How Do I Solve This Differential Equation for a Contagious Disease?

  • Context: Graduate 
  • Thread starter Thread starter kanki
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around solving a differential equation related to the spread of a contagious disease. Participants explore the mathematical formulation of the problem, the implications of a constant rate of infection, and the interpretation of results over time. The scope includes mathematical reasoning and conceptual clarification regarding differential equations and their applications in modeling disease spread.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents the differential equation \(\frac{dx}{dt}=k(x)(1-x)\) and attempts to solve it, raising concerns about the behavior of the solution as \(t\) approaches 24 days.
  • Another participant questions the correctness of the integral calculations and the interpretation of the constant \(k\), suggesting that \(k\) should not be treated as a function of \(x\).
  • Some participants discuss the implications of a constant rate of infection, noting that a continuous solution cannot reach a population of 1, and propose approximations instead.
  • A later reply suggests treating the entire population as "within 1/2 person of entire population" to address the limitations of the model.
  • One participant proposes a solution using a logistic model, while another challenges the interpretation of the problem as a logistic equation versus a linear increase.
  • There is a mention of a potential solution \(x = \frac{e}{e+1}\) after some participants clarify their understanding of the problem's conditions.
  • Some participants express confusion about the problem's requirements and whether it is solvable under the given assumptions.

Areas of Agreement / Disagreement

Participants express differing interpretations of the problem, particularly regarding the nature of the rate of infection and the applicability of the logistic model. There is no clear consensus on the correct approach or solution, and multiple competing views remain.

Contextual Notes

Participants highlight limitations in the model, such as the assumption of a constant rate of infection and the implications of population size on the solution. The discussion reflects uncertainty about the mathematical steps and the interpretation of results.

Who May Find This Useful

This discussion may be useful for those interested in differential equations, mathematical modeling of infectious diseases, and the nuances of interpreting mathematical problems in a biological context.

kanki
Messages
29
Reaction score
0
Another DE problem I've encoutered...

A contagious disease spreads at a rate directly proportional to the product of the number of population infected and the remaining population that is not infected. Initially, one-half of the population is infected and if the rate of infection is kept constant, the whole population will be infected in 24 days. Find the proportion of the population that will be infected after 12 days.

ok i got the differential equation:
\frac{dx}{dt}=k(x)(1-x) (x is the proportion of the population)

\int \frac{dx}{x(1-x)} = \int k dt

and change it into partial fraction:

\int [\frac {1}{x} + \frac {1}{1-x}]dx = \int k dt

ln|x|-ln|1-x|=kt+c

After i substitute the t=0, x=0.5, and got

ln|\frac{0.5}{1-0.5}|=k \cdot 0 + c

ln 1 = c

c = 0

\therefore ln|\frac {x}{1-x}|=kt

The problem is when i tried to find the constant k, as i substituted t=24 and x=1, i got:

ln|\frac{1}{0}|=24k

So, how to solve \frac{1}{0}=e^{24k}?

:rolleyes: thanks for your time here.
 
Last edited:
Physics news on Phys.org
You might not have done the rest of it right, but 1 - 0.5 is not 0, and kt + c is not equal to kt when k = 0.
 
But is k a function of x or do you just mean k (x-x^2) ?

I hope it is the second one, otherwise there is a problem.

Eitherway, the calculated integral is incorrect.

marlon
 
Nope, "k" is a constant.It was a misuse of round brackets, that's all.

Daniel.
 
Last edited:
I've edited the post to add some working inside.
Please correct me if my integral is not correct.
Did i miss something when i change it into partial fraction?
 
Your calculations are all correct. The problem you have is that, given that the "rate of infection" is a constant, as stated, the continuous function solution will never equal 1- it approaches 1 in the limit as t goes to infinity. In particular, you can never have x(24)= 1 which is required by your problem. Of course, population is integer valued so the continuous solution can't be true anyway. If you knew the actual population, you might use the statistician's "half-integer" correction- treat "entire population" as "within 1/2 person of entire population". That is set x(24)= (population- 1/2)/population.

If you don't know the actual size of the population, you should be able to get a good approximation by setting x(24)= 0.99 or some other number close to 1. For example, if x(24)= 0.99, then your equation becomes ln(.99/.01)= ln(99)= 24k so that k= ln(99)/24= 0.191. Then the answer to "proportion infected after 12 days" requires solving ln(x/(1-x))= (0.191)(12)= 2.30 so that x/(1-x)= e2.30= 9.95.
Then x= 9.95- 9.95x so x= 9.95/10.95= 0.91 or 91% of the population.

If instead you use 0.95, you get ln(.95/.05)= ln(19)= 24k so that k= ln(19)/24= 0.123. Now to find "proportion infected after 12 days", solve ln(x/(1-x))= (0.123)(12)= 1.47. 1/(1-x)= e1.47= 4.36 so x= 4.36- 4.36x. x= 4.36/5.36= 0.81 or 81% of the population. Those are not the same answer but they are not wildly off.
 
HallsofIvy said:
treat "entire population" as "within 1/2 person of entire population". That is set x(24)= (population- 1/2)/population.

I don't understand how to treat "entire population" as 1/2 person of entire population.
So there's no solution to this question? Because the rate is constant and the disease is not spread to whole population?
 
kanki said:
and if the rate of infection is kept constant, the whole population will be infected in 24 days.

The question means that if rate of infection is kept constant, ie dx/dt is constant and equal to dx/dt at t=0, the entire population will be infected in 24 days.
Substituting x=0.5 and t=0, the value of dx/dt at t=0 is 0.25k
So taking x=0.25kt+0.5 (x=0.5 at t=0) and substituting x=1 at t=24, we get k=1/12

However dx/dt is not constant and as obtained by you
ln|\frac {x}{1-x}|=kt
Substituting k=1/12 and t=12
ln|\frac {x}{1-x}|=1
x/1-x=e
x=e/e+1

Therefore the proportion infected after 12 days will be e/e+1
 
oh... mustafa u solved the question, the answer is e/e+1 after i checked with my book.
Oh, i missed the word if. So the initial rate of infection is kept constant, whole population will be infected in 24 days. Therefore i just have to use this to solve the value of k right?

Thanks!
 
  • #10
That's a very strange interpretation! You're interpreting the problem as a "logistic equation" problem but then interpreting "rate of infection is kept constant" to mean linear increase in order to find the k used in the logistic equation! If that is what the book meant then I am amazed.
 
  • #11
The teachers asked us to skip the question.
But I'm curious whether i can solve the problem, so i post it here.
I'm not expert in maths, so is this the book's problem?
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K