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February 11th, 2016, 07:53 AM  #1 
Newbie Joined: Feb 2016 From: Toronto Posts: 4 Thanks: 2  Please help me with definite integral
Hello everyone, my question looks like this dS/dt = A/(B + t)  CS/(D+S) 1) A, B, C, D are all constants; 2) Integrate S from S (t=0) to S (t=t) and from t=0 to t=t; 3) S (t=0) >0 and S (t=t) >0;  I tried with similar techniques as the below example. However, I don't think it is correct. Here is how I tired: dS = (A/(B+t)) dt  (CS/(D+S)) dt Integrate from t=0 to t=t for only right hand side: dS = A*ln((B+t)/B)  (CS *t)/(D+S) Then, (CS *t)/(D+S) + dS = A*ln((B+t)/B) (integrate just left hand side?? but the 1st term does not have dS??)  I have an example of this question but simpler: dS/dt = ES/(F+S) 1) E, F are all constants; 2) Integrate S from S (t=0) to S (t=t) and from t=0 to t=t; 3) S (t=0) >0 and S (t=t) >0; (F+S)/S dS = E dt Integrate both side: (F ln(abs(S))+S) [from S(t=0) to S(t=t)] = E*t [from t=0 to t=t] therefore, F ln(S(t=t)/S(t=0))+(S(t=t)S(t=0)) = E*t  Thank everyone in advance!! Last edited by Johnny223; February 11th, 2016 at 08:06 AM. 
February 11th, 2016, 08:55 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
How did this problem arise? Is it from a book?

February 11th, 2016, 12:47 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,197 Thanks: 898 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Mathematica won't solve the equation. It's apparently a really really nasty integral. Dan  
February 11th, 2016, 04:49 PM  #4 
Newbie Joined: Feb 2016 From: Toronto Posts: 4 Thanks: 2  Thank you for your question. When I was trying to describe my situation, I found several mistakes in my previous question. However, I am still not able to solve the definite integral after corrections. My area of study is wastewater treatment. In my equation, there are 2 core concepts which we need to agree with. 1) Mass Balance in wastewater treatment https://books.google.ca/books?id=GJe...ewater&f=false 2) Monod equation for microbial kinetics https://en.wikipedia.org/wiki/Monod_equation  In the book mentioned above, equation 2.17 is the fundamental equation to solve my problem: d(C*V)/dt = Q*C0  Q*C + rp*V  rc*V (eq. 2.17) Pg. 32 C*(dV/dt) + V*(dC/dt) = Q*C0  Q*C + rp*V  rc*V (eq. 2.18 ) my equation C*(dV/dt) + V*(dC/dt) = Qin*Cin  Qout*C + r*V (my equation) where, C = target compound concentration in the reactor (mass/volume) Cin = influent target compound concentration (mass/volume) V = volume of reactor (volume) Qin = influent flow rate (volume/time) Qout = effluent flow rate (volume/time) r = Monod equation (mass/volumetime)  In my situation, there are 2 major differences from the book: 1) V = function(time) = V0 + Qin*t (where, V0 = Volume of Reactor @ t=0) 2) Qout = 0  Substitute the following to my equation: 1) C to become S 2) V = V0 + Qin*t 3) dV/dt = Qin 4) Qout = 0 therefore, S*Qin + (V0 + Qin*t)*(dS/dt) = Qin*Sin + r*(V0 + Qin*t) Divide Qin (Qin>0) in both sides S + (V0/Qin + t)*(dS/dt) = Sin + r*(V0/Qin + t) Let constantA = Sin constantB = V0/Qin Therefore, S + (constantB + t)*(dS/dt) = constantA + r*(constantB + t) (constantB + t)*(dS/dt) = constantA  S + r*(constantB + t) Divide (constantB + t) in both sides (constantB >0) dS/dt = (constantA  S)/(constantB + t) + r Let r = (constantC * S)/(constantD + S) dS/dt = (constantA  S)/(constantB + t)  (constantC * S)/(constantD + S) Omit "constant" character dS/dt = (AS)/(B+t)  (C*S)/(D+S)  Now the new equation still looks similar as the previous one. I still do not have any idea how to tackle this equation. However, thank everyone for your interest!!! Last edited by Johnny223; February 11th, 2016 at 04:56 PM. 
February 12th, 2016, 11:50 AM  #5 
Newbie Joined: Feb 2016 From: Toronto Posts: 4 Thanks: 2 
So, is there any chance that we can solve this integral? dS/dt = (AS)/(B+t)  (C*S)/(D+S) from S(t=0) to S(t=t), and from t=0 to t=t ?? A, B, C, D are constants 
February 12th, 2016, 12:19 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
As Mathematica fails to solve this, it's certainly not easy in general. However, various special cases can be solved, some of them leading to a linear solution, some to a more complicated solution. As a first step, using simple substitutions to simplify the equation would be a good idea. It might be possible to determine all the special cases that can be solved.

February 17th, 2016, 09:27 AM  #7  
Newbie Joined: Feb 2016 From: Toronto Posts: 4 Thanks: 2  Quote:
I will use the simpler example for illustration.  Given S(t=0) = S0 At t=0, S(t=0)=S0 dS/dt (t=0, S=S(t=0)) = E(S(t=0))/(F + S(t=0)) = some number  Let t = t + increment  Therefore At t= increment S(t=increment) = S(t=0) + (increment)*(dS/dt (t=0, S=S(t=0)) dS/dt (t=increment, S=S(t=increment)) = E(S(t=increment))/(F + S(t=increment)) = some number2  At t = increment*2 S(t=increment*2) = S(t=increment) + (increment)*(dS/dt (t=increment, S=S(t=increment)) dS/dt (t=increment*2, S=S(t=increment*2)) = E(S(t=increment*2))/(F + S(t=increment*2)) = some number3  As increment > 0, then I can approximate S(t=t+increment) to be the integral form of dS/dt = ES/(F+S) I confirm my answer with the known integrated form F ln(S(t=t)/S(t=0))+(S(t=t)S(t=0)) = E*t  As this is a realistic question, I do have expectations of the magnitude of my increment and maximum t. Therefore, I can solve the harder equation by this method. Do you guys think if this is reasonable? Last edited by Johnny223; February 17th, 2016 at 09:31 AM.  

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