Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at x = 1. Compare this approximation with the one obtained in Problem 6 using the Taylor method of order 4.

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

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(1-0)=0.25 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.21875 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.222656 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.184336 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0+0.25=0.25 \\ \mathrm{y}=0+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =0+\frac{1}{6}(0.25-2(0.21875)-2(0.222656)-0.194336) \\ =0.22119 \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(1-0.22119)=0.194703 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.170365 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.173407 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.151351 \end{gathered}$$

Now

$$\begin{gathered} \mathrm{x}=0.25+0.25=0.5 \\ \mathrm{y}=0.22119+\frac{1}{6}(0.194703+2(0.170365)+2(0.173407)+1.151351) \\ =0.39345 \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(1-0.39345)=0.151637 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.132683 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.135052 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.117874 \\ \mathrm{x}=0.5+0.25=0.75 \\ \mathrm{y}=0.39345+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =0.39345+\frac{1}{6}(0.151637+2(0.132683)+2(0.1350520)+0.117874) \\ =0.52 \end{gathered}$$

And

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25(1-0.52761)=0.118098 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=0.103335 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=0.105781 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=0.0918024 \\ \mathrm{x}=0.75+0.25=1 \\ \mathrm{y}=0.52761+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =0.52761+\frac{1}{6}(0.118098+2(0.103335)+2(0.105781)+0.0918024) \\ =0.632099 \end{gathered}$$

Therefore, $\varphi \mathbf{(}1\mathbf{)}\mathbf{=}0.6321$

Hence the solution is$\mathit{\varphi }\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6321}$