Question

Repeat Problem 13 using the improved Euler's method. Its global truncation error is $\mathbf{O}\left(h^2\right)$.

Short answer

1. The approximate value of $\mathrm{y}\left(0.1\right)$for the initial value problem $\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$is 1.22.
2. The bound for local truncation error in ${\mathrm{y}}_1$for the initial value problem$\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$ is 0.001628.
3. The actual error of 0.0014 is less than the obtained truncation error of 0.00168 in subpart (b).
4. The approximate value of $\mathrm{y}\left(0.1\right)$using two steps of Euler's method for the initial value problem $\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$is 1.221025.
5. The actual error when h = 0.05 is approximately half the error when h = 0.1.

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 $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{f}\left(x,y\right)$ is given as follows:

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

Euler's method:

As per Euler's method, the solution of the equation of the form is given as follows:

${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hf}\left(x_n,y_n\right)$

a) Finding the approximate value for the equation h = 0.1

${\mathrm{y}}_1$The initial value problem is$\mathrm{y}\text{'}=2\mathrm{y}$such that the value of y(0) is 1. The analytic solution is$\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$.

The initial value problem is given as follows:$\mathrm{y}\text{'}=2\mathrm{y}$

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$\mathrm{y}\text{'}=\mathrm{f}\left(\mathrm{x}\right)$.

$\mathrm{f}(\mathrm{x},\mathrm{y})=2\mathrm{y}$

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

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

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

Substitute 0 for n in equation (1) to obtain the equation of$y_1$as shown below.

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

Substitute 0 for $x_0,1$for ${\mathrm{y}}_0$and 0.1 for h in the above equation.

$$\begin{gathered} {\mathrm{y}}_1=1+(0.1)\mathrm{f}(0,1) \\ {\mathrm{y}}_1=1+(0.1)·\left(2\right) \\ {\mathrm{y}}_1=1+0.2 \\ {\mathrm{y}}_1=1.2 \end{gathered}$$

Substitute 0 for n in equation (2).

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

Substitute 0 for${\mathrm{x}}_0,1$for${\mathrm{y}}_0,0.1$for${\mathrm{x}}_1,1.2$for${{\mathrm{y}}_1}^{*}$and $0.1$ for h in the above equation.

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{0.1}{2}\left(\mathrm{f}\right(0,1)+\mathrm{f}(0.1,1.2\left)\right) \\ {\mathrm{y}}_1=1+(0.05)(2+2(1.2\left)\right) \\ {\mathrm{y}}_1=1+0.22 \\ {\mathrm{y}}_1=1.22 \end{gathered}$$

Thus, the approximate value of $\mathrm{y}\left(0.1\right)$for initial value problem $\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$is 1.22.

b) Finding the upper bound value for local truncation error

The bound for local truncation error in${\mathrm{y}}_1$for the initial value problem$\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$.

Consider the equation$\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$.

Differentiate the equation$\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$with respect to x as follows:

$y\text{'}=2e^{2x}$

Further, differentiate the equation$y\text{'}=2e^{2x}$with respect toas follows:

role="math" localid="1664207592271" $\mathrm{y}"=4{\mathrm{e}}^{2\mathrm{x}}$

The local truncation error of the function$\mathrm{y}"=4{\mathrm{e}}^{2\mathrm{x}}$is as follows:

$\mathrm{y}"\left(\mathrm{c}\right)\frac{{\mathrm{h}}^2}{2}=4{\mathrm{e}}^{2\mathrm{c}}\left(\frac{{\mathrm{h}}^2}{2}\right)$----------(2)

The value of c is defined as follows:${\mathrm{x}}_{\mathrm{n}}<\mathrm{c}<{\mathrm{x}}_{\mathrm{n}+1}$

Now substitute 0 for n in the above equation.

$$\begin{gathered} {\mathrm{x}}_0<\mathrm{c}<{\mathrm{x}}_1<0.1 \\ 0<\mathrm{c}\left({\mathrm{x}}_0\right.\left.=0\mathrm{and}{\mathrm{x}}_1=0.1\right) \end{gathered}$$

Substitute 0.1 for c and 0.1 for h in equation (2) to obtain the upper bound for the local truncation error.

$$\begin{gathered} \mathrm{y}\text{'}\text{'}\left(\mathrm{c}\right)\frac{{\mathrm{h}}^2}{2}=8{\mathrm{e}}^{2(0.1)}\left(\frac{{(0.1)}^2}{2}\right) \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{c}\right)\frac{{\mathrm{h}}^2}{2}=8{\mathrm{e}}^{0.2}\left(\frac{0.01}{2}\right) \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{c}\right)\frac{{\mathrm{h}}^2}{2}=0.00133\left({\mathrm{e}}^{0.2}\right) \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{c}\right)\frac{{\mathrm{h}}^2}{2}=0.001628 \end{gathered}$$

Thus, the upper bound of the local truncation error for the initial value problem $\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$is 0.01628.

c) The comparison of the error in y1 with obtained error bound in subpart (b).

From subpart (a), the obtained value of${\mathrm{y}}_1$is 1.2.

From subpart (b), the considered function for truncation error is$\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$.

Now substitute 0.1 for x in the equation$\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$and solve.

$$\begin{gathered} \mathrm{y}(0.1)={\mathrm{e}}^{2(0.1)} \\ \mathrm{y}(0.1)={\mathrm{e}}^{0.2} \\ \mathrm{y}(0.1)=1.2214 \end{gathered}$$

The actual error is calculated by the equation as follows:

actual error$=y(0.1)-y_1$-----(3)

Substitute 1.2214 for y(0.1) and 1.22 for${\mathrm{y}}_1$in equation (3).

$$\begin{gathered} Actualerror=1.2214-1.22 \\ Actualerror=0.0014 \end{gathered}$$

Thus, the actual error of 0.0014 is less than the obtained truncation error of 0.001628 in subpart (b).

d) Approximate y(0.1) using two steps and Euler's method.

The initial value problem is given as follows:

$\mathrm{y}\text{'}=2\mathrm{y}$

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$\mathrm{y}\text{'}=\mathrm{f}\left(\mathrm{x}\right)$.

$\mathrm{f}(\mathrm{x},\mathrm{y})=2\mathrm{y}$

Obtain the solution of the given differential equation by Euler's method for h = 0.05.

The solution to the initial value problem of the form$\mathrm{y}\text{'}=\mathrm{f}(\mathrm{x},\mathrm{y})$by the Euler's method is given as follows:

${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}\left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)\dots \dots \left(4\right)$

Substitute 0 for n in equation (4) to obtain the equation of${\mathrm{y}}_1$as shown below.

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

Substitute 0 for $x_0,1$for $y_0$and 0.05 for h in the above equation.

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

Substitute 0 for n in equation (2).

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

Substitute 0 for${\mathrm{x}}_0,1$for${\mathrm{y}}_0,0.05$for${\mathrm{x}}_1,1.1$for${{\mathrm{y}}_1}^{*}$and 0.05 for h in the above equation.

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{0.05}{2}\left(\mathrm{f}\right(0,1)+\mathrm{f}(0.05,1.1\left)\right) \\ {\mathrm{y}}_1=1+(0.025)(2+2(1.1\left)\right) \\ {\mathrm{y}}_1=1+0.105 \\ {\mathrm{y}}_1=1.105 \end{gathered}$$

Substitute 1 for n in equation (1) to obtain the equation of${\mathrm{y}}_2^{*}$as shown below.

${\mathrm{y}}_2^{*}={\mathrm{y}}_1+\mathrm{hf}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$

Substitute 0.05 for ${\mathrm{x}}_1,1.105$for ${\mathrm{y}}_1$and 0.05 for h in the above equation.

$$\begin{gathered} {\mathrm{y}}_2^{*}=1.105+(0.05)\mathrm{f}(0.05,1.105) \\ {\mathrm{y}}_2^{*}=1.105+(0.05)·\left(2\right(1.105\left)\right) \\ {\mathrm{y}}_2^{*}=1.105+0.1105 \\ {\mathrm{y}}_2^{*}=1.2155 \end{gathered}$$

Substitute 1 for n in equation (2).

${\mathrm{y}}_2={\mathrm{y}}_1+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)+\mathrm{f}\left({\mathrm{x}}_2,{\mathrm{y}}_2^{*}\right)\right)$

Substitute 0.05 for ${\mathrm{x}}_1,1.105$for ${\mathrm{y}}_1,0.1$for ${\mathrm{x}}_2,1.2155$for ${{\mathrm{y}}_2}^{*}$and 0.05 for h in the above equation.

Substitute 1 for n in equation (2).

${\mathrm{y}}_2={\mathrm{y}}_1+\frac{\mathrm{h}}{2}\left(\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)+\mathrm{f}\left({\mathrm{x}}_2,{\mathrm{y}}_2^{*}\right)\right)$

Substitute 0.05 for${\mathrm{x}}_1,1.105$for$y_1,0.1$for$x_2,1.2155$for${{\mathrm{y}}_2}^{*}$and 0.05 for h in the above equation.

$$\begin{gathered} {\mathrm{y}}_1=1.105+\frac{0.05}{2}\left(\mathrm{f}\right(0.05,1.105)+\mathrm{f}(0.1,1.2155\left)\right) \\ {\mathrm{y}}_1=1.105+(0.025)\left(2\right(1.105)+2(1.2155\left)\right) \\ {\mathrm{y}}_1=1.105+0.116025 \\ {\mathrm{y}}_1=1.221025 \end{gathered}$$

Thus, the approximate y(0.1) value for the initial value problem $\mathrm{y}\text{'}=2\mathrm{y},\mathrm{y}\left(0\right)=1$is 1.221025.

e) Verify that the global truncation error for Euler's method is O(h2) by comparing the errors in parts (a) and (d).

From subpart (d), the obtained value of${\mathrm{y}}_1$is 1.221025

From subpart(b), the considered function for truncation error isrole="math" localid="1664210283285" $\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$.

Now substitute $0.1$ for x in the equation $\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}$and solve.

$\mathrm{y}(0.1)={\mathrm{e}}^{2(0.1)}={\mathrm{e}}^{0.2}=1.2214$

The actual error is calculated by the equation as follows:

actual error$=y(0.1)-y_1$(4)

Substitute 1.2214 for $\mathrm{y}(0.1)$and 1.22 for $y_1$in equation (4).

$$\begin{gathered} Actualerror=1.2214-1.221025 \\ Actualerror=0.000375 \end{gathered}$$

Thus, the actual error in subpart (d) when step size is halved (h = 0.05) is 0.000375. The actual error when the step size (h=0.1) is 0.00168. The actual error when h = 0.05 is approximately half the actual error when h = 0.1.