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 from: Raven on May 25, 2004, 03:04 PM
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. :)
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. :)