91影视

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

Apply the chain rule.

${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=\frac{鈭�\mathrm{f}}{鈭�\mathrm{x}}(\mathrm{x},\mathrm{y})+\frac{鈭�\mathrm{f}}{鈭�\mathrm{y}}(\mathrm{x},\mathrm{y})\mathrm{f}(\mathrm{x},\mathrm{y})$

Since${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=1-y$

$$\begin{gathered} \frac{鈭�\mathrm{f}}{鈭�\mathrm{x}}(\mathrm{x},\mathrm{y})=0 \\ \frac{鈭�\mathrm{f}}{鈭�\mathrm{y}}(\mathrm{x},\mathrm{y})=-1 \end{gathered}$$

So, the equation is${\mathrm{f}}_2(\mathrm{x},\mathrm{y})=\mathrm{y}-1$

Apply the same procedure as step 1

$$\begin{gathered} {\mathrm{f}}_3(\mathrm{x},\mathrm{y})=\mathrm{y}-1 \\ {\mathrm{f}}_4(\mathrm{x},\mathrm{y})=1-\mathrm{y} \end{gathered}$$

The recursive formula is

$$\begin{gathered} {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h} \\ {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+hf({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})+\frac{{\mathrm{h}}^2}{2!}{\mathrm{f}}_2({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})+.....\frac{{\mathrm{h}}^{\mathrm{p}}}{\mathrm{p}!}{\mathrm{f}}_{\mathrm{p}}({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}}) \\ {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+0.25 \\ {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+0.25(1-{\mathrm{y}}_{\mathrm{n}})+\left(\frac{{\displaystyle {\left(0.25\right)}^2}}{2}\right)({\mathrm{y}}_{\mathrm{n}}-1) \end{gathered}$$

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_0=0$

$$\begin{gathered} {\mathrm{x}}_1=0.25 \\ {\mathrm{y}}_1=0.21875 \end{gathered}$$

Put all these values in recursive formulas for the other values.

$$\begin{gathered} {\mathrm{x}}_2=0.5 \\ {\mathrm{y}}_2=0.389648 \\ {\mathrm{x}}_3=0.75 \\ {\mathrm{y}}_3=0.523163 \\ {\mathrm{x}}_4=1 \\ {\mathrm{y}}_4=0.627471 \end{gathered}$$

Therefore the approximation of the solution by the Taylor method of order 2 at point x=1

${蠒}_2\left(1\right)=0.6274$

$$\begin{gathered} {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+0.25 \\ {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+0.25(1-{\mathrm{y}}_{\mathrm{n}})+\left(\frac{{\displaystyle {\left(0.25\right)}^2}}{2}\right)\left({\mathrm{y}}_{\mathrm{n}}-1)+\frac{{\displaystyle {\left(0.25\right)}^3}}{6}\right({\mathrm{y}}_{\mathrm{n}}-1)+\frac{{\displaystyle {\left(0.25\right)}^4}}{24}(1-{\mathrm{y}}_{\mathrm{n}}) \end{gathered}$$

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_0=0$.

$$\begin{gathered} {\mathrm{x}}_1=0.25 \\ {\mathrm{y}}_1=0.216309 \end{gathered}$$

Put all these values in recursive formulas for the other values.

$$\begin{gathered} {\mathrm{x}}_2=0.5 \\ {\mathrm{y}}_2=0.385828 \\ {\mathrm{x}}_3=0.75 \\ {\mathrm{y}}_3=0.518679 \\ {\mathrm{x}}_4=1 \\ {\mathrm{y}}_4=0.622793 \end{gathered}$$

Thus, the approximation of the solution by the Taylor method of order 4 at point x = 1

${蠒}_4\left(1\right)=0.6227$

The actual solution at x = 1

$$\begin{gathered} \mathrm{y}\left(\mathrm{x}\right)=1-{\mathrm{e}}^{-\mathrm{x}} \\ \mathrm{y}\left(1\right)=0.632121 \end{gathered}$$

Now combining the approximation with the actual solution at x = 1

$$\begin{gathered} \left|\mathrm{y}\left(1\right)-{蠒}_2\left(1\right)\right|=0.00465 \\ \left|\mathrm{y}\left(1\right)-{蠒}_4\left(1\right)\right|=0.00933 \end{gathered}$$

Hence the solution is

$$\begin{gathered} {\mathit{蠒}}_{\mathbf{2}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6274} \\ {\mathit{蠒}}_{\mathbf{4}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6227} \end{gathered}$$