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[Homework Help] Solving Seperable Differential Equations

  1. Dec 13, 2009 #1
    Hi, this is my first time posting a homework help problem, so please excuse any mistakes I happen to make :) I actually have two problems I'm stuck on. I have been trying to figure it out, and I'm sure it's a very simple problem, but I guess I'm just not that bright. Anyways, here they are.

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

    1) Find the particular solution of the differential equation for:

    [tex]11x-6y\sqrt{x^2+1}\frac{dy}{dx}=0[/tex]

    2) Find all values of k for which the function [tex]y=sin(kt)[/tex] satisfies the differential equation [tex]y''+16y=0[/tex]

    2. Relevant equations

    For the first question, you're told to use the following initial condition: [tex]y(0)=4[/tex]

    3. The attempt at a solution

    So for the first question, what I did was bring the x's to the right side and then got stuck at the integral part, which is this:

    [tex]\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx[/tex]

    The left hand side is easy to integrate, however how would I go about integrating the right hand side? That's where I'm getting stuck at. A quick rundown would be very appreciated.


    As for the second question, what I think I'm supposed to do is take the derivative of [tex]y=sin(kt)[/tex] twice, so that I have the second derivative of y, then substitute it into the original equation. Would that be right?

    Thank you very much for any help. Anything would be appreciated :)
     
    Last edited: Dec 14, 2009
  2. jcsd
  3. Dec 13, 2009 #2
    I'd use a u-substitution to integrate the right side.
     
  4. Dec 14, 2009 #3
    as for the second one, if I am understanding you right, that is the right approach
     
  5. Dec 14, 2009 #4
    Alright. Thanks guys :) I think I can do the first question now. But the second question is a tad bit confusing. So let's say I take the first derivative of [tex]y=sin(kt)[/tex] which becomes [tex]y'=kcos(kt)[/tex] right? And then I take the second derivative of that, which is [tex]y''=-k^2sin(kt)[/tex] right?

    After finding the second derivative, I plug in [tex]y=sin(kt)[/tex] and [tex]y''=-k^2sin(kt)[/tex] into [tex]y''+16y=0[/tex] which then makes it into [tex]-k^2sin(kt)+16sin(kt)=0[/tex]. What do I do at this point to find all values of k that would satisfy the differential equation [tex]y''+16y=0[/tex]?

    Sorry if it's confusing to read. But thanks a lot for the help :)
     
  6. Dec 14, 2009 #5
    Factor out the sin(kt) term on the left hand side and it comes down to finding the zeros of both sin(kt) and 16-k^2, which is easy.
     
  7. Dec 14, 2009 #6
    Oh wow I can't believe I missed something so obvious. Lol thanks a lot for pointing that out to me. I got the right answer now :)
     
  8. Dec 14, 2009 #7
    Really sorry for double posting, but I'm still having troubles with the first question. This is what I've done so far after moving the x's to the right side:

    [tex]\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx[/tex]

    [tex]-3y^2=\int\frac{-11x}{\sqrt{x^2+1}}dx[/tex]

    U-substitution for right side
    [tex]u=x^2+1[/tex]
    [tex]du=2xdx[/tex]
    [tex]-5.5du=-11xdx[/tex]

    [tex]-3y^2=-5.5\int(u^\frac{-1}{2})du[/tex]

    [tex]-3y^2=-5.5\frac{u^\frac{1}{2}}{\frac{1}{2}}[/tex]

    [tex]-3y^2=-11(u^\frac{1}{2})[/tex]

    [tex]y=\sqrt{\frac{-11(x^2+1)^\frac{1}{2}}{-3}[/tex]

    That's what I'm getting. I know it's wrong, because the intitial condition given was [tex]y(0)=4[/tex] and when I plug in 0 into the x's, I'm not getting 4. What am I doing wrong? (Sorry if I'm bothering you guys by now. I just really need to understand this stuff as I have an exam soon...)
     
  9. Dec 14, 2009 #8
    all your integration steps are missing the "+C"

    use y(0)=4 to solve for "C", and add it to your equation!
     
  10. Dec 14, 2009 #9
    Ah, right, forgot about the C. But I dont think that's what the question is asking for. Here's the question (My homework is posted on the web, :))

    6jj5zs.jpg

    EDIT: Woops, just saw the last line of your last post just now. Let me do that and I'll get back to you.
     
    Last edited: Dec 14, 2009
  11. Dec 14, 2009 #10
    Okay, so I got the C which is 2.085145784. So what I get now is [tex]y=\sqrt{\frac{-11(x^2+1)^\frac{1}{2}}{-3}}+2.085145784[/tex] yet when I submit it online, the webworks says I am wrong. I really don't get what I'm doing wrong.
     
  12. Dec 14, 2009 #11
    edit:///

    i dunno.
     
    Last edited: Dec 14, 2009
  13. Dec 14, 2009 #12
    lol its fine. thank you for the help though, much appreciated.
     
  14. Dec 14, 2009 #13
    The C should be inside the square root because it comes in integration when y is still squared:

    [tex]
    \int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx \Leftrightarrow 3\int d(y^2)= 11\int{d(\sqrt{x^2+1)} \Leftrightarrow y^2 =\frac{11}{3} \sqrt{x^2+1} + C
    [/tex]

    Now solve for C using the initial condition.
     
  15. Dec 14, 2009 #14
    Ah, thanks to everyone! I got the answer now :)
     
  16. Jan 25, 2010 #15
    An airplane is flying with a velocity of 82.0 at an angle of 24.0 above the horizontal. When the plane is a distance 108 directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. Help please!!!!!!!!!!!
     
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