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

2nd order linear homogeneous DE

  1. Feb 3, 2014 #1
    when you solve a 2nd order linear non-homogeneous DE, where it is equal to a constant as in Kirchoff's 2nd Law and the roots of the auxiliary equation are imaginary then you have superposition of 2 solutions. so the particular solution is equal to a constant k and you can solve for this by taking q(0), q'(0), q''(0) and it will have a constant form? I'd like to spend some time studying the theory of this. is it always necessary to use Euler's identity?

    [tex]q(t)=\frac{-1}{25(25e^{20t})}cos(10\sqrt{46}t)+\frac{21}{50\sqrt{46}e^{20t}}sin(10 \sqrt{46}t)+\frac{1}{25}[/tex]


    on another question in the DE which is equal to xsinx can only have derivatives with terms xsinx, xcosx, sinx, cosx by the product rule. so in this case you guess the solution as Axsinx+Bxcosx+Csinx+Dcosx... how do you prove then that these functions are linearly independent?
     
    Last edited: Feb 3, 2014
  2. jcsd
  3. Feb 3, 2014 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    By definition, a homogeneous ODE is equal to zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 2nd order linear homogeneous DE
  1. 2nd order Linear DE (Replies: 1)

Loading...