# Thread: Setting up a Growth Problem (ODE, probably separable or linear)

1. ## Setting up a Growth Problem (ODE, probably separable or linear)

I'm doubtful whether I've set up the equations correctly just because the computations are getting ugly:

By natural increase a city, the population of which is 40,000, will double in 50 years. There is also a net addition of 400 people per year because of people moving to and from the city.

I figger to solve for the natural increase function, where $P=$ population and $t$ = years: $\frac{dP}{dt} = kP \Rightarrow$

$\displaystyle \int_{40,000}^{80,000}\frac{dP}{P} = k\int_{0}^{50}dt = 50k = \ln 2 \Rightarrow k = \frac{\ln 2}{50}$

Now taking into consideration the additional 400 people per year,

$\frac{dP}{dt} = \frac{\ln 2}{50}P + 400t$

When I compute this I get something ridiculously wrong.

2. The population due to net addition of 400 people per year is

$
P=400t
$

and

$
\frac{dP}{dt}=400.
$

3. Ah, right, if I catch your drift: $\frac{dP}{dt} = \frac{\ln 2}{50}P + 400 \Rightarrow$. So if that's right, to find the population in ten years,

$\displaystyle \int_{40,000}^{x}\frac{dP}{\frac{\ln 2}{50}P + 400} = \int_{0}^{10}dt = 10 = \frac{50}{\ln 2}\ln \Big( \frac{\ln 2}{50}P + 400} \Big) \Rightarrow$

$\frac{\ln 2}{5} = \ln \Big( \frac{\ln 2}{50}P + 400} \Big) \Rightarrow$

$2^{\frac{1}{5}} = \frac{\ln 2}{50}P + 400 \Rightarrow ...$

I'll stop here because this computation gets me something egregiously wrong and so I'm guessing I've done something wrong by now.

4. May be you need not dif. equation.
The double of population in 50 years is

$
\displaystyle {
P=P_0 \: 2^{{t}/{\tau}}
}
$

and net addition of 400 people per year is

$
\displaystyle {
P=400t.
}
$

So all population change is

$
\displaystyle {
P=P_0 \: 2^{{t}/{\tau}}+400t
}
$

or

$
\displaystyle {
P=40000 \: \; 2^{{t}/{50}}+400t.
}
$