Solving Population Growth Diff Eq: Is My Approach Faulty?

[cango91]

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

 

[Defennder]

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.

 

[cango91]

Hmm... You're right...

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

P_new = P₀e^kt - 50000*t

 

[Defennder]

I think you forgot to divide 50,000 by k

 

[cango91]

But regardless of the growth rate 50000 will die per hour.

P₀e^kt 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?

 

[Defennder]

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).

 

[cango91]

I tried to find the equation without solving the D.E... Is my solution still wrong then?

 

[HallsofIvy]

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.

 

[lukaszh]

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 \mu\geq0
extent of deaths \sigma\geq0
I can write an increment of P:
\text{d}P(t)=\mu P(t)\text{d}t-\sigma P(t)\text{d}t
So it is a differential equation and its solution is function:
P(t)=P(0)\text{e}^{t(\mu-\sigma)}

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