Question

In Problems, 1-10, use the improved Euler's method to obtain a four-decimal approximation of the indicated value. First, use h = 0.1, and then use

$$\begin{gathered} \mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{y}{\mathbf{)}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)} \end{gathered}$$

Short answer

The four decimal approximation for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 0.5503$; and for $\mathrm{h}=0.05,{\mathrm{y}}_{10}\approx 0.5495$.

Step-by-step solution

Definition of Euler’s method

• The Euler method is a first-order approach, meaning the local error (error per step) is proportional to the square of the step size, and the global error (error over time) is proportionate to the step size.
• The Euler technique is frequently used to build more complicated approaches, such as the predictor-corrector method.

Determine the four decimal approximation of the indicated value

Given that$\mathrm{y}\text{'}={\left(\mathrm{x}-\mathrm{y}\right)}^2$.

Note that, the formula for improving Euler's method is written as:

${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{h}\frac{\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)+\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}+1},{\mathrm{y}}_{\mathrm{n}+1}^{*}\right)}{2}$

Where ${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h}$and ${\mathrm{y}}_{\mathrm{n}+1}^{*}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}\left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)$.

First, find ${\mathrm{y}}_{\mathrm{n}+1}^{*}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}\left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)$for $\mathrm{f}(\mathrm{x},\mathrm{y})=(\mathrm{x}-\mathrm{y}{)}^2,\mathrm{h}=0.1,{\mathrm{x}}_0=0$and ${\mathrm{y}}_0=0.5$as follows:

$$\begin{gathered} {\mathrm{y}}_1^{*}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right) \\ {\mathrm{y}}_1^{*}=0.5+0.1(-0.5{)}^2 \\ {\mathrm{y}}_1^{*}=0.5+0.025 \\ {\mathrm{y}}_1^{*}=0.525 \end{gathered}$$

Find${\mathrm{x}}_1$using${\mathrm{x}}_1={\mathrm{x}}_0+\mathrm{h}$as:

$$\begin{gathered} {\mathrm{x}}_1=0+0.1 \\ {\mathrm{x}}_1=0.1 \end{gathered}$$

Find ${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{h}\frac{\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)+\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}+1},{\mathrm{y}}_{\mathrm{n}+1}^{*}\right)}{2}$for $\mathrm{f}(\mathrm{x},\mathrm{y})=(\mathrm{x}-\mathrm{y}{)}^2,\mathrm{h}=0.1,{\mathrm{x}}_0=0$, ${\mathrm{x}}_1=0.1$and ${\mathrm{y}}_0=0.5$as:

role="math" localid="1664181157514" $$\begin{gathered} {\mathrm{y}}_1={\mathrm{y}}_0+\mathrm{h}\frac{\mathrm{f}\left({\mathrm{x}}_0.{\mathrm{y}}_0\right)+\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1^{*}\right)}{2} \\ {\mathrm{y}}_1=0.5+0.1\frac{\left(0.25+{(0.1-0.525)}^2\right)}{2} \\ {\mathrm{y}}_1=0.5+0.1\frac{\left(0.25+{(0.425)}^2\right.}{2} \\ {\mathrm{y}}_1=0.5+0.1\frac{(0.25+0.180625)}{2} \end{gathered}$$

Simplify the equation as:

$$\begin{gathered} {\mathrm{y}}_1=0.5+0.1(0.2153125) \\ {\mathrm{y}}_1=0.5+0.02153125 \\ {\mathrm{y}}_1=0.52153125 \\ {\mathrm{y}}_1\approx 0.5215 \end{gathered}$$

Plot the table

The continuation of the given calculation is written in Table 1 as follows:

Table 1 . Improved Euler’s method with h = 0.1
$\mathrm{n}$	${\mathrm{x}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}+1}$
0	0	0.5	0.5215
1	0.1	0.5215	0.5362
2	0.2	0.5362	0.5449
3	0.3	0.5449	0.5490
4	0.4	0.5490	0.5503
5	0.5	0.5503	0.5505

Using this method for h = 0.05 is shown in table 2 as follows:

Table 2. Improved Euler’s method with h = 0.05
$\mathrm{n}$	${\mathrm{x}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}+1$
0	0	0.5	0.5116
1	0.05	0.5116	0.5214
2	0.10	0.5214	0.5294
3	0.15	0.5294	0.5359
4	0.20	0.5359	0.5408
5	0.25	0.5408	0.5444
6	0.30	0.5444	0.5469
7	0.35	0.5469	0.5484
8	0.40	0.5484	0.5492
9	0.45	0.5492	0.5495
10	0.5	0.5495	0.5496

Therefore, for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 0.5503$; and for $\mathrm{h}=0.05,{\mathrm{y}}_{10}\approx 0.5495$.