1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 2nd order nonhomogenous DEs

  1. May 27, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations


    3. The attempt at a solution
    First issue is I was under the impression that a particular solution is the final solution to a DE; a solved DE with initial conditions applied, but it would be weird for that to be the first part of the question so I'm interpreting a) as asking for the particular integral (maybe they're the same thing :S). Anyway for the first part of the question I've done:


    so dyPI/dt = iCeit and d2yPI/dt2= -Ceit


    which gives: C=1/3

    so yPI=(1/3)eit

    Using a trial solution for the complementary function gives yCF=Ae2it+Be-2it

    and a general solution of:


    At this point I have no idea how to start applying initial conditions as I have an expression that's full of 'i's, so maybe I don't understand the first part of the question and have gone about the whole thing the wrong way. Some help would be really appreciated!
    Last edited: May 27, 2014
  2. jcsd
  3. May 27, 2014 #2


    User Avatar
    Gold Member

    here is a hint that may help.
    Euler's formula

  4. May 27, 2014 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Sorry. You need to brush up on your ODEs.

    The 'final solution' to a linear ODE, to use that unfortunate term, is composed of the solution to the homogeneous ODE plus the solution to the particular ODE. Because there are various constants associated with these solutions, application of the initial values is required to determine the exact solution to the specified ODE and initial conditions.

    This problem is a little inverted, in that it asks you to find a particular solution first, then find the homogeneous solution.

    To apply your two initial conditions, take your general solution y(t) = MESS and set t = 0, since y(0) = 0.
    Then, for y'(0) = 0, calculate the derivative y'(t) and set t = 0 there. You'll wind up with two equations for the two unknown constants A and B. Solve for same.
  5. May 27, 2014 #4
    I tried using this, but when they say y(0)=0, shouldn't they give the value as a complex number? Is it just the real part that =0, or is the complex part 0 as well?
  6. May 27, 2014 #5


    User Avatar
    Gold Member

    ##0 + 0i = 0##
    if: ##y(t) = e^{it} + C##,
    then ##y(0) = e^{0i} + C = 1+C##
    or: ##y(t) = e^{it} + C = \text{cos}(t) + i \text{sin}(t) + C##,
    and ##y(0) = \text{cos}(0) + i \text{sin}(0) + C = 1 + 0i + C##
  7. May 27, 2014 #6
    In this problem its not y(t)=eit+c but ceit. I'm stuck because I can get an expression for the coefficients with y(0)=0 as below:


    But if I use Euler's method and differentiate the expression, all the real terms become sine terms, so saying that dy/dt(0)=0 doesn't help.
  8. May 27, 2014 #7


    User Avatar
    Gold Member

    but you have no dy/dt expression in your diff eq?
  9. May 27, 2014 #8


    User Avatar
    Gold Member

    oh I didn't see that in your initial conditions
  10. May 27, 2014 #9
    That's where you mistake is! Don't use Euler method Find the complete complex form for dy/dt and set it to zero at t=0.
  11. May 27, 2014 #10
    Ah I got it, didn't need to use Euler's at all, had a massive brain fart and forgot that I could just equate real and complex terms anyway to solve for constants.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted