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chipotleaway
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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
D2+a(x)D+b(x))y(x)=0,
then factors it
(D+A(x))(D+B(x))y=0
then multiplies it out to determine B(x). I thought we would get
(D2+DB+AD+AB)y=0
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
(D2+BD+B'+AD+AB)y=0
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?)
Thanks
He first rewrites it in terms of differential operators
D2+a(x)D+b(x))y(x)=0,
then factors it
(D+A(x))(D+B(x))y=0
then multiplies it out to determine B(x). I thought we would get
(D2+DB+AD+AB)y=0
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
(D2+BD+B'+AD+AB)y=0
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?)
Thanks
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