Solving Population Growth Diff Eq: Is My Approach Faulty?

In summary, this conversation discusses different approaches to modeling the growth of a population, taking into account a replicating rate of 160% every four hours and 50,000 deaths per hour. One method involves solving a differential equation, while another involves using statistics and incremental changes. The final equation for the population at any given time is P(t)=P(0)e^(kt)-D/k(e^(kt)-1), where P(0) represents the initial population, k is the growth rate, and D is the extent of deaths.
  • #1
cango91
14
0
I am trying to model the growth of a population which replicates at a rate of 160% every four hours. Also 50 000 members die every hour. so t denoting time in hours, P denoting the population, k being 0.4 I wrote:

dP/dt=kp-50 000 *t

Is my approach to solving the differential faulty?

dP/dt = kP- 50 000*t

P=kPt - (50 000*t^2)/2
P=(50000*t^2)/(2kt-2)

I don't know very much about differential equations so I might be doing really really wrong...

Thanks in advance
 
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  • #2
cango91 said:
I am trying to model the growth of a population which replicates at a rate of 160% every four hours. Also 50 000 members die every hour. so t denoting time in hours, P denoting the population, k being 0.4 I wrote:

dP/dt=kp-50 000 *t
Why do you mutiply it by t? If you do so, you are saying that for every hour which elapses, the population does not decrease constantly but incrementally, ie. more more die per hour as time elapses.
 
  • #3
Hmm... You're right...

So would this be a correct equation for the problem I stated?

Pnew = P0ekt - 50000*t
 
  • #4
I think you forgot to divide 50,000 by k
 
  • #5
But regardless of the growth rate 50000 will die per hour.

P0ekt is the number of individuals if there were no deaths but there are 50 000 deaths per hour so I added the expression -50000*t ..

Why should I divide by k?
 
  • #6
You shouldn't have to modify the final expression for P(t) after solving the DE, only to find the constant of integration C. Dividing by k is not an extra step you must perform after re-arranging to find an expression for P(t); it's what you must do to get P(t).
 
  • #7
I tried to find the equation without solving the D.E... Is my solution still wrong then?
 
  • #8
I presume that you now realize that your differential equation is dP/dt=kP-50 000.

If it were dP/dt= kP, could you solve that?

Suppose P were equal to some constant, A. What would A have to be to satisfy that equation.

The solution to the entire equation is the sum of those.
 
  • #9
I would like to write here some examples of my models. I would be more happy if you could check my methods. Very easy way is to have a statistics. If you have a percentage of birth and deaths, you can write:
extent of births [tex]\mu\geq0[/tex]
extent of deaths [tex]\sigma\geq0[/tex]
I can write an increment of P:
[tex]\text{d}P(t)=\mu P(t)\text{d}t-\sigma P(t)\text{d}t[/tex]
So it is a differential equation and its solution is function:
[tex]P(t)=P(0)\text{e}^{t(\mu-\sigma)}[/tex]

There is also another possibility as wrote cango91. But I use incremental change. I presume there is an extent of deaths [tex]D[/tex] (number of died people per t) and there is an extent of births [tex]\mu\geq0[/tex]. I can write:
[tex]\text{d}P(t)=\mu P(t)\text{d}t-D\text{d}t[/tex]
I try to solve this:
[tex]\begin{array}{rcl}\text{d}P(t)&=&\mu P(t)\text{d}t-D\text{d}t\\\text{d}&=&(\mu P(t)-D)\text{d}t\\\dfrac{\text{d}P(t)}{\mu P(t)-D}&=&\text{d}t\end{array}[/tex]
The integral:
[tex]\int\dfrac{\text{d}P(t)}{\mu P(t)-D}=\left|\begin{array}{rcl}\mu P(t)-D&=&\varphi\\\text{d}P(t)&=&\dfrac{1}{\mu}\,\text{d}\varphi\end{array}\right|=\frac{1}{\mu}\int\frac{\text{d}\varphi}{\varphi}=\frac{1}{\mu}\ln|\mu P(t)-D|[/tex]
I use this into my equation:
[tex]\begin{array}{rcl}\dfrac{1}{\mu}\ln|\mu P(t)-D|&=&t+\ln|C|\\\mu P(t)-D&=&C^{\mu}\text{e}^{\mu t}\\P(t)&=&\dfrac{1}{\mu}C^{\mu}\text{e}^{\mu t}+\dfrac{D}{\mu}\end{array}[/tex]
The constant C for t = 0 is:
[tex]C=\sqrt[\mu]{\mu P(0)-D}[/tex]
So the final function is:
[tex]P(t)=P(0)\text{e}^{\mu t}-\frac{D}{\mu}(\text{e}^{\mu t}-1)\,;\;\mu\geq0,D\geq0, t\geq0[/tex]
 

FAQ: Solving Population Growth Diff Eq: Is My Approach Faulty?

Why is population growth important to study?

Population growth is important to study because it affects resources, economic development, and the environment. Understanding and managing population growth can help prevent issues such as overpopulation, depletion of resources, and environmental degradation.

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. In the context of population growth, a differential equation can be used to model the rate of change of a population over time.

How can differential equations be used to solve population growth?

Differential equations can be used to solve population growth by creating a mathematical model that represents the change in population over time. This model can then be solved using mathematical techniques to determine the population size at any given time.

What are the limitations of using differential equations to solve population growth?

One limitation of using differential equations to solve population growth is that they are based on assumptions and simplifications, so they may not fully capture the complexities of real-world situations. Additionally, the accuracy of the results depends on the accuracy of the initial conditions and the parameters used in the model.

How can we ensure the accuracy of our approach to solving population growth using differential equations?

To ensure the accuracy of our approach, it is important to verify the initial conditions and parameters used in the model with real-world data. Additionally, sensitivity analysis can be performed to test the robustness of the model and identify any potential sources of error. Collaboration and peer review can also help improve the accuracy of the approach.

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