Review 6: Initial Value Problems
Let's go back to thinking about our first example,
%252C%250A%255Cend%257Bequation%257D "\begin{equation}\label{eq:ant1}
\frac{dA}{dt} = kA(t),
\end{equation}")
dAdt=kA(t),
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 %2520%253D%2520Ce%255E%257Bkt%257D%252C "A(t) = Ce^{kt},")
A(t)=Cekt, for any constant 
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 %2520%253D%2520100%252C "A(t) = 100,")
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 
t=0 into our solution, we have %2520%253D%2520Ce%255E%257Bk%255Ccdot%25200%257D%2520%253D%2520Ce%255E0%2520%253D%2520C.%2520 "A(0) = Ce^{k\cdot 0} = Ce^0 = C.")
A(0)=Cek⋅0=Ce0=C. Thus, we must set 
C=100 so that the initial condition %253D100 "A(0)=100")
A(0)=100 is satisfied. The solution %2520%253D%2520100%2520e%255E%257Bkt%257D "A(t) = 100 e^{kt}")
A(t)=100ekt is called the particular solution to the initial value problem
%255C%255C%2520%2520%2520%2520%2520%2520%250A%2520A(0)%2520%253D%2520100.%250A%255Cend%257Bcases%257D%250A%255Cend%257Bequation*%257D "\begin{equation*}
\begin{cases}
\frac{dA}{dt} = kA(t)\\
A(0) = 100.
\end{cases}
\end{equation*}")
{dAdt=kA(t)A(0)=100.