91Ó°ÊÓ

Here,$\mathrm{y}\text{'}=1-\mathrm{y}+{\mathrm{y}}^3,\mathrm{y}\left(0\right)=0$

For $\mathrm{ξ}=0.01$ , x=0, y₀ = 0 , $\mathrm{c}=\mathrm{Ï€}$, M = 10, h = 1

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{x},\mathrm{y}) \\ =1-\mathrm{y}+{\mathrm{y}}^3 \\ \mathrm{G}=\mathrm{f}(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}) \\ =1-(\mathrm{y}+\mathrm{hF})+(\mathrm{y}+\mathrm{hF}{)}^3 \\ =1-\sin (\mathrm{y}+\mathrm{h}(1-\sin \mathrm{y}\left)\right) \\ 1-(\mathrm{y}+\mathrm{h}(1-\mathrm{y}+{\mathrm{y}}^3\left)\right)+(\mathrm{y}+\mathrm{h}(1-\mathrm{y}+{\mathrm{y}}^3){)}^3 \end{gathered}$$

Apply initial points $\mathrm{x}=0,\mathrm{y}=0,\mathrm{h}=1$

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=1 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=(\mathrm{x}+\mathrm{h}) \\ \mathrm{y}=\mathrm{x}+\frac{\mathrm{h}}{2}(\mathrm{F}+\mathrm{G}) \end{gathered}$$

x = 1

y = 1

Hence $\mathrm{Ï•}(1)=\mathrm{y}(1,1)=1$

$\mathrm{x}=0,\mathrm{y}=0,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}(0,0)=1 \\ \mathrm{G}(0,0)=0.625 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0+0.5 \\ =0.5 \\ \mathrm{y}=0+0.25(1+0.625) \\ =0.40625 \end{gathered}$$

$\mathrm{Ï•}(0.5)=0.40625$

$\mathrm{x}=0.5,\mathrm{y}=0.40625,\mathrm{h}=0.5$

$$\begin{gathered} \mathrm{F}=0.660798 \\ \mathrm{G}=0.663095 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0.5+0.5 \\ =1 \\ \mathrm{y}=0.40625+0.25(0.660798+0.663095) \\ =0.737223 \end{gathered}$$

$\mathrm{Ï•}(1)=\mathrm{y}(1,0.5)=0.737223$

Apply the same procedure for all other values and the values are

$$\begin{gathered} \mathrm{Ï•}(1)=\mathrm{y}(1,0.25) \\ =0.719412 \\ \mathrm{Ï•}(1)=\mathrm{y}(1,0.125) \\ =0.716984 \end{gathered}$$

Since,

$$\begin{gathered} \left|\mathrm{y}(1,0.196350)-\mathrm{y}(1,0.392699)\right| \\ =\left|1.09580-1.09229\right| \\ =0.00351<0.01 \end{gathered}$$

$\mathrm{Ï•}(1)=0.716984$

Hence the solution is role="math" localid="1664310510661" $\mathit{Ï•}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{716984}$