Use the Taylor methods of orders 2 and 4 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 these approximations to the actual solution$\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$ evaluated at x = 1.

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

Apply the chain rule.

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

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

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

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

Apply the same procedure as step 1

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

The recursive formula is

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

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_0=1$.

$$\begin{gathered} {\mathrm{x}}_1=0.25 \\ {\mathrm{y}}_1=1.03125 \end{gathered}$$

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

$$\begin{gathered} {\mathrm{x}}_2=0.5 \\ {\mathrm{y}}_2=1.11035 \\ {\mathrm{x}}_3=0.75 \\ {\mathrm{y}}_3=1.22684 \\ {\mathrm{x}}_4=1 \\ {\mathrm{y}}_4=1.11035 \end{gathered}$$

Therefore, the approximation of the solution by the Taylor method of order 2 at point

x = 1

${\varphi }_2\left(1\right)=1.3725$

$$\begin{gathered} {\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+0.25 \\ {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+0.25({\mathrm{x}}_{\mathrm{n}}+1-{\mathrm{y}}_{\mathrm{n}})+\left(\frac{{\displaystyle {\left(0.25\right)}^2}}{2}\right)({\mathrm{y}}_{\mathrm{n}}-{\mathrm{x}}_{\mathrm{n}})+\frac{\left(0.{25}^3\right)}{6}({\mathrm{x}}_{\mathrm{n}}-{\mathrm{y}}_{\mathrm{n}})+\frac{{\left(0.25\right)}^4}{24}({\mathrm{y}}_{\mathrm{n}}-{\mathrm{x}}_{\mathrm{n}}) \end{gathered}$$

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_0=1$.

$$\begin{gathered} {\mathrm{x}}_1=0.25 \\ {\mathrm{y}}_1=1.02881 \end{gathered}$$

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

$$\begin{gathered} {\mathrm{x}}_2=0.5 \\ {\mathrm{y}}_2=1.10654 \\ {\mathrm{x}}_3=0.75 \\ {\mathrm{y}}_3=1.22238 \\ {\mathrm{x}}_4=1 \\ {\mathrm{y}}_4=1.36789 \end{gathered}$$

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

X = 1

${\varphi }_4\left(1\right)=1.3679$

The actual solution at x = 1

$$\begin{gathered} \mathrm{y}\left(\mathrm{x}\right)=\mathrm{x}+{\mathrm{e}}^{-\mathrm{x}} \\ \mathrm{y}\left(1\right)=1.3678794 \end{gathered}$$

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

$$\begin{gathered} \left|\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}-{\varphi }_2\left(1\right)\right|=0.00462 \\ \left|\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}-{\varphi }_4\left(1\right)\right|=0.00002 \end{gathered}$$

Hence the solution is$$\begin{gathered} {\mathit{\varphi }}_{\mathbf{2}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3725} \\ {\mathit{\varphi }}_{\mathbf{4}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3679} \end{gathered}$$