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Solve differential equation using power series

  1. Nov 16, 2011 #1
    Find y(x) as a power series satisfying:
    y'-2xy=0 , y(0) = 1

    attempt:

    y=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] an(x-x0)n

    y'=[itex]\sum[/itex][itex]^{∞}_{n=0}[/itex] (n+1)an+1(x-x0)n

    y' - 2xy = y' - 2y(x-x0) - 2x0y

    substituting the power series into above formula and simplifying eventually gives:

    a1-2x0a0 + [itex]\sum[/itex][itex]^{∞}_{n=1}[/itex](x-x0)n[(n+1)an+1 - 2an-1 - 2x0an] = 0

    so when n=0 and n=1 we get:
    a1=2x0a0

    when n>1,
    an = [itex]\frac{2}{n}[/itex](x0an-1 + an-2)

    but what does my professor mean by y(0) = 1? I think he was trying to explain that this means that x0 = 0 ?



    Also, another problem is that y' - 2xy = x3ex2, y(0)=1
    In this case, any hint as to what should I do with the e^x^2????
     
    Last edited by a moderator: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2

    Mark44

    Staff: Mentor

    This means that y = 1 when x = 0.
     
  4. Nov 16, 2011 #3
    yes I know that but what I meant was, how can I use that to solve the problem? If I already have the recursive formula giving me the coefficients in terms of an, of what use is that y(0)=1? On my hw it says y(0)=1 but when my professor was talking about it, he wrote y(x0)=1, but they are not the same thing?? Is there some relation between them I am not seeing?

    I mean if x=0, then
    y(0)=[itex]\sum^{∞}_{n=1}[/itex](0-x0)n=1
    so:
    1-x0+x02-x03+... = 1
    so this implies x0=0 ? but then all of my coefficients for odd n's become 0..idk if thats supposed to happen...
     
    Last edited: Nov 16, 2011
  5. Nov 16, 2011 #4
    Nothing wrong with the odd ones being zero if they have to. And as far as that [itex]e^{x^2}[/itex], just expand it out in it's power series form and equate coefficients on both sides just like you did above except in this case, all the coefficients on the right side are no longer zero.
     
  6. Nov 16, 2011 #5
    So but then y(0)=1 means x0=0 and that a0 =1, right?
     
  7. Nov 16, 2011 #6

    Mark44

    Staff: Mentor

    Yes, x0 = 0, so your series should be expanded in powers of x, meaning that your series solutions is y = a0 + a1x + a2x2 + .... The initial condition y(0) = 1 determines a0.
     
  8. Nov 16, 2011 #7
    That is correct. Also, compute four or so of the coefficients and then compare them to that series you wrote for the right side of the other one.
     
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