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Multiplying out differential operators

  1. Apr 16, 2014 #1
    In this video at around 9:00 , Carl Bender demonstrates a method of solving y''+a(x)y'+b(x)y=0.

    He first rewrites it in terms of differential operators


    then factors it


    then multiplies it out to determine B(x). I thought we would get


    but at 15:29, he says that D, when it acts on B, either it acts on B or it 'goes past B' and acts on y and because of that, we get two terms, BD and D', so the result is


    Why doesn't the operator just act on B?
    If it only acts on B, then shouldn't BD disappear somehow (and vice-versa)?
    Also, this would mean for D to act on something, it has to be the right? (DA≠AD?)

    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Apr 16, 2014 #2
    it's because of the chain rule. Note that there is a difference between
    [itex]\partial(b)y[/itex] and [itex]\partial(by)[/itex]. First write it like this, it makes it more clear:
    [itex](\partial + a)(\partial y +by)=0[/itex]
    In this case, the differential operator acts on both b and y, so you have [itex]\partial(by)=\partial(b)y + \partial(y)b[/itex]
  4. Apr 17, 2014 #3
    Thanks, I think I get it - you mean the product rule, yeah?

    = D2y+D(By)+ADy+ABy
    = D2y+yB'+By'+ADy+ABy
    = D2y+yB'+BDy+ADy+ABy
    = (D2+B'+BD+AD+AB)y
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