calculus - Using partial fractions to integrate dy/(y(y-2))

This is a solution given for a practice exam I'm working through. However, I don't get where the 1/(-2)-(-0) (the part with the circled numbers in the denominator) came from.

For instance, when I try to do the partial fraction integration, I get something like:

dy/(y(y-2)) = A/y + B/y-2

With solutions y = 2 and y = 0. So I don't really get where the 1/y-0 - 1/y-2 really even came from at all, why they were circled and brought over to the 1/-2 - -0, and how this method even relates to partial fractions. Any help is greatly appreciated.

Answer

They are using a shortcut to get the partial fraction expansion:

$\displaystyle\frac{1}{(y-a)(y-b)}=\frac{1}{a-b}\left(\frac{1}{y-a}-\frac{1}{y-b}\right)=\frac{1}{-b-(-a)}\left(\frac{1}{y-a}-\frac{1}{y-b}\right)$

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The usual way to do this, as you suggest, is to write

$\displaystyle\frac{1}{(y-a)(y-b)}=\frac{C}{y-a}+\frac{D}{y-b}$ and then solve to get $1=C(y-b)+D(y-a)$, so

$\displaystyle C=\frac{1}{a-b}$ and $\displaystyle D=\frac{1}{b-a}$.