Only attempt this question if you have a doctrate in INTEGRATION. Insert ? here:
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Only attempt this question if you have a doctrate in INTEGRATION. Insert ? here:

[From: ] [author: ] [Date: 11-10-06] [Hit: ]
then this isnt terribly hard.Then integrate both sides.The left requires simplifies using partial fractions.1 / [z(z - m)] = A/z + B/(z - m) .........
dZ/dt=kZ(M-Z)

My frustration level: 7/10

Please show me how to find the value of Z as a function of t and m. If this is not possible, please do something to the equation to make it more pretty. It looks so damn ugly.

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If k and M are constants, then this isn't terribly hard. That's a separable differential equation that can be written as (switching to lowercase for easier typing):

dz / [z(m - z)] = kdt
dz / [z(z - m)] = -kdt

Then integrate both sides. The left requires simplifies using partial fractions. Let:

1 / [z(z - m)] = A/z + B/(z - m) .... where A and B are constants to be determined.

Convert the right side to a common denominator and add:
1 / [z(z-m)]= [A(z - m) + Bz] / [z(z-m)]
1 / [z(z-m)] = [(A + B)z - Am] / [z(z-m)]

For both sides to be identical, A+B must be 0 and -Am must be 1. That means:

A = -1/m
B = 1/m
1 / [z(z-m)] = (1/m)[ -1/z + 1/(z - m)]

That makes the original equation
(1/m)[ 1/(z - m) - 1/z] dz = -k dt
[1/z - 1/(z - m)] dz = mk dt

Integrating gives:

ln(z) - ln(z - m) = mkt + C
z/(z - m) = C' e^(mkt) .... where C' = e^C is an arbitrary *positive* constant

Presumably, you want this as a formula for z, so solve:

z = C' e^(mkt) (z - m)
mC' e^mkt = C' e^(mkt) z - z = [C' e^(mkt) - 1] z
z = m [C' e^(mkt)] / [C' e^(mkt) - 1]

A little messy, but not hard if you remember partial fractions.

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lol I wonder how many people with Ph.Ds roam around on this website.
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