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Better way of finding particular integral

  1. Jan 4, 2007 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the particular integral for a second order linear differential equation:
    given a<<c and b<<c

    3. The attempt at a solution
    It's straight forward to rewrite the right hand side as the sum of two sines, consider each seperately and then add the solutions since it's linear, but this leads to 4 seperate terms and the algebra's a bit messy. So I was just wondering is there a better way to deal with products of functions on the right hand side?

    Last edited: Jan 4, 2007
  2. jcsd
  3. Jan 5, 2007 #2


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    No there isn't. There are, basically, two ways to find a particular solution for a nonhomogeneous linear equation: undetermined coefficients and variation of parameters. In order to use undetermined coefficients, it is necessary that the right hand side be one of the types of functions one can get as a solution to a linear equation with constant coefficients, exponential, polynomial, sine or cosine, or products of those. In order to use that for this problem you have to do just as you say here: write sin(x) cos(x) as a sum. Variation of Parameters works for any right hand side but is typically much messier.
  4. Jan 5, 2007 #3


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    You could slice and dice some other ways, but you will finish up with 4 parts however you go.

    For example you could start by assuming the solution was of the form B cos(bx) sin(cx) (B is an unknown constant) since you obviously need that for the "...+k^2y = A cos(bx)sin(cx)" part of the equation. When you differentite you'll find you also need terms like sin(bx)sin(cx), sin(bx)cos(cx), cos(bx)cos(cx)

    Or you could use exponential functions with 4 terms like e^i(b+c)x, e^i(b-c)x, etc.

    Take your pick - one form may be nicer than the others for the rest of your problem.
  5. Jan 5, 2007 #4
    Thank you,
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