Suppose we have the following differential equation using operator notation: (D−x)(D+x)y=0 where D=ddx
Now I could rewrite (1) as (D−x)(D+x)y=(ddx−x)(y′+xy)=y′′+(xy)′−xy′−x2y=y′′+y+xy′−xy′−x2y⟹y′′−x2y+y=0
Or, by switching the order of the brackets; I could rewrite (1) as (D−x)(D + x)y=(D+x)(D - x)y=(ddx+x)(y′−xy)=y′′−(xy)′+xy′−x2y=y′′−y−xy′+xy′−x2y⟹y′′−x2y−y=0
Lastly, I could rewrite (1) as (D−x)(D+x)y=(D2−x2)y=(d2dx2−x2)y=y′′−x2y⟹ y′′−x2y=0
There's no doubt there's most probably a simple explanation for it; but how can the same differential equation (1) be written in three different ways: (a), (b), (c)?
Many thanks.
Answer
D+x and D−x do not commute. Your (b) is a demonstration of this.
In (c) you are equating (D+x)(D−x) and D2−x2, but that is wrong for a similar reason - in D2−x2 the first D is no longer acting on the second x.
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