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Nonhomogeneous Power Series Solution

  1. Sep 19, 2007 #1
    For the fun of it, my DE book threw in a couple of problems involving nonhomogenous second order DE's in the section I'm currently going through. Although I have solved for the complementary solution, any suggestions on how to find the particular solution?

    For example, the one I'm looking at right now is y''-xy = 1.
     
  2. jcsd
  3. Sep 20, 2007 #2

    HallsofIvy

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    [tex]1= 1+ 0x+ 0x^2+ 0x^3+ \cdot\cdot\cdot[/tex]
    After you have written the left side of the equation as a power series in x, do the same on the right. Coefficients of powers on the left must equal coefficients of corresponding powers on the right. Here, your coefficient of x0 must be equal to 1 and all others equal to 0.

    That was too easy. Suppose your equation were y"- xy= ex?

    Expand ex in a power series:
    [tex]e^x= 1+ x+ (1/2)x^3+ \cdot\cdot\cdot+ (1/n!)x^n+ \cdot\cdot\cdot[/itex]
    Now the coefficient of xn, on the right, will be equal to 1/n!.
     
    Last edited: Sep 25, 2007
  4. Sep 25, 2007 #3
    Thanks; makes good sense.

    I have another question. This time it involves indicial roots. Working with the Frobenius method I find that in the DE: xy'' + 2y' -xy = 0 both of the indicial roots come to give me the same series solution. I used r = 0 to attain it and the book used r = -1. (or so I suppose since r=-1 is in their solution) The problem comes in that when I use r = 0 it doesn't solve the DE while r = -1 does. Is there any reason why there should be this difference between roots? Any rule of thumb to work with when picking which roots to use?

    The book uses the larger root in its first solution when giving the general way of solving this case of the Frobenius method. When doing an actual example it uses the smaller root for the first solution. I take it that the smaller root should always be applied first?
     
  5. Sep 25, 2007 #4

    HallsofIvy

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    To quote from a differential equations text book,
    In that case, you can use the known solution, corresponding to the larger root of the indicial equation, to reduce the order of the equation and solve that for an independent solution. Of course, since you only know the first solution as an infinite series, that may be very difficult!
     
  6. Sep 27, 2007 #5
    I think you have got the indicial roots wrong. The correct roots are r=0 and r=1.
     
  7. Jul 20, 2011 #6
    quick question. So for this question, y" - xy = 1......is the particular solution y=1?


    Also, how would i find the particular solution to y"+y=x?
     
    Last edited: Jul 20, 2011
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