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Integration of ... Given dx/dt

Date: 7/5/96 at 5:48:56
From: Anonymous
Subject: Integration of ... Given dx/dt

I am attempting to integrate  da/dt = k.(a)^m.(1-a)^n
where (a) is a fraction. Therefore (a) could be written as (x/A).
When time = 0 , x = 0 and when time is infinite x = A.


dx/dt = k.A.(x/A)^m.(1-(x/A))^n 

What I wish to find is the value of x (or x/A) as a function of 

I can solve the related problem:

dx/dt = k.(A-x)^m

which results in:

dx/(A-x)^m = k.dt                    let u = (A-x)

k.t = u^(-m) . dx    

	= u^(-m).(-du)

	=   [-u^(1-m) / (1-m) ] + c

k.t = c - [ A-x^(1-m) / (1-m)]

When t=0, x=0 and when t = infinite x = A, 
therefore    c = A^(1-m) / (1-m)

k.t = [A^(1-m) / (1-m) ] - [ (A-x)^(1-m) / (1-m)

So   (A-x) = [A^(1-m) - k.t.(1-n)]^(1/1-m)

However I am unable to solve the equation when there are two power 
terms (m and n). I will be very greatful if you can find a solution to 
this problem.


Date: 7/5/96 at 8:19:36
From: Doctor Anthony
Subject: Re: Integration of ... Given dx/dt

The equation you started with can be more conveniently expressed as
dx/dt = k*x^m*(1-x)^n,  so separating the variables we get

dx/(x^m*(1-x)^n) = k*dt

Now use partial fractions on the left hand side of this equation:

   1/(x^m*(1-x)^n) = A/x^m  +  B/(1-x)^n

                 1 = A(1-x)^n  +  B*x^m

Put x=0 you get 1 = A,   Put x=1 you get 1 = B.  So we have A=1,  B=1

Left hand side of integral becomes:

INT[{1/x^m  +  1/(1-x)^n}*dx] = {x^(1-m)}/(1-m) - {(1-x)^(1-n)}/(1-n) 

and the full integral can be written:

{x^(1-m)}/(1-m) - {(1-x)^(1-n)}/(1-n) = kt + Const.

In general you will not be able to express x explicitly as a function 
of t. Even the relatively simple expression like x^3 - (1-x)^5 = t  
could not be manipulated to make x the subject of the formula.  To 
find values of x for given values of t you should use the Newton-
Raphson method, Maple or a TI-92 calculator. 

-Doctor Anthony,  The Math Forum
 Check out our web site!   
Associated Topics:
College Calculus

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