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Differential equation question (I think)

  1. Oct 7, 2015 #1

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

    It's to do with mirages, but I don't think the physics context is too important. It's also possible that the solution doesn't involve differential equations, and my method is completely wrong. I've been given that:

    ##A = \frac{n(1+ay)}{ sqrt(1+(y')^2)}##

    where y' is dy/dx. I have to show that

    ##y = -\frac{1}{a} + \frac{A}{na}## cosh( ##\frac{na}{A}##(x-x0))

    A, n and a are real constants.
    2. Relevant equations


    3. The attempt at a solution
    I tried rearranging it to get dy/dx = ... , by multiplying by the denominator then dividing by A, squaring both sides, subtracting 1 and then square rooting. I ended up with:

    ##y'=\sqrt{\frac{n^2 (1+ay)^2}{A^2} -1}##

    And that's separable. So
    ##dx = \frac{1}{\sqrt{\frac{n^2 (1+ay)^2}{A^2} -1}}dy##
    And then if I integrate both sides by doing substitutions like u =1+ay, I get ln of something. Nowhere near the show that result.
    Checked the rearrangement so many times. What did I do?
     
  2. jcsd
  3. Oct 7, 2015 #2

    SteamKing

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    It's not clear how your integration got LN (something). Please post your work to get this result.
     
  4. Oct 7, 2015 #3
    Sorry. So I substituted u=1+ay, du/a = dy so that the integration becomes the integral of

    ##\frac{1}{a} \frac{1}{\sqrt{ \frac{n^2 u^2}{A^2} -1}} du##

    Which is ln. I checked with wolfram alpha by replacing n^2/A^2 with just b, because it's just a constant, like the 1/a which is ignored in the integral. So I think it's still the same thing.
     
    Last edited: Oct 7, 2015
  5. Oct 7, 2015 #4

    epenguin

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    From what you say you get x = LN(some function of y).
    This is not at all a sign that you have done anything wrong - when you work on your result (invert the equation to get y = some function of x) this may very well have a cosh in it.
     
  6. Oct 7, 2015 #5
    I don't think I know how to do that. To invert the ln, wouldn't that involve e?
     
  7. Oct 7, 2015 #6
    Oh, wait, there's a definition of cosh involving e! 1/2 e-x+ex?

    Haven't actually done the integral yet, just assumed because of the ln it would be wrong. So I'll try that integral now...
     
    Last edited: Oct 7, 2015
  8. Oct 7, 2015 #7

    HallsofIvy

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    Well, 1/2 (e-x+ex)- the parentheses are important. If B= (1/2)(e-x+ ex) then e-x+ ex= 2B. Multiply on both sides by ex to get 1+ (ex)2= 2Bex. Let y= ex and that becomes the quadratic equation 1+ y2= 2By or y2- 2By+ 1= 0. Use the quadratic formula to find y and then take the logarithm to solve y= ex for x.
     
  9. Oct 7, 2015 #8
    Oh, forgot about the brackets. So my integral is equal to
    1/a ln(##\sqrt{a^2x^2-1}+ax) + c

    If a is the n^2/A^2.

    And that integral is equal to x. So I don't think I end up with two terms of e to the power of something, just two es multiplied together.

    Hardest 'show that' question I've ever been given!
     
  10. Oct 8, 2015 #9
    I think I still haven't got a cosh. Although I haven't posted my full working, as far as I can tell I don't get two e^something terms when I invert to find y. I'm fairly sure my integration is correct, checked it with wolfram alpha!
     
  11. Oct 8, 2015 #10

    epenguin

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    I'm not working on it but idea where to look: it might have to do with when you write your first √ you have to write ± so maybe you get two |solutions you have to combine in one? Or if you integrate to get a LN there a tricky thing about LN | | , maybe? - check your textbook. Maybe these two things are related.
     
  12. Oct 12, 2015 #11

    epenguin

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    Forget most of what I said. We haven't seen working :oldfrown: and if we had I dare say someone would have spotted quite soon.

    Now I had time to look at it (then was cut off by 48 h Internet blackout) it is looking like a simple error in calculation. The OP probably overlooked a square rooting at some point - he would have got this log result integrating an inverse quadratic, instead the problem gives an inverse square root of a quadratic whose standard integral is an inverse cosh.
     
    Last edited: Oct 12, 2015
  13. Oct 12, 2015 #12

    HallsofIvy

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    Since the problem is to show that [itex]y = -\frac{1}{a} + \frac{A}{na} cosh( naA \frac{na}{A} (x-x0))[/itex] is a solution to the given differential equation, I don't understand why you are integrating at all! Just differentiate y, the substitute the given y and its derivative into the equation and show that the equation is satisfied.
     
  14. Oct 12, 2015 #13

    epenguin

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    Well yes people should always do that when they read physics instead of the so-called 'solving', they will get through college them faster and with less distraction from the physics. (Seriously).
    You
     
  15. Oct 12, 2015 #14
    Yes, I did try that first.

    So dy/dx is -sinh(na/A (x-x0) ) from the given equation, so y'2 is then sinh2(...), and sub into equation for A to get n(1+ay)/ sqrt (1+ sinh2(...) ) = n(1+ay) / sqrt ( cosh2(...) )
    which is just n(1+ay) / cosh(...) = A. Then substitute the y? That'll be the step I forgot! Oops!

    Edit: And the step which, having just completed, gave me the answer in a really quick and easy way! What a stupid mistake to make. Thanks for solving my problem! :D
     
    Last edited: Oct 12, 2015
  16. Oct 12, 2015 #15
    I got my ln result from integrating ## \frac{1}{ \sqrt{\frac{n^2(1+ay)^2}{A^2} -1}}##. Not that I actually know how to integrate that, I had to use wolfram alpha... Yeah, I suppose that might be the main problem really, I don't know where to even start with an integral like that!

    Because even if I try and condense the thing under the square root into something manageable, which for me means replacing multiple constants with a single constant, I end up with
    1 / ##\sqrt{a(1+by)^2 -1}##
    Which isn't a standard kind of integral that I know...

    Also an integral I no longer need to solve! :) I do feel very silly though!
     
    Last edited: Oct 12, 2015
  17. Oct 12, 2015 #16
    I know, I did realise afterwards that not posting workings must have made it more difficult. I do appreciate you replying despite that :) Thank you for helping me everyone! :D
     
  18. Oct 12, 2015 #17

    epenguin

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    We were taught to do.these integrations at school. Somewhat pointless knowledge for most, I think. Or not even knowledge, at least I remember being asked for it or something like it a few months after school - similar things have some slight application in chemical kinetics - and I couldn't do it.

    You should however be able to reduce it to standard form by substitution, otherwise you may not be able to find or recognize it in a table of standard forms. You only needed to make a new variable Y = n(1 + ay)/A and you will get to integrating 1/√(Y2 - 1), a standard form found in halfway decent books on calculus.
     
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