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

4th oder ODE

  1. Oct 6, 2009 #1
    how can i solve d4y/dt4 - λ4 y= 0
     
  2. jcsd
  3. Oct 6, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since that is a linear equation with constant coefficients, you can immediately write down its "characteristic equation", [itex]x^4- \lambxa^4= 0[/itex]. That factors as [itex](x^2- \lambda^2)(x^2+ \lambda^2)= (x- \lambda)(x+ \lambda)(x- i\lambda)(x+ i\lambda)= 0[/itex] and so has characteristic values [itex]\lambda[/itex], [itex]-\lambda[/itex], [itex]i\lambda[/itex], and [itex]-i\lambda[/itex]. Do you know what to do with those?
     
  4. Oct 7, 2009 #3
    hmm... i dont really understand. can u show me a clearer working? wat you mean by x^4- \lambxa^4= 0? thanks
     
    Last edited: Oct 7, 2009
  5. Oct 7, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What kind of explanation would you understand? Do you know anything at all about linear differential equations? I need to know what you do understand before I can explain much more.
     
  6. Oct 8, 2009 #5
    oh how did you solve x^2 + λ^2?
     
  7. Oct 8, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I didn't solve [itex]x^2+ \lamba^2[/itex]. That is not an equation. I did solve [itex]x^2+ \lambda^2= 0[/itex] by the obvious method: I subtracted [itex]\lambda^2[/itex] from both sides to get [itex]x^2= -\lambda^2[/itex] and then took the square root of both sides. But now, in addition to my previous questions, which you still haven't answered, why are you trying to do differential equations if you don't know how to solve a simple quadratic equation?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook