$8$ billion people in the world, increase each year by $2\%$, how many people will be in $9$ months?

Question

I have the following ODE problem :

$8$ billion people in the world, increase each year by $2\%$, how many people will be in $9$ months?

What I did :

The change each year is : $$x'(t)=0.02(x(t))$$

Therefore $$x(t)=Ce^{0.02t}$$

In $t_0$ we know that $x(t_0)=8$, therefore $C=8*e^{-0.02t_0}$

So I got that $$x(t)=8*e^{(0.02(t-t_0))}$$

The number of people in $9$ months are $t-t_0=3/4$

Therefore I got : $$x(t)=8*e^{0.02(0.75)}=8.1209$$

In the text book the right answer is $8.1197$

I don't understand what's wrong with my solution.

Any help will be appreciated, thanks.

Answer 1

You don't have to assume the exponential form to begin with. In fact, if this is an exercise on ordinary differential equations, then you shouldn't in fact assume the form.

Instead, recognise the basic behaviour - this is a continuous growth model where the rate of increase of population $\frac {dP}{dt}$ is proportional to the present population $P$. Let the constant of proportionality be an unknown $k$ (which is to be determined).

So you start with $\frac{dP}{dt} = kP$, solve it to get $\ln P = kt + C$, where $C$ is another constant (of indefinite integration), giving $P = Ce^{kt}$.

Now, you know that when $t=0$ (at the start), the population is $P_0$, the initial population. So $P_0 = C$, and the solution can be written $P = P_0e^{kt}$.

You're also told that when $t=1$, $P = 1.02P_0$ (that's how you use the $2 \%$ increase given in the question).

So $1.02P_0 = P_0e^{k}$, giving $k = \ln{1.02}$.

Your final equation is therefore $P = P_0e^{t\ln{1.02}} = P_01.02^t = (8 \times 10^9)1.02^t$, into which you can substitute $t=0.75$ to get the correct required value.

So, pro tip: don't start out assuming the constants in your ODE, get a general form first then work out the constants using what's given.

Answer 2

The function that you're after is an exponential one; to be more precise, it is of the type $f(t)=8e^{kt}$. What is the value of $k$? It musto be such that $f(1)=8\times1.02$, which means that $k=\log(1.02)$. So, $f(t)=8e^{\log(1.02)t}$ and $f(1)$ is indeed about $8.1197$.

Answer 3

This is wrong:

The change each year is : $$x′(t)=0.02(x(t))$$

Plug in you your numbers:

$$x(t)=8∗e^{0.02(t−t_0)}$$

At $t=1+t_0$ year we get

$$x(1)=8∗e^{0.02}$$

or

$$x(1) = 8161610720.21$$

The question clearly stated that over 1 year, the people would increase by 2%. 2% of 8 billion is 160 million, not 161.6 million.

The core of the problem is:

$$x′(t)=0.02(x(t))$$

does not correspond to an annual growth rate of 2%. It corresponds to an annual growth rate a bit over 2%.

You want

$$x′(t)=\ln(1.02)(x(t))$$

for an annual growth rate of 2%.

That gives you an equation of

$$x(t)=8∗e^{\ln(1.02)(t−t_0)}$$

and $x(1) = 8 * e^{\ln(1.02)} = 8 *1.02 $ as the problem requires.

Your answer is close because $\ln(1+x)$ ~ $x$ for small $x$.

$x'(t) = 0.02x(t)$ means that the derivative at a point t is 0.02 times the value at point t. That isn't quite the right value to produce a growth factor of 1.02 from $t=0$ to $t=1$, because that growth compounds.