Question

Consider the initial-value problem ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$. Use the improved Euler's method with $\mathbf{h}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1}$and $\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$to obtain approximate values of the solution at $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$. At each step, compare the approximate value with the actual value of the analytic solution.

Short answer

The approximated value of$\mathrm{y}\left(0.5\right)$ for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 3.8254$; and for $\mathrm{h}=0.05,{\mathrm{y}}_{10}\approx 3.8840$. The actual value of$\mathrm{y}\left(0.5\right)$ is 3.9082.

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:

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

Find the approximate value for the equation h = 0.1

The linear differential equation is given as follows:

$\mathrm{y}\text{'}=(\mathrm{x}+\mathrm{y}-1{)}^2$

The initial value is given as $\mathrm{y}\left(0\right)=2$.

This implies that${\mathrm{x}}_0=0$and ${\mathrm{y}}_0=2$.

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

Then, $\mathrm{f}(\mathrm{x},\mathrm{y})=(\mathrm{x}+\mathrm{y}-1{)}^2$.

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

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${\mathrm{x}}_0,2$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:

$$\begin{gathered} {\mathrm{y}}_1^{*}=2+(0.1)\mathrm{f}(0,2) \\ {\mathrm{y}}_1^{*}=2+(0.1)·(0+2-1{)}^2 \\ {\mathrm{y}}_1^{*}=2+0.1 \\ {\mathrm{y}}_1^{*}=2.1 \end{gathered}$$

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 ${\mathrm{x}}_0,2$for ${\mathrm{y}}_0,0.1$for ${\mathrm{x}}_1,2.1$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:

$$\begin{gathered} {\mathrm{y}}_1=2+\frac{0.1}{2}\left(\mathrm{f}\right(0,2)+\mathrm{f}(0.1,2.1\left)\right) \\ {\mathrm{y}}_1=2+(0.05)\left({(0+2-1)}^2+{(0.1+2.1-1)}^2\right) \\ {\mathrm{y}}_1=2+(0.05)(2.44) \\ {\mathrm{y}}_1=2.122 \end{gathered}$$

Therefore, the approximated value of${\mathrm{y}}_1$ or$\mathrm{y}(0.1)$ is 2.1220.

Finding the actual value for the equation

Now, to obtain the exact solution of the differential equation $y\text{'}=(x+y-1{)}^2$, assume$(\mathrm{x}+\mathrm{y}-1)$as a variable t then the expression becomes as shown below:

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dt}}-1={\mathrm{t}}^2 \\ \frac{\mathrm{dy}}{\mathrm{dt}}={\mathrm{t}}^2+1 \end{gathered}$$

Now integrate the above expression by rearranging the similar variable terms on one side of the equation as follows:

$$\begin{gathered} \int \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}=\int \mathrm{dx} \\ {\tan }^{-1}\mathrm{t}=\mathrm{x}+\mathrm{c} \end{gathered}$$

Back-substitute the value of t as $(\mathrm{x}+\mathrm{y}-1)$in the above expression and simplify as:

$$\begin{gathered} {\tan }^{-1}\mathrm{t}=\mathrm{x}+\mathrm{c} \\ {\tan }^{-1}(\mathrm{x}+\mathrm{y}-1)=\mathrm{x}+\mathrm{c} \\ \mathrm{x}+\mathrm{y}-1=\tan (\mathrm{x}+\mathrm{c}) \\ \mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+\mathrm{c}) \end{gathered}$$

Substitute the initial value$\mathrm{y}(0)=2$into$\mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+\mathrm{c})$and solve for integration constant c as:

$$\begin{gathered} 2=1-0+\tan (0+\mathrm{c}) \\ 2-1=\mathrm{tanc} \\ \mathrm{tanc}=1 \\ \mathrm{c}\approx 0.7854 \end{gathered}$$

Hence, the solution of the given differential equation is obtained as $\mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+0.7854)$

Substitute 0.1 for x into$\mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+0.7854)$to obtain the actual value of$\mathrm{y}(0.10)$as:

$$\begin{gathered} \mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+0.7854) \\ \mathrm{y}\approx 1-0.1+\tan (0.1+0.7854) \\ \mathrm{y}\approx 0.9+1.2230 \\ \mathrm{y}=2.1230 \end{gathered}$$

Therefore, the actual value of ${\mathrm{y}}_1$or $\mathrm{y}(0.1)$is 2.1230.

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

xn	Approximated yn	Actual yn
0.00	2.0000	2.0000
0.10	2.1230	2.1230
0.20	2.3049	2.3085
0.30	2.5858	2.5958
0.40	3.0378	3.6050
0.50	3.8254	3.9082

Find the approximate value for the equation h = 0.05

Now for step size h = 0.05, calculate for 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${\mathrm{x}}_0,2$ 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:

$$\begin{gathered} {\mathrm{y}}_1^{*}=2+(0.05)\mathrm{f}(0,2) \\ {\mathrm{y}}_1^{*}=2+(0.05)·(0+2-1) \\ {\mathrm{y}}_1^{*}=2+0.05 \\ {\mathrm{y}}_1^{*}=2.05 \end{gathered}$$

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 ${\mathrm{x}}_0,2$for ${\mathrm{y}}_0,0.05$for ${\mathrm{x}}_1,2.05$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:

$$\begin{gathered} {\mathrm{y}}_1=2+\frac{0.05}{2}\left(\mathrm{f}\right(0,2)+\mathrm{f}(0.05,2.05\left)\right) \\ {\mathrm{y}}_1=2+(0.025)\left({(0+2-1)}^2+{(0.05+2.05-1)}^2\right) \\ {\mathrm{y}}_1=2+(0.025)(2.21) \\ {\mathrm{y}}_1=2.05525 \end{gathered}$$

Therefore, the approximated value of${\mathrm{y}}_1$ or$\mathrm{y}(0.05)$ is 2.0553.

Find the actual value for h = 0.05

Substitute 0.05 for x into$\mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+0.7854)$ to obtain the actual value of$\mathrm{y}(0.05)$ as:

$$\begin{gathered} \mathrm{y}=1-\mathrm{x}+\tan (\mathrm{x}+0.7854) \\ \mathrm{y}=1-0.05+\tan (0.05+0.7854) \\ \mathrm{y}=0.95+1.1054 \\ \mathrm{y}=2.0554 \end{gathered}$$

Therefore, the actual value of${\mathrm{y}}_1$ or$\mathrm{y}(0.05)$ is 2.0554.

Similarly, the value of $y(0.1),y(0.15),y(0.20),y(0.25),y(0.30),y(0.35),y(0.40)$, are obtained as shown$y(0.45)$ and$y(0.45)$ in Table 2.

xn	Approximated yn	Actual yn
0.00	2.0000	2.0000
0.05	2.0553	2.0554
0.10	2.1228	2.1230
0.15	2.2056	2.2061
0.20	2.3075	2.3085
0.25	2.4342	2.4358
0.30	2.5931	2.5958
0.35	2.7953	2.7997
0.40	3.0574	3.0650
0.45	3.4507	3.4189
0.50	3.8840	3.9082

Hence, The approximated value of$\mathrm{y}\left(0.5\right)$ for $\mathrm{h}=0.1,{\mathrm{y}}_5\approx 3.8254$; and for $\mathrm{h}=0.05,{\mathrm{y}}_{10}\approx 3.8840$. The actual value of$\mathrm{y}\left(0.5\right)$ is 3.9082.