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

  • Thread starter kanki
  • Start date
In summary, 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.
  • #1
kanki
29
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:
[tex]\frac{dx}{dt}=k(x)(1-x)[/tex] (x is the proportion of the population)

[tex]\int \frac{dx}{x(1-x)} = \int k dt[/tex]

and change it into partial fraction:

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

[tex]ln|x|-ln|1-x|=kt+c[/tex]

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

[tex]ln|\frac{0.5}{1-0.5}|=k \cdot 0 + c[/tex]

[tex]ln 1 = c[/tex]

[tex]c = 0[/tex]

[tex]\therefore ln|\frac {x}{1-x}|=kt [/tex]

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

[tex]ln|\frac{1}{0}|=24k[/tex]

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

:uhh: thanks for your time here.
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
But is k a function of x or do you just mean [tex]k (x-x^2)[/tex] ?

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

Eitherway, the calculated integral is incorrect.

marlon
 
  • #4
Nope, "k" is a constant.It was a misuse of round brackets, that's all.

Daniel.
 
Last edited:
  • #5
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?
 
  • #6
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.
 
  • #7
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?
 
  • #8
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
[tex]ln|\frac {x}{1-x}|=kt [/tex]
Substituting k=1/12 and t=12
[tex]ln|\frac {x}{1-x}|=1 [/tex]
x/1-x=e
x=e/e+1

Therefore the proportion infected after 12 days will be e/e+1
 
  • #9
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?
 

1. What does "Ln|0|=kx+c? DE problem" mean?

The expression "Ln|0|=kx+c? DE problem" refers to a logarithmic differential equation problem that involves finding the value of the variable x given the constants k and c.

2. How do you solve "Ln|0|=kx+c? DE problem"?

To solve this type of problem, you can use separation of variables, which involves isolating the logarithmic expression on one side of the equation and the remaining terms on the other side. Then, you can integrate both sides and solve for x.

3. What is the purpose of using "Ln|0|=kx+c? DE problem" in science?

"Ln|0|=kx+c? DE problem" is used in science to model various real-life situations, such as population growth, radioactive decay, and chemical reactions. It allows scientists to study and predict the behavior of these systems and make informed decisions based on the results.

4. Can "Ln|0|=kx+c? DE problem" have multiple solutions?

Yes, "Ln|0|=kx+c? DE problem" can have multiple solutions depending on the initial conditions and the range of the variable x. Each solution represents a different possible outcome of the system being modeled.

5. What are some common applications of "Ln|0|=kx+c? DE problem" in research?

"Ln|0|=kx+c? DE problem" is commonly used in fields such as biology, chemistry, physics, and economics to study and understand complex systems. It is also used in engineering and technology to design and improve systems and processes.

Similar threads

  • Differential Equations
Replies
16
Views
818
  • Differential Equations
Replies
2
Views
919
  • Differential Equations
Replies
18
Views
2K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
2
Views
2K
  • Differential Equations
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
534
Replies
8
Views
2K
Replies
2
Views
1K
Back
Top