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First and second order ODE.

  1. Sep 29, 2008 #1
    The Problem

    You are given: [​IMG]

    Where [​IMG] is constant (taken as B).

    a) Differentiate both sides to produce a second order ODE for y(x)
    b) Show that it can be written as a first order ODE for u=dy/dx
    [​IMG]
    c) Find the general solution for part b), you should have two arbitrary constants.
    2. Relevant equations

    The fundamental theorem of calculus: [​IMG]

    3. The attempt at a solution

    a) Using fundamental theorem of calculus, d2y/dx2 = Bf(x)
    => d2y/dx2 = B(1 + (dy/dx)2)1/2

    b) Let u = dy/dx => du/dx = B(1 + (dy/dx)2)1/2 = B(1 + u2)1/2

    c) Not a clue.. I have actually got no idea where to start for this. I would solve du/dx first, but wouldn't that just result in what I was given, the equation for part a)?

    Thanks in advance.
     
  2. jcsd
  3. Sep 29, 2008 #2

    Dick

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    The differential equation for u is separable. Separate it and integrate both sides. Then once you have u, integrate again to get y.
     
  4. Sep 29, 2008 #3

    Defennder

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    (a) and (b) looks correct.

    (c)Just do as it says. You will not end up with what you have in part a, if you denote dy/dx with u. You just have to find the GS.
     
  5. Sep 30, 2008 #4
    Okay, thanks for the fast replies.

    I seperated and integrated both sides of the ODE for u like this:

    [​IMG]

    Think it's right so far, but I need to get theta back into terms of u. I don't exactly know how I'm going to go about this..
    After that I'm going to need to integrate again I think, however my equation won't be in the form of a differential equation will it? How will I know what to integrate in terms of.
    Thanks.
     
  6. Sep 30, 2008 #5

    HallsofIvy

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    Gosh, that's so hard to read!

    You have [itex]du/dx= B\sqrt{1+ u^2}[/itex] and then
    [itex]\sqrt{1+ u^2}du= Bdx[/itex]

    Well, that's wrong right there. You have to divide both sides by [itex]\sqrt{1+ u^2}[/itex], not multiply.
     
  7. Sep 30, 2008 #6
    Wow. I cannot believe that I made such a blatant error myself. Blimey!!

    Okay, I shall withdraw myself for sometime and make sure that next time I put something up here it isn't so horribly flawed haha...

    Also, so that it isn't so hard to read next time, it would be great if someone could show me how to type all those math symbols in normal posts (I'm pretty new here).

    Cheers
     
  8. Sep 30, 2008 #7

    Dick

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    If you click the mouse on some of those symbols, like the ones in Hall's post, a window should pop up with the TeX in it. You then enclose the stuff between [ tex] and [ /tex] (without the space). Like this,
    [tex]
    \int \frac{1}{\sqrt{1+ u^2}}du= \int Bdx
    [/tex]
     
  9. Sep 30, 2008 #8
    I did the tex thing, but for some reason when I write more than one line of it I get a page load error. Been trying different combinations for about 40 minutes so I decided I'll try again later and post this for now.

    [​IMG]

    The problem I have now is getting theta back in terms of u.

    Thanks

    EDIT: Hang on....... theta = arcsinh(u)?
    if it's that simple I'm gonna be kicking myself!
     
    Last edited: Sep 30, 2008
  10. Sep 30, 2008 #9

    Dick

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    Ok, arcsinh(u)=B*x+C. u=y'. You still have to get back to y. Sometimes the tex thing can be frustrating. But I'm not using tex, and you understand me, right?
     
    Last edited: Sep 30, 2008
  11. Sep 30, 2008 #10
    so [tex] y = \int sinh(Bx+C)dx = \int sinh(Bx)dx + \int Cdx = \frac{1}{B}cosh(Bx) + Cx + D
    [/tex]

    The only thing I'm not sure about here is whether or not I'm allowed to take the C out of the sinh function, I think I am allowed to because it's an arbitrary constant. However if I'm not allowed to, I might need a helping hand in doing the integral because I'm not sure how I'd go about it.

    Pretty sure this is correct though, and I do have the correct number of constants!
     
  12. Sep 30, 2008 #11

    Dick

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    You aren't allowed to take C outside of the sinh. You have to live with that. Can't you still integrate sinh(Bx+C) without making that mistake? Just substitute u=Bx+C and integrate.
     
  13. Sep 30, 2008 #12
    Okay hopefully this is the last time..


    [tex] dx = \frac{du}{B} [/tex]

    [tex] y = \int sinh(u)dx
    \int \frac{1}{B} sinh(u)du
    = \frac{1}{B} cosh(u) + D
    = \frac{1}{B} cosh(Bx + C) + D
    [/tex]

    Thanks for being so patient, I know this is really frustrating!!
     
  14. Sep 30, 2008 #13

    Dick

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    It's not half as frustrating as most. In fact, it's not frustrating at all. Yes, y=cosh(Bx+C)/B+D. If you have any doubts, put it back into the ODE and try it. The only thing frustrating here is that I have to keep scrolling my pages back and forth because of your oversized scans.
     
    Last edited: Sep 30, 2008
  15. Oct 2, 2008 #14
    Referring to part (a) , I dont quite get it. If F'(x) = f(x), how do you solve the integral f(t) dt to get f(x)? Mine giving me a headstart to this? Thanks
     
  16. Oct 2, 2008 #15

    Dick

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    To get the integral of f(t) from 0 to x, you find an antiderivative F (such that F'=f) and evaluate F(x)-F(0), right? So the derivative of that integral is f(x).
     
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