Integration of ... Given dx/dtDate: 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. So: 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 time(t). 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. Richard. 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! http://mathforum.org/dr.math/ |
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