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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


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    By definition, a homogeneous ODE is equal to zero.
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