Multiplying out differential operators

[chipotleaway]

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

 

[bigfooted]

it's because of the chain rule. Note that there is a difference between
\partial(b)y and \partial(by). First write it like this, it makes it more clear:
(\partial + a)(\partial y +by)=0
In this case, the differential operator acts on both b and y, so you have \partial(by)=\partial(b)y + \partial(y)b

 

[chipotleaway]

Thanks, I think I get it - you mean the product rule, yeah?

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