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ODE Help!

  1. Nov 14, 2011 #1
    Hi all,

    I've been having trouble answering these two ODE problems. Hopefully someone can help me out.

    1. Compute the solution of y"' - xy' = 0 which satisfies y(0) = 1, y'(0) = 0, and y"(0) = 0.

    I've tried using the power series expansion for y and substituting it in and getting the recurrence relation, but when I substitute back the initial conditions I'm getting the two series cancelling out which I don't think is right.

    2. Solve the initial value problem 3y" - y' + (x+1)y = 1 with y(0) = y'(0) = 0

    For this one, I know you have to compute both the power series expansion as well as a particular solution through substitution to be able to apply the initial conditions, but I'm seriously stuck in even thinking about what to substitute, let alone how to tackle this. I tried to start developing out the recurrence relationship, but it got really messy and I don't think I'm doing it right.

    Thanks a lot to whoever can help, I really appreciate it!
     
  2. jcsd
  3. Nov 15, 2011 #2

    HallsofIvy

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    Yes, that's exactly what has to happen! The first thing I would do is let u= y' so the equation becomes u''= xu, u(0)= 0, u'(0)= 0. That satisfies all the conditions for "existence and uniqueness" of solutions and u(x)= 0 for all x is an obvious solution. That means that y is the constant function: y(x)= 1 for all x satisfies the differential equation and initial conditions.


    Doesn't look too bad. Letting [itex]y= \sum_{n=0}^\infty a_nx^n[/itex], [itex]y'= \sum_{n=1}^\infty na_nx^n[/itex] and [itex]y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}[/itex] so the equation becomes
    [tex]3\sum_{n=2}^\infty n(n-1)a_nx^{n-2}- \sum_{n=1}^\infty na_mx^{n-1}+ \sum_{n=0}^\infty a_mx^{n+1}+ \sum_{n=0}^\infty a_nx^n= 1[/tex]
    To get the same powers, change the dummy indices: in the first sum, let j= n-2 so that n= j+ 2 and it becomes
    [tex]3\sum_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j[/tex]
    In the second sum, let j= n- 1 so it becomes
    [tex]\sum_{j= 0}^\infty (j+1)a_{j+1}x^j[/tex]
    In the third sum, let j= n+ 1 so that n= j- 1 and it becomes
    [tex]\sum_{j= 1}^\infty a_{j-1}x^j[/tex].
    Finally, in the fourth sum, let j= n so it becomes
    [tex]\sum_{j= 0}^\infty a_jx^j[/tex]

    Note that the third sum does not start until j= 1 (since it starts with [itex]x^1[/itex]) so for j= 0 we have
    [tex]6a_2- a_1+ a_0= 1[/tex]
    so that [itex]a_2= (a_1- a_0+ 1)/6[/itex]
    since we are given that [itex]a_0= y(0)= 0[/itex] and [itex]a_1= y'(0)= 0[/itex]
    that tells us that [itex]a_2= 1/6[/itex].

    For j> 0, we have
    [tex]3(j+2)(j+1)a_{j+2}- (j+1)a_{j+1}+ a_{j-1}+ a_j= 0[/tex]
    or
    [tex]a_{j+2}= \frac{(j+1)a_{j+1}- a_{j-1}- a_j}{3(j+2)(j+ 1)}[/tex]

    Then [itex]a_3= (2a_2- a_0- a_1)/18)= 1/18[/itex], [itex]a_4= (2a_3- a_1- a_2)/36= -1/648[/itex], [itex]a_5= (2a_4- a_2- a_3)/90[/itex] etc. Yes, it wil probably be difficult to find a general formula. Are you required to?
     
    Last edited: Nov 16, 2011
  4. Nov 15, 2011 #3
    Hi. Thanks a lot for the help!

    For number 2, what happened to the coefficient of 3 in front of the y"? You don't factor this into the series solution?

    And I think we're allowed to keep the answers in a recurrence relation as long as it's correct, so I think that's what I'll do and I'll list out some terms in the series for clarification. Thanks!
     
  5. Nov 16, 2011 #4

    HallsofIvy

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    Yes, I accidently dropped the 3. I've gone back and corrected that. Hope its right this time!
     
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