# Reduction of order

1. Oct 15, 2006

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. Oct 17, 2006

### qbert

partial fractions
dw/w = -2 dx/ (x*(x^2+1))
= (-2/x + 2x/(x^2+1) )dx

3. Oct 18, 2006

### HallsofIvy

Staff Emeritus
qbert used "partial fractions". The fraction can be written
$$\frac{-2}{x(x^2+1)}= \frac{A}{x}+ \frac{Bx+ C}{x^2+ 1}[/itex] Multiplying on both sides by that denominator, [tex]-2= A(x^2+ 1)+ (Bx+ C)(x)$$
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 $\frac{2x}{x^2+1}$ can be done by the substitution u= x2+ 1.

Last edited: Oct 20, 2006