0
$\begingroup$

enter image description here

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.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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

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

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

$\endgroup$
2
  • $\begingroup$ Ah alright, and can this expansion be done for any formula, or only for those in the format of y(y-C)? $\endgroup$
    – zavtra
    Mar 2, 2015 at 0:33
  • 1
    $\begingroup$ This is just for the product of two linear factors, such as y(y-c) or more generally (y-a)(y-c). $\endgroup$
    – user84413
    Mar 2, 2015 at 0:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .