Question 11.4.1: Apply Algorithm 11.4, with h = 0.1, to the nonlinear boundar...
Apply Algorithm 11.4, with h = 0.1, to the nonlinear boundary-value problem
y^{\prime \prime}=\frac{1}{8}\left(32+2 x^{3}-y y^{\prime}\right), \quad \text { for } 1 \leq x \leq 3, \text { with } y(1)=17 \text { and } y(3)=\frac{43}{3}
and compare the results to those obtained in Example 1 of Section 11.2.
The stopping procedure used in Algorithm 11.4 was to iterate until values of successive iterates differed by less than 10^{-8} . This was accomplished with four iterations.
This gives the results in Table 11.5. They are less accurate than those obtained using the nonlinear shooting method, which gave results in the middle of the table accurate on the order of 10^{-5} .
Table 11.5
\begin{array}{cccl}\hline x_{i} & w_{i} & y\left(x_{i}\right) & \left|w_{i}-y\left(x_{i}\right)\right| \\\hline 1.0 & 17.000000 & 17.000000 & \\1.1 & 15.754503 & 15.755455 & 9.520 \times 10^{-4} \\1.2 & 14.771740 & 14.773333 & 1.594 \times 10^{-3} \\1.3 & 13.995677 & 13.997692 & 2.015 \times 10^{-3} \\1.4 & 13.386297 & 13.388571 & 2.275 \times 10^{-3} \\1.5 & 12.914252 & 12.916667 & 2.414 \times 10^{-3} \\1.6 & 12.557538 & 12.560000 & 2.462 \times 10^{-3} \\1.7 & 12.299326 & 12.301765 & 2.438 \times 10^{-3} \\1.8 & 12.126529 & 12.128889 & 2.360 \times 10^{-3} \\1.9 & 12.028814 & 12.031053 & 2.239 \times 10^{-3} \\2.0 & 11.997915 & 12.000000 & 2.085 \times 10^{-3} \\2.1 & 12.027142 & 12.029048 & 1.905 \times 10^{-3} \\2.2 & 12.111020 & 12.112727 & 1.707 \times 10^{-3} \\2.3 & 12.245025 & 12.246522 & 1.497 \times 10^{-3} \\2.4 & 12.425388 & 12.426667 & 1.278 \times 10^{-3} \\2.5 & 12.648944 & 12.650000 & 1.056 \times 10^{-3} \\2.6 & 12.913013 & 12.913846 & 8.335 \times 10^{-4} \\2.7 & 13.215312 & 13.215926 & 6.142 \times 10^{-4} \\2.8 & 13.553885 & 13.554286 & 4.006 \times 10^{-4} \\2.9 & 13.927046 & 13.927241 & 1.953 \times 10^{-4} \\3.0 & 14.333333 & 14.333333 & \\\hline\end{array}