Showing that one physics equation 'satisfies' another 
Background: This is from a Physics 2 Lab. The equations come from a circuit that has a resistor and a capacitor

I'm given these two equations 
$V - \frac{dq}{dt} R - \frac{q}{C} = 0$ <== Eqn(2)
$q(t) = CV(1 - e^{-\frac{t}{RC}})$ <== Eqn(3)

Problem: Show that eqn(3) satisfies eqn(2)

I am very confused on what they mean by satisfies the equation. Are they just wanting me to manipulate eqn(2) into eqn(3)?

My Attempt:

I think that $t = RC$ (though I could be wrong). Using this fact I found that:
$q(t) = CV(1 - e^{-1}) = 0$
I know that $(1 - e^{-1}) = 0.63212...$ but it doesn't $= 0$. Any advice to point me in the right direction would be great
 A: No, they aren't asking you to "manipulate" (3) into (2). 
They are asking you to show that the $q(t)$ given by (3) is a solution to the differential equation (2). You must find $\frac{dq}{dt}$ based on (3), and then substitute that, as well as the expression for $q(t)$, into (2); then verify that the left-hand side of (2) really does come out to be zero.
To get you started: given (3), we have
$$\frac{dq}{dt}=\frac{V}{R}e^{-t/RC}$$
Now substitute this, as well as $q(t)$, into (2), and verify that the left-hand side is zero.
This is analogous to verifying that the number 2 is a solution of the algebraic equation $x+1=3$. To verify 2 is a solution, you just substitute 2 for $x$ and see that the equation is indeed true. The only difference for differential equations is that you now are substituting functions (and their derivatives), not numbers.
A: Equation 2 can be thought of as a condition on the function $q$—in particular, a differential condition.  A function $q$ that satisfies this condition is called a solution to the differential equation.
To check it, you evaluate the derivative $\frac{dq}{dt}$, substitute Equation 3 and its derivative into Equation 2, and see if Equation 2 "checks out".  If it does, the condition is satisfied.
A: The equation to be satisfied is an ordinary differential equation (ODE).
$$
V - R \dot{q} - \frac{1}{C} q = 0
$$
You take your solution candidate for $q(t)$,
$$
q(t) = CV \left(1 - e^{-t/(RC)} \right)
$$
and calculate the needed derivatives (here $\dot{q}(t)$) 
$$
\dot{q}(t) 
= CV \frac{1}{RC} e^{-t/(RC)}
= \frac{V}{R} e^{-t/(RC)}
$$
and plug all of those into the ODE and check if the resulting equation is true.
\begin{align}
0 
&= V - R \dot{q} - \frac{1}{C} q \\
&= V - R \left( \frac{V}{R} e^{-t/(RC)} \right) 
- \frac{1}{C} \left( CV \left(1 - e^{-t/(RC)} \right) \right) \\
&= V - V e^{-t/(RC)} - V \left(1 - e^{-t/(RC)} \right) \\
&= V - V e^{-t/(RC)} - V + V e^{-t/(RC)} \\
&= 0
\end{align}
which is a true equation. Your candidate is a solution of the ODE.
Usually one has to check as well if some initial or boundary conditions are fulfilled, as these are needed to single out a specific solution from the set of solutions.
