Let's go back to thinking about our first example,

$LaTeX: \begin{equation}\label{eq:ant1} \frac{dA}{dt} = kA(t), \end{equation}$

which models the population of ants in your ant colony.

Naturally, it makes a difference how many ants you start out with. If you start out with 10 ants, you will have a much smaller colony than if you start out with 1000 ants.

We already know from the last Now You Try that $LaTeX: A(t) = Ce^{kt},$ for any constant LaTeX: C, is a solution to equation above (it turns out there are no other solutions). This is called the general solution to the equation. If we now set the initial condition LaTeX: A(t) = 100, (that is, that you started out the colony with 100 ants), we will be able to find a unique solution that satisfies both the equation above and the initial condition.

Plugging in LaTeX: t= 0 into our solution, we have $LaTeX: A(0) = Ce^{k\cdot 0} = Ce^0 = C.$ Thus, we must set LaTeX: C =100 so that the initial condition LaTeX: A(0)=100 is satisfied. The solution $LaTeX: A(t) = 100 e^{kt}$ is called the particular solution to the initial value problem

$LaTeX: \begin{equation*} \begin{cases} \frac{dA}{dt} = kA(t)\\ A(0) = 100. \end{cases} \end{equation*}$