Acos(lnx) + Bsin(lnx) = y, where A and B are constants.
Prove: (x^2)y'' + (xy)' + y = 0.

I hate having to figure out how to go about these proofs. Thanks. :)

[quote author=Raven link=board=36;threadid=6956;start=0#msg61812 date=1085515471]
Acos(lnx) + Bsin(lnx) = y, where A and B are constants.
Prove: (x^2)y'' + (xy)' + y = 0.

I hate having to figure out how to go about these proofs. Thanks. :)
[/quote]

Looks easy, all you need to do is take the derivative of a few things?

An attempt:

y = Acos(lnx) + Bsin(lnx)
y' = (-Asin(lnx) + Bcos(lnx))/x
y'' = (-Acos(lnx) - Bsin(lnx) + Asin(lnx) - Bcos(lnx))/ x^2
(xy)' = xy' + y = -Asin(lnx) + Bcos(lnx) + Acos(lnx) + Bsin(lnx)

(x^2)y'' + xy' + y = 0 :

(x^2)((-Acos(lnx) - Bsin(lnx) + Asin(lnx) - Bcos(lnx))/ x^2) + x(-Asin(lnx) + Bcos(lnx))/x + Acos(lnx) + Bsin(lnx) = - Acos(lnx) - Bsin(lnx) + Asin(lnx) - Bcos(lnx) - Asin(lnx) + Bcos(lnx) + Acos(lnx) + Bsin(lnx) = 0

(x^2)y'' + (xy)' + y = 0 : (false)

(x^2)y'' + (xy)' + y = (x^2)y'' + xy' + y + y = y

(x^2)((-Acos(lnx) - Bsin(lnx) + Asin(lnx) - Bcos(lnx))/ x^2) - Asin(lnx) + Bcos(lnx) + Acos(lnx) + Bsin(lnx) + Acos(lnx) + Bsin(lnx) = - Acos(lnx) - Bsin(lnx) + Asin(lnx) - Bcos(lnx) - Asin(lnx) + Bcos(lnx) + Acos(lnx) + Bsin(lnx) + Acos(lnx) + Bsin(lnx) = Acos(lnx) + Bsin(lnx) = y

Yeah, that looks about accurate. I'll try and go over it later just incase. Thanks alot, you saved me what'd likely be a headache. :)