partial fraction help

[samus101]

Hi, can someone please help me with this partial fraction.

dp/(P-P_u)(P_s-P)

1/(P-P_u)(P_s-P) = a/(P-P_u) + b/(P_s-P)

1 = a(P_s-P) + b(P-P_u)

1 = aP_s - aP + bP - bP_u

1 =aP_s - bP_u- P(a-b)

a - b = 0

a = b

1 = aP_s - bP_u
This is where I get stuck. I cant get a or b by themselves so I can put back into the equation.
Thanks.

 

[tkhunny]

Have you considered substituting a = b into the last equation?

 

[Deleted member 4993]

Hi, can someone please help me with this partial fraction.

dp/(P-P_u)(P_s-P)

1/(P-P_u)(P_s-P) = a/(P-P_u) + b/(P_s-P)

1 = a(P_s-P) + b(P-P_u)

1 = aP_s - aP + bP - bP_u

1 =aP_s - bP_u- P(a-b)

a - b = 0

a = b

1 = aP_s - bP_u
This is where I get stuck. I cant get a or b by themselves so I can put back into the equation.
Thanks.

\(\displaystyle \dfrac{1}{ (P_s - P) (P_u - P)} \ = \ \dfrac{A}{P_s - P} \ - \ \dfrac{B}{P_u - P} \)

\(\displaystyle \dfrac{1}{(P_u - P)} \ = \ A \ - \ \dfrac{B(P_s - P)}{P_u - P}\)

Find the limit of both sides as P → P_s

\(\displaystyle \dfrac{1}{(P_u - P_s)} \ = \ A \ \)

similarly

\(\displaystyle \dfrac{1}{(P_s - P)} \ = \ \dfrac{A(P_u - P)}{P_s - P}- \ B \)

Find the limit of both sides as P → P_u

\(\displaystyle \dfrac{1}{(P_u - P_s)} \ = \ B \ \)

 

[soroban]

Hello, samus101!

The subscripts and the "reversals" are confusing.
Let \(\displaystyle a = P_u,\; b = P_s\)

\(\displaystyle \dfrac{1}{(P-a)(b-P)}\)

We have: .\(\displaystyle \dfrac{-1}{(P-a)(P-b)} \;=\;\dfrac{A}{P-a} + \dfrac{B}{P-b}\)

. . . . . . . . . . . . . . . . . \(\displaystyle -1 \;=\;A(P-b) + B(P-a)\)


Let \(\displaystyle P = a\!:\;\;-1 \;=\;A(a-b) \quad\Rightarrow\quad A \,=\,\dfrac{1}{b-a}\)

Let \(\displaystyle P = b\!:\;\;-1 \;=\;B(b-a) \quad\Rightarrow\quad B \,=\,\dfrac{-1}{b-a}\)


Therefore: .\(\displaystyle \dfrac{-1}{(P-a)(P-b)} \;=\;\dfrac{\frac{1}{b-a}}{P-a} - \dfrac{\frac{1}{b-a}}{P-b} \;=\;\dfrac{1}{b-a}\left(\dfrac{1}{P-a} - \dfrac{1}{P-b}\right) \)

 

[samus101]

Thanks for your help guys.