\section{(General explicit 3-step Method)}
The order of the method is 3, $\therefore$ $p=3$. And
\begin{eqnarray*}
C_{0} & = & C_{1} = C_{2} = C_{3} = 0 \\
C_{4} & \neq & 0 \\
C_{4} & = & \textrm{error constant of method}
\end{eqnarray*}
Now, the given method is
\begin{eqnarray}
&&y_{n+3} + \alpha y_{n+2} + ay_{n+1} + by_{n} = h\left[ \beta_{2}f_{n+2} + \beta_{1}f_{n+1} + \beta_{0}f_{n}\right] \\
&&\sum_{j=0}^{k} \alpha_{j}y_{n+j} = h\sum_{j=0}^{k}\beta_{j}f_{n+j}
\end{eqnarray}
Comparing (2) and (3), we get
\[\begin{array}{ll}
	\alpha_{0} = b & \beta_{0} = ? \\
	\alpha_{1} = a & \beta_{1} = ? \\
	\alpha_{2} = ? & \beta_{2} = ? \\
	\alpha_{3} = 1 & \beta_{3} = 0
\end{array}\]
Throughout we will be using the equations for the local trucation error.  These are
\begin{eqnarray}
C_{0} & = & \sum_{j=0}^{k}\alpha_{j} \\
C_{1} & = & \sum_{j=1}^{k}j\alpha_{j} - \sum_{j=0}^{k}\beta_{j} \\
C_{p} & = & \frac{1}{p!}\sum_{j=1}^{k}j^{p}\alpha_{j} - \frac{1}{(p-1)!}\sum_{j=1}^{k}j^{p-1}\beta_{j}
\end{eqnarray}
From (4) we get
\[a + b + \alpha_{2} + 1 = 0\]
\begin{eqnarray}
	\alpha_{2} = -(1 + a + b)
\end{eqnarray}
From (5) we get
\begin{eqnarray*}
&&	\alpha_{1} + 2\alpha_{2} + 3\alpha_{3} - \left[ \beta_{0} + \beta_{1} + \beta_{2} \right] = 0 \\
&&	a + 2\alpha_{2} + 3 - \left[ \beta_{0} + \beta_{1} + \beta_{2} \right] = 0 \\
&&	\beta_{0} + \beta_{1} + \beta_{2} = a + 2(-1-a-b) + 3
\end{eqnarray*}
And, finally
\begin{eqnarray}
	\beta_{0} + \beta_{1} + \beta_{2} = -a - 2b - 1
\end{eqnarray}
Then, from (6) we get
\begin{eqnarray}
&&	C_{2} = \frac{1}{2!}\sum_{j=1}^{k}j^{2}\alpha_{j} - \frac{1}{1!}\sum_{j=1}^{k}j^{1}\beta_{j} = 0 \nonumber \\
&&	\frac{1}{2} \left[\alpha_{1} + 4\alpha_{2} + 9\alpha_{3} \right] - \left\{ \beta_{1} + 2\beta_{2}\right\} = 0 \nonumber \\
&&	\beta_{1} + 2\beta_{2} = \frac{1}{2} \left[\alpha_{1} + 4\alpha_{2} + 9\alpha_{3} \right] \nonumber \\
&&	\beta_{1} + 2\beta_{2} = \frac{1}{2} \left[-3a - 4b + 5 \right]
\end{eqnarray}
Again, from (6) we get
\begin{eqnarray}
&&	C_{3} = \frac{1}{3!}\sum_{j=1}^{k}j^{3}\alpha_{j} - \frac{1}{2!}\sum_{j=1}^{k}j^{2}\beta_{j} = 0 \nonumber \\
&&	\frac{1}{6} \left[\alpha_{1} + 8\alpha_{2} + 27\alpha_{3} \right] - \frac{1}{2}\left[ \beta_{1} + 4\beta_{2}\right] = 0 \nonumber \\
&&	\frac{1}{3} \left[a + 8\alpha_{2} + 27 \right] = \beta_{1} + 4\beta_{2}\nonumber \\
&&	\frac{1}{3} \left[-7a - 8b + 19 \right] = \beta_{1} + 4\beta_{2}
\end{eqnarray}
Solving equations (8), (9), (10), we get
\begin{eqnarray*}
\beta_{0} & = & -1.583 + 0.0833a - 0.33b \\
\beta_{1} & = & -1.33 - 0.66a - 1.33b \\
\beta_{2} & = & 1.9166 - 0.4166a - 0.33b
\end{eqnarray*}
We get the error constant by solving (6) for $C_{4}$
\begin{eqnarray*}
C_{4} & = & \frac{1}{4!}\sum_{j=1}^{k}j^{4}\alpha_{j} - \frac{1}{3!}\sum_{j=1}^{k}j^{3}\beta_{j} \\
& = & \frac{1}{4!} ( \alpha_{1} + 2^{4}\alpha_{2} + 3^{4}\alpha_{3} ) - \frac{1}{3!} (\beta_{1} + 2^{3}\beta_{2}) \\
& = & \frac{1}{24} ( a + 16[-a -b -1] + 243 ) - \frac{1}{6} (\beta_{1} + 8\beta_{2}) \\
& = & \frac{1}{24} (-15a - 16b + 227) - \frac{1}{6} (\beta_{1} + 8\beta_{2}) \\
%& = & \frac{1}{24} (-15a - 16b + 227) - \frac{1}{6} (-1.33 - 0.66a - 1.33b + 8(1.9166 - 0.4166a - 0.33b))
& = & \frac{1}{24} (-15a - 16b + 227) - \frac{1}{6} (13.9998 - 3.9988a - 3.973b) \\
C_{4} & = & 0.04166a - 4.5\times 10^{-3} b + 7
\end{eqnarray*}
