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

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

Short answer

The four decimal approximation for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 1.7557$; and for $\mathrm{h}=0.05,{\mathrm{y}}_{10}\approx 1.7560$.

Step-by-step solution

Improved Euler's method

As per the Improved Euler's method, the solution of a linear differential equation of the form $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ is given as follows:

$\begin{array}{l}{\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}^{\mathbf{*}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hf}\left(x_n,y_n\right)\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\\ {\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\frac{\mathbf{h}}{\mathbf{2}}\left(f\left(x_n,y_n\right)+f\left(x_{n+1},y_{n+1}^{*}\right)\right)\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Solution for the linear differential equation

The linear differential equation is given as follows:

${\mathrm{y}}^{\text{'}}=\mathrm{xy}+\sqrt{\mathrm{y}}$

The initial value is given as $\mathrm{y}\left(0\right)=1$.

This implies that${\mathrm{x}}_0=0$and ${\mathrm{y}}_0=1$.

The given differential equation is of the form $\mathrm{y}\text{'}=\mathrm{f}\left(\mathrm{x}\right)$.

Then, $\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{xy}+\sqrt{\mathrm{y}}$.

Obtain the solution of the given differential equation by Euler's method for h = 0.1.

Substitute 0 for n into equation (1) to obtain the equation of${\mathrm{y}}_1^{*}$as:

${\mathrm{y}}_1^{*}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$

Substitute 0 for ${\mathrm{x}}_0,1$for ${\mathrm{y}}_0$and 0.1 for h into ${\mathrm{y}}_1^{*}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$as:

$$\begin{gathered} {\mathrm{y}}_1^{*}=1+(0.1)\mathrm{f}(0,1) \\ {\mathrm{y}}_1^{*}=1+(0.1)·\left(0\right(1)+\sqrt{1}) \\ {\mathrm{y}}_1^{*}=1+0.1 \\ {\mathrm{y}}_1^{*}=1.1 \end{gathered}$$

Substitute 0 for n into equation (2) as follows:

${\mathrm{y}}_1={\mathrm{y}}_0+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)+\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1^{*}\right)\right)$

Substitute 0 for${\mathrm{x}}_0,1$ for${\mathrm{y}}_0,0.1$ for$x_1,1.1$ for${{\mathrm{y}}_1}^{*}$ and 0.1 for h into${\mathrm{y}}_1={\mathrm{y}}_0+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)+\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1^{*}\right)\right)$ as:

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{0.1}{2}\left(\mathrm{f}\right(0,1)+\mathrm{f}(0.1,1.1\left)\right) \\ {\mathrm{y}}_1=1+(0.05)\left(0\right(1)+\sqrt{1}+0.1(1.1)+\sqrt{1.1}) \\ {\mathrm{y}}_1=1+0.10794 \\ {\mathrm{y}}_1=1.1079 \end{gathered}$$

Therefore, the value of${\mathrm{y}}_1$ or$\mathrm{y}(0.1)$ is 1.1079.

Values obtained from improved Euler method for h = 0.1

Similarly, the value of $\mathrm{y}(0.2),\mathrm{y}(0.3),\mathrm{y}(0.4)$and$\mathrm{y}(0.5)$are obtained as shown in Table 1:

${\mathrm{x}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}$
0.00	1.0000
0.10	1.1079
0.20	1.2337
0.30	1.3806
0.40	1.5529
0.50	1.7557

Now for step size h = 0.05, calculate the value of y(0.5) as shown below:

Substitute 0 for n into equation (1) to obtain the equation of${\mathrm{y}}_1^{*}$as:

${\mathrm{y}}_1^{*}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$

Substitute 0 for${\mathrm{x}}_0,1$for${\mathrm{y}}_0$and 0.05 for h into${\mathrm{y}}_1^{*}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$as follows:

$$\begin{gathered} {\mathrm{y}}_1^{*}=1+(0.05)\mathrm{f}(0,1) \\ {\mathrm{y}}_1^{*}=1+(0.05)·\left(0\right(1)+\sqrt{1}) \\ {\mathrm{y}}_1^{*}=1+0.05 \\ {\mathrm{y}}_1^{*}=1.05 \end{gathered}$$

Substitute 0 for n into equation (2) as:

${\mathrm{y}}_1={\mathrm{y}}_0+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)+\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1^{*}\right)\right)$

Substitute 0 for ${\mathrm{x}}_0,1$for ${\mathrm{y}}_0,0.05$for ${\mathrm{x}}_1,1.05$for ${{\text{y}}_1}^{*}$and 0.05 for h into ${\mathrm{y}}_1={\mathrm{y}}_0+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)+\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1^{*}\right)\right)$as follows:

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{0.05}{2}\left(\mathrm{f}\right(0,1)+\mathrm{f}(0.05,1.05\left)\right) \\ {\mathrm{y}}_1=1+(0.025)\left(0\right(1)+\sqrt{1}+0.05(1.05)+\sqrt{1.05}) \\ {\mathrm{y}}_1=1+0.051929 \\ {\mathrm{y}}_1\approx 1.0519 \end{gathered}$$

Therefore, the value of ${\mathrm{y}}_1$ or$\mathrm{y}\left(0.05\right)$ is 1.0519.

Values obtained from improved Euler method for h = 0.05

Similarly, the value of $\mathrm{y}(0.1),\mathrm{y}(0.15),\mathrm{y}(0.20),\mathrm{y}(0.25),\mathrm{y}(0.30),\mathrm{y}(0.35),\mathrm{y}(0.40)$, $\mathrm{y}(0.45)$and $\mathrm{y}\left(0.5\right)$are obtained as shown in Table 2 as:

${\mathrm{x}}_{\mathrm{n}}$	${\mathrm{y}}_{\mathrm{n}}$
0.00	1.0000
0.05	1.0519
0.10	1.1079
0.15	1.1648
0.20	1.2337
0.25	1.3043
0.30	1.3807
0.35	1.4634
0.40	1.5530
0.45	1.6503
0.50	1.7560

Thus, the four decimal approximation of the value$\mathrm{y}\left(0.5\right)$for$\mathrm{h}=0.1$ and$\mathrm{h}=0.05$ is obtained as 1.7557 and 1.7560, respectively.