$8$ billion people in the world, increase each year by $2\%$, how many people will be in $9$ months? 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.
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.
There are two types of exponential functions:
$$\text{ordinary: } S=P(1+r)^t \ \ \text{and} \ \ \text{natural: } S=Pe^{rt}.$$
You used natural:
$$S=8\cdot e^{0.02\cdot \frac{9}{12}}=8.1209045,$$
whereas the textbook used ordinary:
$$S=8\cdot (1+0.02)^{\frac{9}{12}}=8.119702.$$
The population grows continuously, so the natural exponential function is more realistic.