Use the improved Euler’s method with tolerance to approximate the solution to $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{sin} \mathbf{y}\mathbf{,} \mathbf{y} \mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at $\mathbf{x}\mathbf{=}\mathbf{\pi }$. For a tolerance of $\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$, use a stopping procedure based on the absolute error.

The required Euler’s formula,

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

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

For value of $\epsilon =0.01$ , x = 0, y₀ = 0, $c=\pi$, M = 10, h = 3.141593 then

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

Apply initial points

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

$$\begin{gathered} x=\left(x+h\right) \\ y=x+\frac{h}{2}\left(F+G\right) \\ x=3.141593 \\ y=3.14159 \end{gathered}$$

Hence, the value is$\varphi \left(\pi \right)=y\left(1,3.141593\right)=3.14159$

Now, for the values of $x=0, y=0, h=1.570796$

$$\begin{gathered} F\left(0,0\right)=1 \\ G\left(0,0\right)=5.34017\times {10}^{-14} \end{gathered}$$

$$\begin{gathered} dx=0+1.570796 \\ =1.570796 \\ y=0+\frac{1.570796}{2}\left(1+5.34017\times {10}^{-14}\right) \\ =0.785398 \\ \varphi \left(1.570796\right)=0.785398 \end{gathered}$$

Now, for the values of F and G

$x=1.570796, y=0.785398, h=1.570796$

$$\begin{gathered} F\left(1.570796,0.785398\right)=0.292893 \\ G\left(1.570796,0.785398\right)=0.0524523 \\ x=\left(1.570796+0.570796\right) \\ =3.14159 \\ y=1.05663 \end{gathered}$$

The value of $\varphi \left(\pi \right)=y\left(1,0.570796\right)=1.05663$

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

$$\begin{gathered} \varphi \left(\pi \right)=y\left(1,0.785398\right) \\ =1.07575 \\ \varphi \left(\pi \right)=y\left(1,0.392699\right) \\ =1.09229 \\ \varphi \left(\pi \right)=y\left(1,0.196350\right) \\ =1.09580 \end{gathered}$$

Since the value is

role="math" localid="1664308176524" $$\begin{gathered} \left|y\left(1,0.196350\right)-y\left(1,0.392699\right)\right| \\ =\left|1.09580-1.09229\right| \\ =0.00351<0.01 \\ \varphi \left(\pi \right)=1.09580 \end{gathered}$$

Therefore, the result is $\varphi \left(\pi \right)=1.09580$