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

I Constructing a 2nd order homogenous DE given fundamental solution

  1. May 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Given a set of fundamental solutions {ex*sinx*cosx, ex*cos(2x)}

    2. Relevant equations
    y''+p(x)y'+q(x)=0
    det W(y1,y2) =Ce-∫p(x)dx

    3. The attempt at a solution
    I took the determinant of the matrix to get
    e2x[cos(2x)cosxsinx-2sin(2x)sinxcosx-cos(2x)sinxcosx- cos2xcos(2x)+sin2xcos(2x)

    Then using the identities sin2x+cos2x = 1 (for the last 2 terms) and sin(2x) = 2sinx*cosx (for the second term) and cancelling the 2 "cos(2x)cosxsinx" (first and third terms) I got
    -e2x(sin2x+cos(2x))

    Setting this equal to Ce-∫p(x)dx and trying to solve I got as far as
    ln(-e2x(sin2x+cos(2x))/C) = -∫p(x)dx

    and now I'm a little bit stuck, I also don't know how to solve for q(x) here. Thanks for the help!
     
  2. jcsd
  3. May 13, 2017 #2

    Mark44

    Staff: Mentor

    Your work seems like the long way around -- there's a much simpler way. Your fundamental set could also be ##\{e^x\sin(2x), e^x\cos(2x) \}##. The fundamental set as given and this one both span identical function spaces, and are bases for the same space.

    It's also helpful to recognize that yet another basis would suffice: ##\{e^xe^{2ix}, e^xe^{-2ix} \} = \{e^{(1 + 2i)x}, e^{(1 - 2i)x} \}##. All three of these fundamental sets span the same solution space, and are bases for it.

    Finally, if your fundamental set were ##\{ e^{r_1t}, e^{r_2t}\}##, do you understand that ##r_1## and ##r_2## are solutions to the quadratic characteristic equation? If so, you can work backwards from the characteristic equation to the homogeneous diff. equation with very little work.
     
  4. May 13, 2017 #3
    I actually realized this shortly after posting the problem, but wouldn't sinx*cosx end up with the solution being 1/2*ex*sin(2x)? That's the next part I got stuck on. I completed the problem using only 1+/-2i to get the quadratic in hopes for partial credit and got y''-2y'+3/2
     
  5. May 13, 2017 #4

    Mark44

    Staff: Mentor

    Yes, it would, but it wouldn't matter. Constant multiples of the two original functions will still be a fundamental set (i.e., a basis for the solution space), so the constant 1/2 doesn't need to be included.
    That's not the correct DE. Its characteristic equation would be ##r^2 - 2r + 3/2 = 0##, and the roots of that equation aren't 1 +/- 2i.
     
    Last edited: May 14, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Constructing a 2nd order homogenous DE given fundamental solution
  1. 2nd Order De Solution (Replies: 2)

Loading...