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@nincompoop , I'm waiting.

y''+p(x)y'+q(x)y=0
I'll start with homogenous first? And find two linearly independent solutions?

Mm - I should've clarified and used constant coefficients. >.<

http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html
wait, lemme read this >.>

this is because the operators are commutative.
(D-a)[(D-b)y(x)] = (D-b)[(D-a)y(x)]

at least for constant coefficients .. since there is no 'x' factor to differentiate.

@nincompoop , I don't need your copy pasting it's lagging my post.

I see @experimentX - so in case of so in other cases you can't simply factor the operator?

No ... check this out
(D + a)(x^2D + b) y(x)

?

(D + a)(x^2D + b) y(x) =/= (x^2D + b)(D + a) y(x)

ohhh

sure