Question

Partial fractions trick for linear factors?

Answer 1

Our prof gave the example 1/(y(y-2)) and said something that by inspection, you'd get -1/(2y) and 1/(2(y-2)) ... How does he find -1/2 and 1/2 so easily by inspection? He said the trick only worked for non-repeated linear factors that don't have a constant in front of the y either

Answer 2

\[\frac{1}{y(y-2)}\] put your hand over the \(y\) in the denominator (so you are looking only at \(\frac{1}{y-2}\)) the replace \(y\) by \(0\) get \(-\frac{1}{2}\)

Answer 3

then ignore the \(y-2\) in the denominator, so you are looking only at (\frac{1}{y}\) replace \(y\) by \(2\) get \(\frac{1}{2}\)

Answer 4

\[\frac{ 1 }{ y(y-2) }=\frac{ \frac{ -1 }{ 2} }{y }+\frac{ \frac{ 1 }{ 2 } }{ y-2 }\]

Answer 5

typo i mean put your hand over the \(y-2\) so you are looking only at \(\frac{1}{y}\) replace \(y\) by 2 get \(\frac{1}{2}\)

it is more or less by inspection

Answer 6

it is a slick easy trick when you have linear factors in the denominator

Answer 7

hope it is clear

Answer 8

Ohhh I see that's what he was trying to say but went too fast >.< haha Thank you!!

Answer 9

yw
try it for yourself
make up one, and then see how easy it works

Answer 10

\[\frac{ 1 }{ y(y-2) }=\frac{A }{y }+\frac{ B }{ y-2 }\]
\[1=A(y-2)+b(y)\]
for constant: 1=-2A____A=-1/2
When y=2
1=2B
B=1/2