Chapter 7.2, Problem 19E

a.

To determine

To estimate: the value of y (0.4) with each of the following step sizes using Euler’s method.

Answer to Problem 19E

  $y(0.4)=1.4$ .

  $y(0.4)=\mathrm{1.44.}$

  $y(0.4)=1.4641$ .

Explanation of Solution

Given information: Where y is the solution of the initial − value problem y’=y .y (0) =1.

Stem sizes are:

(i) h =0.4 (ii) h =0.2 (iii) h =0.1.

Formula used:

According to Euler’s method:

  $y(x+h)\approx y(x)+h\to 1$

Calculation:

y (0) = 1 (Given)

1. Need to find y (0.4) with h =.04.

$\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0,\frac{dy}{dx}=y\text{'}(0)=y(0)=1.\\ \text{To}\text{get}y(0+0.4)\approx y(0)+(0.4)(1)=1.4\\ y(0.4)=1.4\end{array}$

2. Need to find y (0.4) with h =.02

$\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0,\frac{dy}{dx}=y\text{'}(0)=y(0)=1.\\ \text{To}\text{get}y(0+0.2)\approx y(0)+(0.2)(1)\approx 1+0.2=1.2\\ y(0.2)=1.2\end{array}$

  $\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0.2,\frac{dy}{dx}=y\text{'}(0.2)=y(0.2)=\mathrm{1.2.}\\ \text{To}\text{get}y(0.2+0.2)\approx y(0.2)+(0.2)y(0.2)\approx {1.2}^2=1.44\\ y(0.4)=1.44\end{array}$

3. Need to find y (0.4) with h =.01

$\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0,\frac{dy}{dx}=y\text{'}(0)=y(0)=1.\\ \text{To}\text{get}y(0+0.1)\approx y(0)+(0.1)(1)\approx 1+0.1=1.1\\ y(0.1)=1.1\end{array}$

  $\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0.1,\frac{dy}{dx}=y\text{'}(0.1)=y(0.1)=\mathrm{1.1.}\\ \text{To}\text{get}y(0.1+0.1)\approx y(0.1)+(0.1)y(0.1)\approx {1.1}^2=1.21\\ y(0.2)=1.21\end{array}$

  $\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0.2,\frac{dy}{dx}=y\text{'}(0.2)=y(0.2)=\mathrm{1.21.}\\ \text{To}\text{get}y(0.2+0.1)\approx y(0.2)+(0.1)y(0.2)\approx (1.21)(1.1)=\mathrm{1.331.}\\ y(0.3)=1.331\end{array}$

  $\begin{array}{c}y(x+h)\approx y(x)+h\frac{dy}{dx}.\\ \text{Replace}x=0.3,\frac{dy}{dx}=y\text{'}(0.3)=y(0.3)=\mathrm{1.331.}\\ \text{To}\text{get}y(0.3+0.1)\approx y(0.3)+(0.1)y(0.3)\approx (1.331)(1.1)=\mathrm{1.4641.}\\ y(0.4)=1.4641\end{array}$

b.

To determine

To draw: the graph of the given equation.

Explanation of Solution

Given information: Given equation is $y=e^x,0\le x\le \mathrm{0.4.}$

Calculation:

Euler’s method under estimates in this case, as shown in the graph $y=e^x,0\le x\le \mathrm{0.4.}$ is the graph in purple.

  

c.

To determine

To find: the errors made in part (a) in using Euler’s method to estimate the true value of y ( 0.4), namely $e^{0.4}$ .

Answer to Problem 19E

1. e =0.0918.
2. e =0.0518.
3. e =0.0277.

Explanation of Solution

Given information: y ( 0.4), namely $e^{0.4}$ .

Calculation:

In part (a)

1. Forh =0.4 $\Rightarrow y(0.4)$=1.4.
2. Forh =0.2 $\Rightarrow y(0.4)$=1.44.
3. Forh =0.1 $\Rightarrow y(0.4)$=1.4641.

For using calculator the real value ofy (0.4):

y (0.4)= $e^{0.4}=1.4918$ .

Using the fact the error is the real value minus the calculated value :

4. e= 1.4918-1.4=0.0918.
5. e= 1.4918-1.44=0.0518.
6. e= 1.4918-1.4641=0.0277.
Looking at the errors above, see that as h is halves so is the error.