# Thread: separable and exact ODE

1. ## separable and exact ODE

I'm unsure of how to set up the differential equation for this problem:

Given that the earth's population on July 1980 was 4,473,000,000 and that on July 1987 it was 5,055,000,000. Assuming that P satisfies model A. When would the earth's population reach 10 billion? Thanks to all in advance for your help!

2. ## Re: separable and exact ODE

What is model A?

3. ## Re: separable and exact ODE

I'm sorry model A is P' = kP where P' is the first derivative.

4. ## Re: separable and exact ODE

Let t represent time measured in years and let July 1980 coincide with $\displaystyle t=0$. Let $\displaystyle P$ represent the population at time t in billions. Using the given model, we have the IVP:

$\displaystyle \frac{dP}{dt}=kP$ where $\displaystyle P(0)=4.473$

Separating the variables in the ODE, we have:

$\displaystyle \frac{1}{P}\,dP=k\,dt$

Integrate:

$\displaystyle \int\frac{1}{P}\,dP=k\int\,dt$

$\displaystyle \ln(c_1\cdot P)=kt$

$\displaystyle t=\frac{\ln(c_1\cdot P)}{k}$

Now, you have two points on the curve, from which you may determine the two constants. Once you have these, then set $\displaystyle P=10$ to answer the question.

5. ## Re: separable and exact ODE

So I'm still a little lost. What is the k value? Is it the rate?

I found k = 5.055 - 4.473 = .582
where .582/7 = .0831 so that's the rate of the increase of people per year.

-plug k into the t equation to solve for c1 which gives me .2235

put c1 back into t equation and my answer is not correct. The answer is between mid 2020 and 2035 according to the book. Please show me where i'm going wrong.

6. ## Re: separable and exact ODE

$\displaystyle t=\frac{\ln(c_1\cdot P)}{k}$

We have the two points (0,4.473) and (7,5.055) so this gives us the system:

$\displaystyle 0=\frac{\ln(c_1\cdot4.473)}{k}$

$\displaystyle 7=\frac{\ln(c_1\cdot5.055)}{k}$

From the first equation, we find:

$\displaystyle c_1\cdot4.473=1\,\therefore\,c_1=\frac{1000}{4473}$

Putting this into the second equation, we find:

$\displaystyle k=\frac{\ln\left(\frac{5055}{4473} \right)}{7}$

And so we have:

$\displaystyle t=\frac{7\ln\left(\frac{1000P}{4473} \right)}{\ln\left(\frac{5055}{4473} \right)}$

Now, let $\displaystyle P=10$ and we find:

$\displaystyle t=\frac{7\ln\left(\frac{10000}{4473} \right)}{\ln\left(\frac{5055}{4473} \right)}\approx46.04$

This is the year 2026.

7. ## Re: separable and exact ODE

Cool I was thinking too hard. Thank you again!