Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

Since $\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}+1-\mathrm{y}$and x = 0, y = 1, and h = 0.25

role="math" localid="1664324055813" $$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(0+1-1)=0 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.03125 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.0273438 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.0556641 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0+0.25=0.25 \\ \mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =1+\frac{1}{6}(0-2(0.031125)-2(0.0273438)-0.0556641) \\ =1.02881 \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(0.25+1-1.02881)=0.0552975 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.0796353 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.0765931 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.0986392 \end{gathered}$$

Now

$$\begin{gathered} \mathrm{x}=0.25+0.25=0.5 \\ \mathrm{y}=1.02881+\frac{1}{6}(0.0552975-2(0.0796353)-2(0.0765931)-0.0986492) \\ =1.10654 \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(0.5+1-1.10654)=0.098365 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.117319 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.11495 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.132127 \\ \mathrm{x}=0.5+0.25=0.75 \\ \mathrm{y}=1.10654+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =1.10654+\frac{1}{6}(0.098365-2(0.117319)-2(0.11495)-0.132127) \\ =1.22238 \end{gathered}$$

And

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(0.75+1-1.22238)=0.131905 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.146667 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.144822 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.1582 \\ \mathrm{x}=0.75+0.25=1 \\ \mathrm{y}=1.22238+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =0.52761+\frac{1}{6}(0.131905-2(0.146667)-2(0.144822)-0.1582) \\ =1.36789 \end{gathered}$$

Therefore$\varphi \mathbf{(}1\mathbf{)}\mathbf{=}1.3679$

Hence the solution is$\mathit{\varphi }\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3679}$