What does $a_1$ plus $a_2$ have to equal $0$, and (other stipulations) for RK2? I am studying Runge Kutta methods using the videos here - http://mathforcollege.com/nm/videos/youtube/08ode/rungekutta2nd/rungekutta2nd_08ode_derivationone.html.
$y_{i+1} = y_i + h(ak_1 + ak_2)h$
where
$k_1 = f(x_i, y_i)$
$k_2 = f(x_i + p_1h, y_i + q_{11}k_1h)$
and we have the following stipulations -
$a_1 + a_2 = 1$
$ap_1 = \frac{1}{2}$
$aq_{11} = \frac{1}{2}$ 
Does anyone know the reasons for the stipulations? I think I saw it mentioned somewhere that $a_1 + a_2 = 1$ means the method is 'consistent', whatever that is?
 A: (Note, what you call $a_i$, I call $b_i$ to keep with standard single-step nomenclature).
This condition is sometimes known as the quadrature condition. Essentially, we can write down a table of constants for a general single-step ODE integrator in the following form:
$$\begin{array}{c|cccc}
c_1 & a_{11} & a_{12} & \cdots & a_{1n} \\
c_2 & a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots  & \vdots & \ddots & \vdots \\
c_n & a_{n1} & a_{n2} & \cdots & a_{nn} \\
\hline
    & b_1 & b_2 & \cdots & b_n \\
\end{array}$$
The matrix $A$ is the matrix of $a_{ij}$ elements, and the vector $\mathbf{b}$ is the column vector of $b_i$ elements.
To ensure $0$-stability, and therefore convergence (via a more involved theorem), we must have
$$\mathbf{b}^TA^{k-1}\mathbf{1} = \frac{1}{k!},\ k = 1,2,\ldots, p$$
for a method of order $p$.
Having this condition allows us to set an estimate of local truncation error to
$$d_n \approx h^p\left(\mathbf{b}^TA^p\mathbf{1}-\frac{1}{(p+1)!}\right).$$
(Much of this was refreshed from the fantastic book Computer Methods for Ordinary Differential Equations by Ascher and Petzold).
Proof of this is a little bit more involved, certainly more than I care to put in an MSE answer, but the short version is that $\sum_i b_i = 1$ is a necessary condition to achieve $0$-stability.
It is also worth mentioning that in the table above, for an explicit method, only the sub-diagonal entries are non-zero. So, for the standard RK4 method, the table looks like this:
$$\begin{array}{c|cccc}
0 & 0 & 0 & 0 & 0 \\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0\\
\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\
1 & 0 & 0 & 1 & 0\\
\hline
    & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6} \\
\end{array}$$
