1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reduction of order

  1. Oct 15, 2006 #1
    hello everyone, im stuck on this problem and i have a hard time figuring out how they went from here dw/w=-(2dx/x(x^2+1)) to here w=c(x^2+1)/x^2. i know they integrated, but can anyone show me the details of the integration along with an explanation?
  2. jcsd
  3. Oct 17, 2006 #2
    partial fractions
    dw/w = -2 dx/ (x*(x^2+1))
    = (-2/x + 2x/(x^2+1) )dx
  4. Oct 18, 2006 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    qbert used "partial fractions". The fraction can be written
    [tex]\frac{-2}{x(x^2+1)}= \frac{A}{x}+ \frac{Bx+ C}{x^2+ 1}[/itex]
    Multiplying on both sides by that denominator,
    [tex]-2= A(x^2+ 1)+ (Bx+ C)(x)[/tex]
    Let x= 0 and that becomes -2= A.
    Let x= 1 and we have -2= (-2)(2)+ B+ C or B+ C= 2.
    Let x= -1 and we have -2= (-2)(2)+ B- C or B- C= 2.
    Adding, 2B= 4 or B= 2 and C= 0.
    That gives the formula qbert wrote.
    Of course, the integral of -2/x is -2 ln(x) and the the integral of [itex]\frac{2x}{x^2+1}[/itex] can be done by the substitution u= x2+ 1.
    Last edited: Oct 20, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Reduction of order
  1. Reduction of Order (Replies: 2)