-1

I have: $ \frac{d\tau}{d Z}=\frac{N}{(P-Z)Z} $

And apparently this function can be rewritten as:

$ \frac{d\tau}{d Z}=\frac NP \left(\frac1{P-Z}+\frac 1Z\right) $ for further integration.

But how must I rewrite this function? I tried

$ \frac{d\tau}{d Z}=\frac{N}{(P-Z)Z}\\ \frac{d\tau}{d Z}=\frac{A}{Z}+\frac{B}{(P-Z)}\\ \frac{d\tau}{d Z} =\frac{A(P-Z)}{Z(P-Z)} + \frac{BZ}{(p-Z)Z}\\ N = A(P-Z) + BZ\\ N= AP - AZ + BZ\\ N = AP - Z(A-B)\\ $

and now am I am stuck.

Question 1. Are the steps I have taken so far correct?
Question 2. If so, how do I proceed?

Geetings Gerard

Gerard de Jong
  • 1
  • Apparently you are asking about how to write $\frac{1}{(P-Z)Z}$ as a sum of the form $\frac{a}{P-Z}+\frac{b}{Z}$. You have unnecessarily confused the essential problem with unrelated notation. – hardmath Apr 14 '18 at 20:00

2 Answers2

1

You could use the Cover up rule

we have ,

$\frac{N}{Z(P-Z)}=\frac{A}{P-Z}+\frac{B}{Z}$

To find A we set $P-Z = 0\implies Z =P$

therefore $A =\frac{N}{P}$

to find B we set $Z =0$

therefore $B = \frac{N}{P}$

Hence we can write $\frac{N}{Z(P-Z)}=\frac{N}{P}\cdot\frac{1}{P-Z}+\frac{N}{P}\cdot\frac{1}{Z} = \frac{N}{P}\bigg[\frac{1}{P-Z}+\frac1{Z}\bigg]$

EDIT:

You can also do it the following way,

you have $N =Z(A)+B(P-Z)$

$N = Z(A-B)+P\cdot B$

treat $N,P$ as constants here , and comparing the coefficents gives

$N =P\cdot B \implies B =\frac NP$

and $A-B = 0\implies A = B \implies A = \frac NP$

The Integrator
  • 3,200
0

You can solve it by treating it like a partial fraction, and considering cases

$\ \frac{dr}{dZ} = N\bigl(\frac{1}{(P-Z)Z}\bigr) $ Factorise out N

$\ \frac{dr}{dZ} = N\bigl(\frac{A}{(P-Z)}+\frac{B}{Z}\bigr) $ Write in partial fraction form

Where, $\frac{A}{(P-Z)}+\frac{B}{P}=\frac{1}{(P-Z)P}$

$\frac{AZ+B(P-Z)}{P(P-Z)}=\frac{1}{(P-Z)P}$

$(A-B)Z+BP=1$

This is the part where you got stuck. But you can get out by considering the cases where this expression is true.

Case 1: $(A - B) = 1/Z$ and $B = 0$ (Trivial case, it will give back the original expression and you won't go anywhere)

Case 2: $(A - B) = 0 $ and $B = \frac{1}{P}$, $\Rightarrow A = \frac{1}{P} $

Considering only case 2: $\frac{1}{P(P-Z)}+\frac{1}{P(P)}=\frac{1}{(P-Z)P}$

Plugging this back into the original equation and factorising:

$\ \frac{dr}{dZ} = N\bigl(\frac{1}{P(P-Z)}+\frac{1}{P(P)}\bigr) $ $\ \frac{dr}{dZ} = \frac{N}{P}\bigl(\frac{1}{P-Z}+\frac{1}{P}\bigr) $ ##

NukeyFox
  • 1