User Name Remember Me? Password

 Calculus Calculus Math Forum

 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,267
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
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
Newbie

Joined: Feb 2016
From: Toronto

Posts: 4
Thanks: 2

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
Newbie

Joined: Feb 2016
From: Toronto

Posts: 4
Thanks: 2

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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top