Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Which solution for this DE is preferred?

  1. Oct 3, 2013 #1
    64ADMbk.png

    Are they both correct? Would the first or second solution be preferred?
     
  2. jcsd
  3. Oct 3, 2013 #2

    phyzguy

    User Avatar
    Science Advisor

    The second one is certainly not a solution. There is a solution in terms of exponentials, but that is not it.
     
  4. Oct 3, 2013 #3
    Why is the second one not a solution? If you convert the original DE into an auxiliary equation, you will get roots: m1=0 and m2=k2
     
  5. Oct 3, 2013 #4

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The first of your solutions can be arrived at from the second. I would say that the second is the fundamental.
     
  6. Oct 4, 2013 #5
    No, the second is not correct.
     

    Attached Files:

  7. Oct 4, 2013 #6

    phyzguy

    User Avatar
    Science Advisor

    No you don't. You get roots of +k and -k. Try plugging your second solution into the DE and see if it works. You'll see that it doesn't.
     
  8. Oct 4, 2013 #7
    That's weird: http://www.wolframalpha.com/input/?i=m^2-k^2m=0

    Going to try plugging it in.

    Edit: Oh wow. You're right. I'm an idiot. Lol
     
  9. Oct 4, 2013 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The characteristic equation for the D.E. you give, [itex]y''- k^2y= 0[/itex] is [itex]r^2- k= 0[/itex] which is equivalent to [itex]r^2= k^2[/itex] and has roots k and -k. You, apparently, miswrote the equation as [itex]y''- k^2y'= 0[/itex], which has characteristic equation [itex]r^2- k^2r= r(r- k^2)= 0[/itex] and has roots 0 and [itex]k^2[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Which solution for this DE is preferred?
  1. 2nd Order De Solution (Replies: 2)

  2. DE solutions (Replies: 1)

Loading...