In Following problem, compare the accuracy of the approximationwith y (x*) using the improved Euler’s method starting at x0 with step sizea. 0.2      b. 0.1       c. 0.05d. Describe what happens to the error as the step size decreases. y' = 2y2(x - 1), y(2) = -1/2, x0 = 2, x* = 3

Given:

$y\text{'}=2y^2\left(x-1\right)$, $y\left(2\right)=-\frac{1}{2}=-0.5$.

To find: $y\left(3\right)$ when
(a) $h=0.2$,
(b) $h=0.1$,
(c) $h=0.05$.

(d) Describe what happens to the error as the step size decreases.

(a) $h=0.2$

$y\text{'}=f\left(x,y\right)=2y^2\left(x-1\right)$

Applying the Improved Euler method:

$y_{m+1}=y_m+\frac{1}{2}h\left[f\left(x_m, y_m\right)+f\left(x_m+h, y_m+hf\left(x_m, y_m\right)\right)\right]$

$f\left(x_0, y_0\right)=f\left(2, -0.5\right)=0.5$

$f\left(x_0+h, y_0+hf\left(x_0, y_0\right)\right)=f\left(2.2, -0.4\right)=0.384$

$y_1=y_0+\frac{1}{2}h\left[f\left(x_0, y_0\right)+f\left(x_0+h, y_0+hf\left(x_0, y_0\right)\right)\right]$

$$\begin{gathered} y_1=-0.5+\frac{0.2}{2}\left[0.5+0.384\right] \\ {y}_1=-0.4116 \end{gathered}$$

Again taking $\left(x_1, y_1\right)$ in place of $\left(x_0, y_0\right)$ and repeating the above process we get,

$f\left(x_1, y_1\right)=f\left(2.2, -0.4116\right)=0.4066$

$f\left(x_1+h, y_1+hf\left(x_1, y_1\right)\right)=f\left(2.4, -0.3303\right)=0.3054$

$y_2=y_1+\frac{1}{2}h\left[f\left(x_1, y_1\right)+f\left(x_1+h, y_1+hf\left(x_1, y_1\right)\right)\right]$

$$\begin{gathered} y_2=-0.4116+\frac{0.2}{2}\left[0.4066+0.3054\right] \\ {y}_2=-0.3404 \end{gathered}$$

Again taking $\left(x_2, y_2\right)$ in place of $\left(x_1, y_1\right)$ and repeating the above process we get,

$f\left(x_2, y_2\right)=f\left(2.4, -0.3404\right)=0.3244$

$f\left(x_2+h, y_2+hf\left(x_2, y_2\right)\right)=f\left(2.6, -0.2755\right)=0.2429$

$y_3=y_2+\frac{1}{2}h\left[f\left(x_2, y_2\right)+f\left(x_2+h, y_2+hf\left(x_2, y_2\right)\right)\right]$

$$\begin{gathered} y_3=-0.3404+\frac{0.2}{2}\left[0.3244+0.2429\right] \\ {y}_3=-0.2837 \end{gathered}$$

Again taking $\left(x_3, y_3\right)$ in place of $\left(x_2, y_2\right)$ and repeating the above process we get,

$f\left(x_3, y_3\right)=f\left(2.6, -0.2837\right)=0.2575$

$f\left(x_3+h, y_3+hf\left(x_3, y_3\right)\right)=f\left(2.8, -0.2322\right)=0.194$

$y_4=y_3+\frac{1}{2}h\left[f\left(x_3, y_3\right)+f\left(x_3+h, y_3+hf\left(x_3, y_3\right)\right)\right]$

$$\begin{gathered} y_4=-0.2837+\frac{0.2}{2}\left[0.2575+0.194\right] \\ {y}_4=-0.2385 \end{gathered}$$

Again taking $\left(x_4, y_4\right)$ in place of $\left(x_3, y_3\right)$ and repeating the above process we get,

$f\left(x_4, y_4\right)=f\left(2.8, -0.2385\right)=0.2048$

$f\left(x_4+h, y_4+hf\left(x_4, y_4\right)\right)=f\left(3, -0.1976\right)=0.1561$

$y_5=y_4+\frac{1}{2}h\left[f\left(x_4, y_4\right)+f\left(x_4+h, y_4+hf\left(x_4, y_4\right)\right)\right]$

$$\begin{gathered} y_5=-0.2385+\frac{0.2}{2}\left[0.2048+0.1561\right] \\ {y}_5=-0.2024 \end{gathered}$$

$\therefore y\left(3\right)=-0.2024$

(b) $h=0.1$

$y\text{'}=f\left(x,y\right)=2y^2\left(x-1\right)$

Applying the Improved Euler method:

$y_{m+1}=y_m+\frac{1}{2}h\left[f\left(x_m, y_m\right)+f\left(x_m+h, y_m+hf\left(x_m, y_m\right)\right)\right]$

$f\left(x_0, y_0\right)=f\left(2, -0.5\right)=0.5$

$f\left(x_0+h, y_0+hf\left(x_0, y_0\right)\right)=f\left(2.1, -0.45\right)=0.4455$

$y_1=y_0+\frac{1}{2}h\left[f\left(x_0, y_0\right)+f\left(x_0+h, y_0+hf\left(x_0, y_0\right)\right)\right]$

$$\begin{gathered} y_1=-0.5+\frac{0.1}{2}\left[0.5+0.4455\right] \\ {y}_1=-0.4527 \end{gathered}$$

Again taking $\left(x_1, y_1\right)$ in place of $\left(x_0, y_0\right)$ and repeating the above process we get,

$f\left(x_1, y_1\right)=f\left(2.1, -0.4527\right)=0.4509$

$f\left(x_1+h, y_1+hf\left(x_1, y_1\right)\right)=f\left(2.2, -0.4076\right)=0.3988$

$y_2=y_1+\frac{1}{2}h\left[f\left(x_1, y_1\right)+f\left(x_1+h, y_1+hf\left(x_1, y_1\right)\right)\right]$

$$\begin{gathered} y_2=-0.4527+\frac{0.1}{2}\left[0.4509+0.3988\right] \\ {y}_2=-0.4102 \end{gathered}$$

Again taking $\left(x_2, y_2\right)$ in place of $\left(x_1, y_1\right)$ and repeating the above process we get,

$f\left(x_2, y_2\right)=f\left(2.2, -0.4102\right)=0.4039$

$f\left(x_2+h, y_2+hf\left(x_2, y_2\right)\right)=f\left(2.3, -0.3698\right)=0.3556$

$y_3=y_2+\frac{1}{2}h\left[f\left(x_2, y_2\right)+f\left(x_2+h, y_2+hf\left(x_2, y_2\right)\right)\right]$

$$\begin{gathered} y_3=-0.4102+\frac{0.1}{2}\left[0.4039+0.3556\right] \\ {y}_3=-0.3723 \end{gathered}$$

Again taking $\left(x_3, y_3\right)$ in place of $\left(x_2, y_2\right)$ and repeating the above process we get,

$f\left(x_3, y_3\right)=f\left(2.3, -0.3723\right)=0.3603$

$f\left(x_3+h, y_3+hf\left(x_3, y_3\right)\right)=f\left(2.4, -0.3362\right)=0.3165$

$y_4=y_3+\frac{1}{2}h\left[f\left(x_3, y_3\right)+f\left(x_3+h, y_3+hf\left(x_3, y_3\right)\right)\right]$

$$\begin{gathered} y_4=-0.3723+\frac{0.1}{2}\left[0.3603+0.3165\right] \\ {y}_4=-0.3384 \end{gathered}$$

Again taking $\left(x_4, y_4\right)$ in place of $\left(x_3, y_3\right)$ and repeating the above process we get,

$f\left(x_4, y_4\right)=f\left(2.4, -0.3384\right)=0.3207$

$f\left(x_4+h, y_4+hf\left(x_4, y_4\right)\right)=f\left(2.5, -0.3064\right)=0.2816$

$y_5=y_4+\frac{1}{2}h\left[f\left(x_4, y_4\right)+f\left(x_4+h, y_4+hf\left(x_4, y_4\right)\right)\right]$

$$\begin{gathered} y_5=-0.3384+\frac{0.1}{2}\left[0.3207+0.2816\right] \\ {y}_5=-0.3083 \end{gathered}$$

Again taking $\left(x_5, y_5\right)$ in place of $\left(x_4, y_4\right)$ and repeating the above process we get,

$f\left(x_5, y_5\right)=f\left(2.5, -0.3083\right)=0.2852$

$f\left(x_5+h, y_5+hf\left(x_5, y_5\right)\right)=f\left(2.6, -0.2798\right)=0.2505$

$y_6=y_5+\frac{1}{2}h\left[f\left(x_5, y_5\right)+f\left(x_5+h, y_5+hf\left(x_5, y_5\right)\right)\right]$

$$\begin{gathered} y_6=-0.3083+\frac{0.1}{2}\left[0.2852+0.2505\right] \\ {y}_6=-0.2815 \end{gathered}$$

Again taking $\left(x_6, y_6\right)$ in place of $\left(x_5, y_5\right)$ and repeating the above process we get,

$f\left(x_6, y_6\right)=f\left(2.6, -0.2815\right)=0.2536$

$f\left(x_7+h, y_7+hf\left(x_7, y_7\right)\right)=f\left(2.7, -0.2562\right)=0.2231$

$y_7=y_6+\frac{1}{2}h\left[f\left(x_6, y_6\right)+f\left(x_6+h, y_6+hf\left(x_6, y_6\right)\right)\right]$

$$\begin{gathered} y_7=-0.2815+\frac{0.1}{2}\left[0.2536+0.2231\right] \\ {y}_7=-0.2577 \end{gathered}$$

Again taking $\left(x_7, y_7\right)$ in place of $\left(x_6, y_6\right)$ and repeating the above process we get,

$f\left(x_7, y_7\right)=f\left(2.7, -0.2577\right)=0.2258$

$f\left(x_7+h, y_7+hf\left(x_7, y_7\right)\right)=f\left(2.8, -0.2351\right)=0.199$

$y_8=y_7+\frac{1}{2}h\left[f\left(x_7, y_7\right)+f\left(x_7+h, y_7+hf\left(x_7, y_7\right)\right)\right]$

$$\begin{gathered} y_8=-0.2577+\frac{0.1}{2}\left[0.2258+0.199\right] \\ {y}_8=-0.2364 \end{gathered}$$

Again taking $\left(x_8, y_8\right)$ in place of $\left(x_7, y_7\right)$ and repeating the above process we get,

$f\left(x_8, y_8\right)=f\left(2.8, -0.2364\right)=0.2013$

$f\left(x_8+h, y_8+hf\left(x_8, y_8\right)\right)=f\left(2.9, -0.2163\right)=0.1778$

$y_9=y_8+\frac{1}{2}h\left[f\left(x_8, y_8\right)+f\left(x_8+h, y_8+hf\left(x_8, y_8\right)\right)\right]$

$$\begin{gathered} y_9=-0.2364+\frac{0.1}{2}\left[0.2013+0.1778\right] \\ {y}_9=-0.2175 \end{gathered}$$

Again taking $\left(x_9, y_9\right)$ in place of $\left(x_8, y_8\right)$ and repeating the above process we get,

$f\left(x_9, y_9\right)=f\left(2.9, -0.2175\right)=0.1798$

$f\left(x_9+h, y_9+hf\left(x_9, y_9\right)\right)=f\left(3, -0.1995\right)=0.1592$

$y_{10}=y_9+\frac{1}{2}h\left[f\left(x_9, y_9\right)+f\left(x_9+h, y_9+hf\left(x_9, y_9\right)\right)\right]$

$$\begin{gathered} y_{10}=-0.2175+\frac{0.1}{2}\left[0.1798+0.1592\right] \\ {y}_{10}=-0.2005 \end{gathered}$$

$\therefore y\left(3\right)=-0.2005$