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

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

Short answer

The four decimal approximation for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 0.6224$; 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:

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

The initial value is given as $y\left(0\right)=0.5$.

This implies that $x_0=0$and ${\mathrm{y}}_0=0.5$.

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

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

Obtain the solution of the given differential equation by Euler's method for $\mathrm{h}=0.1$.

Substitute 0 for n into equation (1) to obtain the equation of as:

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

Substitute 0 for ${\mathrm{x}}_0,0.5$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:

$$\begin{gathered} {\mathrm{y}}_1^{*}=0.5+(0.1)\mathrm{f}(0,0.5) \\ {\mathrm{y}}_1^{*}=0.5+(0.1)·\left(0.5-{(0.5)}^2\right) \\ {\mathrm{y}}_1^{*}=0.5+0.025 \\ {\mathrm{y}}_1^{*}=0.525 \end{gathered}$$

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${\mathrm{x}}_0,0.5$ for${\mathrm{y}}_0,0.1$ for${\mathrm{x}}_1,0.525$ for${{\mathrm{y}}_1}^{*}$ and 0.1 for into as:

$$\begin{gathered} {\mathrm{y}}_1={\mathrm{y}}_0+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)+\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1^{*}\right)\right) \\ {\mathrm{y}}_1=0.5+\frac{0.1}{2}\left(\mathrm{f}\right(0,0.5)+\mathrm{f}(0.1,0.525\left)\right) \\ {\mathrm{y}}_1=0.5+(0.05)\left(0.5-{(0.5)}^2+0.525-{(0.525)}^2\right) \\ {\mathrm{y}}_1=0.5+0.05(0.499375) \\ {\mathrm{y}}_1\approx 0.5250 \end{gathered}$$

Therefore, the value of ${\mathrm{y}}_1$or $\mathrm{y}(0.1)$is 0.5250.

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:

xn	yn
0.00	0.5000
0.10	0.5250
0.20	0.5498
0.30	0.5744
0.40	0.5986
0.50	0.6244

Now for step size h = 0.05, calculate the value of$\mathrm{y}(0.5)$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${\mathrm{x}}_0,0.5$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:

$$\begin{gathered} {\mathrm{y}}_1^{*}=0.5+(0.05)\mathrm{f}(0,0.5) \\ {\mathrm{y}}_1^{*}=0.5+(0.05)·\left(0.5-{(0.5)}^2\right) \\ {\mathrm{y}}_1^{*}=0.5+0.0125 \\ {\mathrm{y}}_1^{*}=0.5125 \end{gathered}$$

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 ${\mathrm{x}}_0,0.5$for ${\mathrm{y}}_0,0.05$for ${\mathrm{x}}_1,0.5125$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:

$$\begin{gathered} {\mathrm{y}}_1=0.5+\frac{0.05}{2}\left(\mathrm{f}\right(0,0.5)+\mathrm{f}(0.05,0.5125\left)\right) \\ {\mathrm{y}}_1=0.5+(0.025)\left(0.5-{(0.5)}^2+0.5125-{(0.5125)}^2\right) \\ {\mathrm{y}}_1=0.5+0.025(0.49984) \\ {\mathrm{y}}_1=0.5125 \end{gathered}$$

Therefore, the value of${\mathrm{y}}_1$ or$\mathrm{y}(0.05)$ is 0.5125.

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$\mathrm{y}(0.5)$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$\mathrm{y}(0.5)$ for$\mathrm{h}=0.1$ and$\mathrm{h}=0.05$ is obtained as 1.6244 and 1.3315, respectively.