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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: 20,968 Thanks: 2217 How did this problem arise? Is it from a book? February 11th, 2016, 12:47 PM   #3
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Quote:
 Originally Posted by Johnny223 Integrate from t=0 to t=t for only right hand side: dS = A*ln((B+t)/B) - (CS *t)/(D+S)
I can't think of any circumstance or method that would do this. It's like squaring just one part of an equation. It isn't a valid technique.

Mathematica won't solve the equation. It's apparently a really really nasty integral.

-Dan February 11th, 2016, 04:49 PM   #4
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Quote:
 Originally Posted by skipjack How did this problem arise? Is it from a book?
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

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/volume-time)
----------------------

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 = (A-S)/(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 = (A-S)/(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: 20,968 Thanks: 2217 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
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Quote:
 Originally Posted by Johnny223 Hello everyone, 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!! I did come up with an analytical solution of integrating this type of equation.

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. Tags definite, integral Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Akcope Calculus 2 April 6th, 2014 02:43 AM cheyb93 Calculus 1 January 13th, 2013 06:37 PM panky Calculus 1 December 10th, 2012 09:36 PM zaidalyafey Calculus 1 August 14th, 2012 02:36 AM panky Calculus 2 October 8th, 2011 07:05 PM

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