/dx, implicit differentiating to get [itex]\frac{-1}{v^2}\frac{dv}{dx} = \frac{d^2y}{dx^2} = \frac{du}{dx}[/itex], removing dx and integrating, only got me back to 1/v = u + C. Hmph. I'm stumped.
     
  2. jcsd
  3. Jul 3, 2010 #2
    [STRIKE]I would try it the same way you did with the other differential equation, rewrite it first as
    y'' + 1/y'' = 1

    Turn it into a quadratic equation and take it from there.[/STRIKE]
     
    Last edited: Jul 3, 2010
  4. Jul 3, 2010 #3

    Hurkyl

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    That's not the same differential equation. While dy/dx dx/dy is one, d²y/dx² d²x/d²y is not.
     
  5. Jul 3, 2010 #4
    That notation with higher derivatives gets me sometimes. :redface: I see
     
  6. Jul 3, 2010 #5
    Bohrok I thought of that at first too, but Hurkyl is right (although there might be a typo with the second denominator). And I still haven't thought of anything :/
     
  7. Jul 3, 2010 #6
    Let:
    [tex]u=\frac{dy}{dx}[/tex],

    then

    [tex]1/u=\frac{dx}{dy}[/tex]

    so that:

    [tex]u'+\frac{d}{dx}\left(\frac{1}{u}\right)=1[/tex]

    isn't that just:

    [tex]u'-\frac{1}{u^2}u'=1[/tex]

    Can you finish it? Also, I'm not entirely sure about this.
     
  8. Jul 3, 2010 #7
    jackmell, if 1/u = dx/dy then wouldn't I have to take d/dy of 1/u?
     
  9. Jul 3, 2010 #8
    Hi. I think what I did is correct but I'm not entirely sure ok. This is how I'd approach it: Either wait til' someone helps us or just for the moment assume what I wrote is ok and work through it to see what happens. Get an answer. Then back-substitute it into the original expression if possible and see if it satisfies the equality. If it does, then there is a chance what I wrote is correct. Then I would proceed to somehow verify that it is indeed valid. Remember, it's very important in mathematics to try something even if it's turns out to be wrong because often the wrong ones lead you to the right answer if you're patient an tolerant of some failures sometimes. :)
     
    Last edited: Jul 3, 2010
  10. Jul 3, 2010 #9
    I can rearrange this into

    [tex]u' = \frac{u^2}{u^2-1}[/tex]

    Maybe I could use some trigonometry from here. Indeed I can let cosθ = 1/u.
    Then, u = secθ, u^2 = secθ, and tanθ = u^2 - 1. However, the u' from above is u'(x) whereas if I take d/dθ of u = secθ then I'll get u'(θ) = secθtanθ. In other words, I've no clue.
     
  11. Jul 3, 2010 #10

    Hurkyl

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    It's a rational function. You've anti-differentiated rational functions before, haven't you?
     
  12. Jul 3, 2010 #11
    So,

    du/dx = u^2 / (u^2 - 1)

    Separating variables:

    (u^2 - 1) du / u^2 = dx

    (1 - 1/u^2) du = dx

    Integrating:

    u + 1/u = x + C

    Multiply through by u to get u^1 + 1 = u(x + C). Should I use the quadratic equation now? It's a mess.
     
  13. Jul 3, 2010 #12
    I get:

    [tex]u^2-u(x+c_1)+1=0[/tex]

    and I would then try the quadratic formula. Gotta go.
     
  14. Jul 5, 2010 #13
    That should be

    [tex]u'+\frac{d}{dy}\left(\frac{1}{u}\right)=1[/tex]

    However, this can be changed to

    [tex]u'+\frac{d}{dx}\left(\frac{1}{u}\right)\frac{dx}{dy}=1[/tex]
     
  15. Jul 5, 2010 #14
    Thanks Phrak! That's what I said:

    Then you said
    This is exactly the step I was just trying today! Thanks a lot, I'll see where I go from here, recalling that dx/dy = 1/u.
     
  16. Jul 5, 2010 #15
    Yes. I noticed that as I was posting. I though you might want some confirmation anyway.
     
  17. Jul 5, 2010 #16
    I simplified it down to a cubic, u^3 - u^2 (x+C) + 1 = 0. Any suggestions on how to proceed?
     
  18. Jul 5, 2010 #17
    I don't think that's correct. What do you have for a first order differential equation in u?
     
  19. Jul 5, 2010 #18
    Here is all the work so far:

    [tex]\frac{d^2y}{dx^2} + \frac{d^2x}{dy^2} = 1[/tex]

    [tex]\frac{d}{dx}\frac{dy}{dx} + \frac{d}{dy}\frac{dx}{dy} = 1[/tex]

    Let
    [tex]u = \frac{dy}{dx}[/tex]

    So the expression becomes
    [tex]\frac{d}{dx} u + \frac{d}{dy} \left(\frac{1}{u}\right) = 1[/tex]

    [tex]u' + \frac{d}{dx}\left( \frac{1}{u}\right) \cdot \frac{dx}{dy} = 1[/tex]

    [tex]u' - \frac{1}{u^2} \cdot \frac{du}{dx} \cdot \frac{1}{u} = 1[/tex]

    [tex]u' - \frac{u'}{u^3} = 1[/tex]

    That is my differential equation in u. I can common factor u' to get
    [tex]u' \left( 1 - \frac{1}{u^3} \right) = 1[/tex]

    And re-writing u' as du/dx,
    [tex]du \left( 1 - \frac{1}{u^3} \right) = dx[/tex]

    Then we can integrate
    [tex]\int 1 - \frac{1}{u^3} du = \int dx[/tex]

    [tex]u + \frac{1}{2u^2} = x + C[/tex]

    Multiply through by 2u^2
    [tex]2u^3 + 1 = 2u^2(x + C)[/tex]

    [tex]2u^3 - 2u^2(x + C) + 1 = 0[/tex]

    So I admit my expression in post #16 was wrong. Is there anything wrong with my math above?
     
    Last edited: Jul 5, 2010
  20. Jul 5, 2010 #19

    HallsofIvy

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    There is a reason for that notation! While [itex]dy/dx[/itex] can be treated "like" a fraction, [itex]d^2y/dx^2[/itex] cannot.
     
  21. Jul 5, 2010 #20
    Ok. Sorry about that Unit. Been workin' on it but haven't come up with a solution I can successfully back-substitute into the original expression and confirm it satisfies it. Perhaps I can with the result from the cubic.
     
  22. Jul 5, 2010 #21
    you need to express the second derivative:

    [tex]
    \frac{d^{2} x}{d y^{2}}
    [/tex]

    In terms of derivatives of y(x) with respect to x. For this you will need the chain rule. Let me show you the first derivative:

    [tex]
    \frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \frac{1}{y'}
    [/tex]
     
  23. Jul 5, 2010 #22
    Since the above discussion went further than where I stopped, I think it is safe to write the second derivative as well:

    [tex]
    \frac{d^{2} x}{dy^{2}} = \frac{d}{d y}\left(\frac{d x}{d y}\right) = \frac{d x}{d y} \, \frac{d}{d x} \left(\frac{d x}{d y}\right)
    [/tex]

    [tex]
    \frac{d^{2} x}{dy^{2}} = \frac{1}{y'} \, \frac{d}{d x} \left(\frac{1}{y'}\right) = \frac{1}{y'} \, \left( -\frac{1}{(y')^{2}}\right) \, y'' = -\frac{y''}{(y')^{3}}
    [/tex]

    Then, your equation reads:

    [tex]
    y'' - \frac{y''}{(y')^{3}} = 1
    [/tex]

    You can reduce the order by one by introducing the substitution:

    [tex]
    z = y'
    [/tex]

    [tex]
    z' - \frac{z'}{z^{3}} = 1
    [/tex]

    This equation can be integrated now.
     
  24. Jul 5, 2010 #23

    Thanks Dickfore! I got this far too, on the same chain rule principle, where my u = your z = dy/dx:

    Then, as you suggest, I integrated:
    The problem is, I don't know where to go from here. Tartaglia's method for finding a real root of a cubic equation does not look inviting...
     
    Last edited: Jul 5, 2010
  25. Jul 5, 2010 #24
    The integrated equation is easily solved in terms of x as a function of u (z). Thus, your solution is given in parametric form x = x(u). You still need to find y = y(u). For this, use:

    [tex]
    \frac{d y}{d u} = \frac{d y}{d x} \, \frac{d x}{d u} = u \, \frac{d x}{d u}, \ u = y' = \frac{dy}{dx}
    [/tex]

    and the already known equation for x = x(u) to express dx/du. Then, integrate again.
     
  26. Jul 5, 2010 #25
    Thanks, Dickfore. I didn't realize I could express my solution in parametric form x = x(u), y = y(u), and u = dy/dx.

    Following your hint, I did dy/du = u dx/du = u (1 - 1/u^3) = u - 1/u^2. Integrating, y(u) = 1/2 u^2 + 1/u + D. I picked D instead of C to distinguish it from the constant in x(u).

    I don't really see where to go from here, though. Or is this as far as the solution should go?
     
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