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

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

Short answer

The four decimal approximation for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 1.3260$; and for $\mathrm{h}=0.05,{\mathrm{y}}_{10}\approx 1.3315$.

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:

$y\text{'}=x{y}^2-\frac{y}{x}$

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

This implies that$x_0=1$and $y_0=1$.

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

Then, $f(x,y)=x{y}^2-\frac{y}{x}$.

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:

$y_1^{*}=y_0+hf\left(x_0,y_0\right)$

Substitute 1 for $x_0,1$for $y_0$and 0.1 for h into ${\mathrm{y}}_1^{*}={\mathrm{y}}_0+\mathrm{hf}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$as:

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

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 1 for $x_0,1$for $y_0,1.1$for $x_1,1$for${{\mathrm{y}}_1}^{*}$and 0.1 for 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:

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{0.1}{2}\left(\mathrm{f}\right(1,1)+\mathrm{f}(1.1,1\left)\right) \\ {\mathrm{y}}_1=1+(0.05)\left(1\left(1\right)-\frac{1}{1}+1.1\left(1\right)-\frac{1}{1.1}\right) \\ {\mathrm{y}}_1=1+(0.05)(0.1909) \\ {\mathrm{y}}_1=1.0095 \end{gathered}$$

Therefore, the value of${\mathrm{y}}_1$ or$\mathrm{y}(1.1)$ is 1.0095.

Values obtained from improved Euler method for h = 0.1

Similarly, the value of and are obtained as shown in Table 1:

xn	yn
1.00	1.0000
1.10	1.0095
1.20	1.0404
1.30	1.0967
1.40	1.1866
1.50	1.3260

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 1 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:

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

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 1 for${\mathrm{x}}_0,1$ for${\mathrm{y}}_0,1.05$ for${\mathrm{x}}_1,1$ for${{\mathrm{y}}_1}^{*}$ 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:

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{0.05}{2}\left(\mathrm{f}\right(1,1)+\mathrm{f}(1.05,1\left)\right) \\ {\mathrm{y}}_1=1+(0.025)\left(1\left(1\right)-\frac{1}{1}+1.05\left(1\right)-\frac{1}{1.05}\right) \\ {\mathrm{y}}_1=1+(0.025)(0.09761) \\ {\mathrm{y}}_1=1.0024 \end{gathered}$$

Therefore, the value of${\mathrm{y}}_1$ or$\mathrm{y}(1.05)$ is 1.0024.

Values obtained from improved Euler method for h = 0.05

Similarly, the value of $\mathrm{y}(1.1),\mathrm{y}(1.15),\mathrm{y}(1.20),\mathrm{y}(1.25),\mathrm{y}(1.30),\mathrm{y}(1.35),\mathrm{y}(1.40)$,$\mathrm{y}(1.45)$ and$\mathrm{y}(1.5)$ are obtained as shown in Table 2 as:

xn	yn
1.00	1.0000
1.05	1.0024
1.10	1.0100
1.15	1.0228
1.20	1.0414
1.25	1.0663
1.30	1.0984
1.35	1.1389
1.40	1.1805
1.45	1.2526
1.50	1.3315

Thus, the four decimal approximation of the value$y(0.5)$ for$\mathrm{h}=0.1$ and$\mathrm{h}=0.05$ is obtained as 1.3260 and 1.3315, respectively.