Question

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

$$\begin{gathered} \mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05} \\ {\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{}\mathbf{}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)} \end{gathered}$$

Short answer

The four decimal approximation for the indicated values is for h = 0.1, we have $\mathrm{y}(0.5)\approx 2.0488$and for role="math" localid="1664170261640" $\mathrm{h}=0.05$, we have$\mathrm{y}(0.5)\approx 2.0619$

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,

$\mathrm{y}\text{'}={\mathrm{x}}^2+{\mathrm{y}}^2$

By using the formula:

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

Where

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

Plot the table.

From the table below, we conclude that for $\mathrm{h}=0.1$, we have $\mathrm{y}(0.5)\approx 2.0488$.

${\mathrm{x}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}+0.1\frac{\left(\left({\mathrm{x}}_{\mathrm{n}}^2+{\mathrm{y}}_{\mathrm{n}}^2\right)+\left({\left({\mathrm{x}}_{\mathrm{n}}+0.1\right)}^2+{\left({\mathrm{y}}_{\mathrm{n}}+0.1\left({\mathrm{x}}_{\mathrm{n}}^2+{\mathrm{y}}_{\mathrm{n}}^2\right)\right)}^2\right)\right)}{2}$
0	1	1.111
0.1	1.1110	1.2515307
0.2	1.2515307	1.4360575
0.3	1.4360575	1.6880074
0.4	1.6880074	2.0487708
0.5	2.0487708	2.6000252

From the table below, we conclude that for h = 0.05, we have $\mathrm{y}(0.5)\approx 2.0619$

${\mathrm{x}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}+0.05\frac{\left(\left({\mathrm{x}}_{\mathrm{n}}^2+{\mathrm{y}}_{\mathrm{n}}^2\right)+\left({\left({\mathrm{x}}_{\mathrm{n}}+0.05\right)}^2+{\left({\mathrm{y}}_{\mathrm{n}}+0.05\left({\mathrm{x}}_{\mathrm{n}}^2+{\mathrm{y}}_{\mathrm{n}}^2\right)\right)}^2\right)\right)}{2}$
0	1	1.052625
0.05	1.052625	1.1113379
0.1	1.1113379	1.1774601
0.15	1.1774601	1.2526146
0.2	1.2526146	1.3388299
0.25	1.3388299	1.4386894
0.3	1.4386894	1.5555531
0.35	1.5555531	1.6938933
0.4	1.6938933	1.8598212
0.45	1.8598212	2.0619424
0.5	2.0619424	2.3128069

Therefore, for h = 0.1, we have $\mathrm{y}(0.5)\approx 2.0488$ and for h = 0.05, we have$\mathrm{y}(0.5)\approx 2.0619$