Question

calculus 2 - partial fractions

Answer 1

@ganeshie8

Answer 2

step1 : factor denominator

Answer 3

8/(y(y+2)(y-2)

Answer 4

step2 : setup the fractions

Answer 5

A/y + B/(y+2) + C/(y-2)

Answer 6

yes

\(\large \frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}\)

Answer 7

there is a short cut for finding A, B, C values
u heard of 'cover up method' before ?

Answer 8

no what is that?

Answer 9

its easy,
let me show u how to work that method by doing this

Answer 10

ok thx

Answer 11

\(\large \frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}\)

to find A, multiply both sides with \(y\) and plug \(y= 0\) :

\(\large \frac{8}{(y+2)(y-2)} = A + y [\frac{B}{y+2} + \frac{C}{y-2}]\)

Answer 12

plug y = 0, and solve A

Answer 13

you can do this for all A,b,c's?

Answer 14

yup, let me walk u thru completely.
first find A

Answer 15

-2?

Answer 16

yes A = -2

Answer 17

\(\large \frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}\)

to find B, multiply both sides with \(y+2\) and plug \(y=-2\) :

\(\large \frac{8}{y(y-2)} = \frac{A(y+2)}{y} + B + \frac{C(y+2)}{y-2}\)

Answer 18

plug y = -2

Answer 19

and solve B

Answer 20

1

Answer 21

yes, find C next..

Answer 22

5/3

Answer 23

try again

Answer 24

\(\large \frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}\)

to find C, multiply both sides with \(y-2\) and plug \(y=2\) :

\(\large \frac{8}{y(y+2)} = \frac{A(y-2)}{y} + \frac{B(y-2)}{y+2} +C\)

Answer 25

plug y = 2

Answer 26

1

Answer 27

Thank you

Answer 28

np :)
remember, after setting up fractions, u can use any of 3 methods below for finding A, B, C... values :
1) comparing coefficients
2) cover up method
3) plug a random point

Answer 29

use which ever is easy to use based on the problem at hand...

Answer 30

btw, cover up method oly works if the fractions are LINEAR

Answer 31

the other two methods will work always... no restrictions for them.