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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

    HallsofIvy

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    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
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