Solving the Gompert Equation: Can You Help?

  • Thread starter Thread starter Gogsey
  • Start date Start date
  • Tags Tags
    Stuck
Click For Summary

Homework Help Overview

The discussion revolves around solving the Gompertz equation, specifically the differential equation dp/dt = c ln(K/p)p. Participants are attempting to derive an expression for p(t) through various substitutions and manipulations of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts substitution and integration but encounters difficulties in evaluating the expression for p(t). Other participants suggest exponentiating to simplify the equation and express p(t) in different forms. There is also a discussion about discrepancies between the original poster's result and a peer's solution.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the manipulation of the equation, but there is no explicit consensus on the correct form of the solution. The original poster expresses a desire to understand the reasoning behind a peer's solution rather than just accepting it.

Contextual Notes

Participants are working under the constraints of needing to submit their answers online, which provides immediate feedback on correctness. There is some confusion regarding the definitions of pt and po, as well as the initial value problem (IVP) context.

Gogsey
Messages
153
Reaction score
0
Can you help me solve this equation? Its the Gompert equation, dp/dt=cln(K/p)p.

I used substution to get -ln(u) = ct+b, where c is a constant, and bi s a constant of integration.

Next we have -ln(ln(k/p)) evaluated from pt to po = ct+b

Then ln(k/p) from pt to po = Bexp(-ct).

This is where I am stuck and don't know how to evaluate the rest of this to get an expression for pt.
 
Physics news on Phys.org
Just exponentiate again and solve for p, right?
 
Then I get k/pt - k/po = exp(bexp(-ct), right? and then solving for pt i get

pt = k/exp(bexp(-ct) + po. But this is different from what my buddy got, and his is correct. We have to enter it online in the space provided, and it tells us if its correct once we submit it.

He got Kexp((ln(b))exp(-ct)) before he solved for b. I want to understand how he got this, not just plug the answer in without knowing.
 
I would say you get k/p(t)=exp(B*exp(-ct)), or p(t)=k/exp(B*exp(-ct)), which you can also write as p(t)=k*exp(-B*exp(-ct)). That's p as a function of t. I'm not very sure what p0 and pt are supposed to be.
 
Don't worry about it. This was initially what I had, but chaged it for some reason, thinking I had to evaluate the integral between pt and po, instead of pt and 0, lol. po is the IVP.

So I actually had the same answer as you got except I left the exp function in the denominator, and left is a a positive exp.

Thanks
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K