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Finding particular solutions of ODEs'

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    He tells us to find the form of the particular solution without having to compute the actual particular solution.

    For Example,
    (D[itex]^{2}[/itex]+1)y = xe[itex]^{-x}[/itex]+3sinx

    2. Relevant equations
    I'm not even 100% sure how to begin...I was kind of hoping someone could explain what the differential operators mean and how I could even start looking for the form of the particular solution.
     
  2. jcsd
  3. Mar 24, 2013 #2

    Simon Bridge

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    How would you go about finding the actual particular solution in that case then?
    The "form" is the step before you go about getting specific - it's where you decide what sort of function the solution is going to be. It may be good enough just to name the type of function - be it hyperbolic, quadratic, exponential, etc.
     
  4. Mar 24, 2013 #3
    For the 3sin(x) would the guess of the particular solution be in the form of A*sin(x)+B*cos(x) and for the other term, would it be in the form of (C*x + D)*(Fe[itex]^{-x}[/itex])?


    EDIT: Changed T to x
     
  5. Mar 24, 2013 #4

    Simon Bridge

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    Rereading - I may have misunderstood your question:

    The differential operator is just a shorthand for the differentiation operation. $$D_x = \frac{d}{dx}$$... so (D2+1)y = xe−x+3sinx would be $$\frac{d^2y}{dx^2}+y = xe^{-x}+3\sin x$$... so this is an inhomogeneous 2nd order DE.

    The overall solution is composed of the general solution for the homogeneous part added to any independent specific solution which you can guess or figure out from the form of the inhomogeneity... or any other means at your disposal.
    There are a bunch of rules for those guesses - you should have them written down somewhere.

    -------------------------

    Aside (JIC):
    $$(D^2+1)y=D^2y+y = \frac{d^2}{dx^2}y+y=\frac{d^2y}{dx^2}+y$$

    If you don't have notes:
    http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx
     
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