3.1: Euler's Method

If an initial value problem

$$\label{eq:3.1.1} y'=f(x,y),\quad y(x_0)=y_0$$

cannot be solved analytically, it is necessary to resort to numerical methods to obtain useful approximations to a solution of Equation $\ref{eq:3.1.1}$. we will consider such methods in this chapter.

We’re interested in computing approximate values of the solution of Equation $\ref{eq:3.1.1}$ at equally spaced points $x_0$, $x_1$, …, $x_n=b$ in an interval $[x_0,b]$. Thus,

$$x_i=x_0+ih,\quad i=0,1, \dots,n, \nonumber$$

where

$$h={b-x_0\over n}.\nonumber$$

we will denote the approximate values of the solution at these points by $y_0$, $y_1$, …, $y_n$; thus, $y_i$ is an approximation to $y(x_i)$. we will call

$$e_i=y(x_i)-y_i \nonumber$$

the error at the $i$th step. Because of the initial condition $y(x_0)=y_0$, we will always have $e_0=0$. However, in general $e_i\ne0$ if $i>0$.

We encounter two sources of error in applying a numerical method to solve an initial value problem:

• The formulas defining the method are based on some sort of approximation. Errors due to the inaccuracy of the approximation are called truncation errors.
• Computers do arithmetic with a fixed number of digits, and therefore make errors in evaluating the formulas defining the numerical methods. Errors due to the computer’s inability to do exact arithmetic are called roundoff errors.

Since a careful analysis of roundoff error is beyond the scope of this book, we will consider only truncation errors.

Euler’s Method

The simplest numerical method for solving Equation $\ref{eq:3.1.1}$ is Euler’s method. This method is so crude that it is seldom used in practice; however, its simplicity makes it useful for illustrative purposes. Euler’s method is based on the assumption that the tangent line to the integral curve of Equation $\ref{eq:3.1.1}$ at $(x_i,y(x_i))$ approximates the integral curve over the interval $[x_i,x_{i+1}]$. Since the slope of the integral curve of Equation $\ref{eq:3.1.1}$ at $(x_i,y(x_i))$ is $y'(x_i)=f(x_i,y(x_i))$, the equation of the tangent line to the integral curve at $(x_i,y(x_i))$ is

$$\label{eq:3.1.2} y=y(x_i)+f(x_i,y(x_i))(x-x_i).$$

Setting $x=x_{i+1}=x_i+h$ in Equation $\ref{eq:3.1.2}$ yields

$$\label{eq:3.1.3} y_{i+1}=y(x_i)+hf(x_i,y(x_i))$$

as an approximation to $y(x_{i+1})$. Since $y(x_0)=y_0$ is known, we can use Equation $\ref{eq:3.1.3}$ with $i=0$ to compute

$$y_1=y_0+hf(x_0,y_0).\nonumber$$

However, setting $i=1$ in Equation $\ref{eq:3.1.3}$ yields

$$y_2=y(x_1)+hf(x_1,y(x_1)),\nonumber$$

which isn’t useful, since we don’t know $y(x_1)$. Therefore we replace $y(x_1)$ by its approximate value $y_1$ and redefine

$$y_2=y_1+hf(x_1,y_1).\nonumber$$

Having computed $y_2$, we can compute

$$y_3=y_2+hf(x_2,y_2).\nonumber$$

In general, Euler’s method starts with the known value $y(x_0)=y_0$ and computes $y_1$, $y_2$, …, $y_n$ successively by with the formula

$$\label{eq:3.1.4} y_{i+1}=y_i+hf(x_i,y_i),\quad 0\le i\le n-1.$$

The next example illustrates the computational procedure indicated in Euler’s method.

Example 3.1.1

Use Euler’s method with $h=0.1$ to find approximate values for the solution of the initial value problem

$$\label{eq:3.1.5} y'+2y=x^3e^{-2x},\quad y(0)=1$$

at $x=0.1,0.2,0.3$.

Solution

We rewrite Equation $\ref{eq:3.1.5}$ as

$$y'=-2y+x^3e^{-2x},\quad y(0)=1, \nonumber$$

which is of the form Equation $\ref{eq:3.1.1}$, with

$$f(x,y)=-2y+x^3e^{-2x}, \, x_0=0, \, \text{and} \, y_0=1. \nonumber$$

Euler’s method yields

$$\begin{align*} y_1 &= y_0+hf(x_0,y_0) \\[4pt] &= 1+(0.1)f(0,1)=1+(0.1)(-2)=0.8,\\[4pt] y_2 & = y_1+hf(x_1,y_1)\\[4pt] & = 0.8+(0.1)f(0.1,0.8)=0.8+(0.1)\left(-2(0.8)+(0.1)^3e^{-0.2}\right)= 0.640081873,\\[4pt] y_3 & = y_2+hf(x_2,y_2)\\[4pt] & = 0.640081873+(0.1)\left(-2(0.640081873)+(0.2)^3e^{-0.4}\right)= 0.512601754. \end{align*}$$

We’ve written the details of these computations to ensure that you understand the procedure. However, in the rest of the examples as well as the exercises in this chapter, we will assume that you can use a programmable calculator or a computer to carry out the necessary computations.

Examples Illustrating The Error in Euler’s Method

Example 3.1.2

Use Euler’s method with step sizes $h=0.1$, $h=0.05$, and $h=0.025$ to find approximate values of the solution of the initial value problem

$$y'+2y=x^3e^{-2x},\quad y(0)=1\nonumber$$

at $x=0$, $0.1$, $0.2$, $0.3$, …, $1.0$. Compare these approximate values with the values of the exact solution

$$\label{eq:3.1.6} y={e^{-2x}\over4}(x^4+4),$$

which can be obtained by the method of Section 2.1. (Verify.)

Table 3.1.1 shows the values of the exact solution Equation $\ref{eq:3.1.6}$ at the specified points, and the approximate values of the solution at these points obtained by Euler’s method with step sizes $h=0.1$, $h=0.05$, and $h=0.025$. In examining this table, keep in mind that the approximate values in the column corresponding to $h=0.05$ are actually the results of 20 steps with Euler’s method. We haven’t listed the estimates of the solution obtained for $x=0.05$, $0.15$, …, since there’s nothing to compare them with in the column corresponding to $h=0.1$. Similarly, the approximate values in the column corresponding to $h=0.025$ are actually the results of 40 steps with Euler’s method.

Table 3.1.1 : Numerical solution of $y'+2y=x^3e^{-2x},\ y(0)=1$, by Euler’s method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$	Exact
0.0	1.000000000	1.000000000	1.000000000	1.000000000
0.1	0.800000000	0.810005655	0.814518349	0.818751221
0.2	0.640081873	0.656266437	0.663635953	0.670588174
0.3	0.512601754	0.532290981	0.541339495	0.549922980
0.4	0.411563195	0.432887056	0.442774766	0.452204669
0.5	0.332126261	0.353785015	0.363915597	0.373627557
0.6	0.270299502	0.291404256	0.301359885	0.310952904
0.7	0.222745397	0.242707257	0.252202935	0.261398947
0.8	0.186654593	0.205105754	0.213956311	0.222570721
0.9	0.159660776	0.176396883	0.184492463	0.192412038
1.0	0.139778910	0.154715925	0.162003293	0.169169104

You can see from Table 3.1.1 that decreasing the step size improves the accuracy of Euler’s method. For example,

$$y_{exact}(1)-y_{approx}(1)\approx \left\{\begin{array}{l} 0.0293 \text{with} h=0.1,\\[4pt] 0.0144\mbox{ with }h=0.05,\\[4pt] 0.0071\mbox{ with }h=0.025. \end{array}\right.\nonumber$$

Based on this scanty evidence, you might guess that the error in approximating the exact solution at a fixed value of $x$ by Euler’s method is roughly halved when the step size is halved. You can find more evidence to support this conjecture by examining Table 3.1.2 , which lists the approximate values of $y_{exact}-y_{approx}$ at $x=0.1$, $0.2$, …, $1.0$.

Table 3.1.2 : Errors in approximate solutions of $y'+2y=x^3e^{-2x},\ y(0)=1$, obtained by Euler’s method.

$x$	$h=0.1$	$h=0.05$	$h=0.0.25$
0.1	0.0187	0.0087	0.0042
0.2	0.0305	0.0143	0.0069
0.3	0.0373	0.0176	0.0085
0.4	0.0406	0.0193	0.0094
0.5	0.0415	0.0198	0.0097
0.6	0.0406	0.0195	0.0095
0.7	0.0386	0.0186	0.0091
0.8	0.0359	0.0174	0.0086
0.9	0.0327	0.0160	0.0079
1.0	0.0293	0.0144	0.0071

Example 3.1.3

Tables 3.1.2 and 3.1.4 show analogous results for the nonlinear initial value problem

$$\label{eq:3.1.7} y'=-2y^2+xy+x^2,\ y(0)=1,$$

except in this case we cannot solve Equation $\ref{eq:3.1.7}$ exactly. The results in the “Exact” column were obtained by using a more accurate numerical method known as the Runge-Kutta method with a small step size. They are exact to eight decimal places.

Table 3.1.3 : Numerical solution of $y'=-2y^2+xy+x^2,\ y(0)=1$, by Euler’s method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$	Exact
0.0	1.000000000	1.000000000	1.000000000	1.000000000
0.1	0.800000000	0.821375000	0.829977007	0.837584494
0.2	0.681000000	0.707795377	0.719226253	0.729641890
0.3	0.605867800	0.633776590	0.646115227	0.657580377
0.4	0.559628676	0.587454526	0.600045701	0.611901791
0.5	0.535376972	0.562906169	0.575556391	0.587575491
0.6	0.529820120	0.557143535	0.569824171	0.581942225
0.7	0.541467455	0.568716935	0.581435423	0.593629526
0.8	0.569732776	0.596951988	0.609684903	0.621907458
0.9	0.614392311	0.641457729	0.654110862	0.666250842
1.0	0.675192037	0.701764495	0.714151626	0.726015790

Since we think it is important in evaluating the accuracy of the numerical methods that we will be studying in this chapter, we often include a column listing values of the exact solution of the initial value problem, even if the directions in the example or exercise don’t specifically call for it. If quotation marks are included in the heading, the values were obtained by applying the Runge-Kutta method in a way that’s explained in Section 3.3. If quotation marks are not included, the values were obtained from a known formula for the solution. In either case, the values are exact to eight places to the right of the decimal point.

Table 3.1.4 : Errors in approximate solutions of $y'=-2y^2+xy+x^2,\ y(0)=1$, obtained by Euler’s method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$
0.1	0.0376	0.0162	0.0076
0.2	0.0486	0.0218	0.0104
0.3	0.0517	0.0238	0.0115
0.4	0.0523	0.0244	0.0119
0.5	0.0522	0.0247	0.0121
0.6	0.0521	0.0248	0.0121
0.7	0.0522	0.0249	0.0122
0.8	0.0522	0.0250	0.0122
0.9	0.0519	0.0248	0.0121
1.0	0.0508	0.0243	0.0119

Truncation Error in Euler’s Method

Consistent with the results indicated in Tables 3.1.1 - 3.1.4 , we will now show that under reasonable assumptions on $f$ there’s a constant $K$ such that the error in approximating the solution of the initial value problem

$$y'=f(x,y),\quad y(x_0)=y_0,\nonumber$$

at a given point $b>x_0$ by Euler’s method with step size $h=(b-x_0)/n$ satisfies the inequality

$$|y(b)-y_n|\le Kh,\nonumber$$

where $K$ is a constant independent of $n$.

There are two sources of error (not counting roundoff) in Euler’s method:

1. The error committed in approximating the integral curve by the tangent line Equation $\ref{eq:3.1.2}$ over the interval $[x_i,x_{i+1}]$.
2. The error committed in replacing $y(x_i)$ by $y_i$ in Equation $\ref{eq:3.1.2}$ and using Equation $\ref{eq:3.1.4}$ rather than Equation $\ref{eq:3.1.2}$ to compute $y_{i+1}$.

Euler’s method assumes that $y_{i+1}$ defined in Equation $\ref{eq:3.1.2}$ is an approximation to $y(x_{i+1})$. We call the error in this approximation the local truncation error at the $i$th step, and denote it by $T_i$; thus,

$$\label{eq:3.1.8} T_i=y(x_{i+1})-y(x_i)-hf(x_i,y(x_i)).$$

we will now use Taylor’s theorem to estimate $T_i$, assuming for simplicity that $f$, $f_x$, and $f_y$ are continuous and bounded for all $(x,y)$. Then $y''$ exists and is bounded on $[x_0,b]$. To see this, we differentiate

$$y'(x)=f(x,y(x))\nonumber$$

to obtain

$$\begin{aligned} y''(x) & = & f_x(x,y(x))+f_y(x,y(x))y'(x)\\[4pt] & = & f_x(x,y(x))+f_y(x,y(x))f(x,y(x)).\end{aligned}\nonumber$$

Since we assumed that $f$, $f_x$ and $f_y$ are bounded, there’s a constant $M$ such that

$$|f_{x}(x,y(x))+f_{y}(x,y(x))y'(x)|\leq M\quad x_{0}<x<b\nonumber$$

which implies that

$$\label{eq:3.1.9} |y''(x)|\leq M,\quad x_{0}<x<b$$

Since $x_{i+1}=x_i+h$, Taylor’s theorem implies that

$$y(x_{i+1})=y(x_i)+hy'(x_i)+{h^2\over2}y''(\tilde x_i),\nonumber$$

where $\tilde x_i$ is some number between $x_i$ and $x_{i+1}$. Since $y'(x_i)=f(x_i,y(x_i))$ this can be written as

$$y(x_{i+1})=y(x_i)+hf(x_i,y(x_i))+{h^2\over2}y''(\tilde x_i), \nonumber$$

or, equivalently,

$$y(x_{i+1})-y(x_i)-hf(x_i,y(x_i))={h^2\over2}y''(\tilde x_i). \nonumber$$

Comparing this with Equation $\ref{eq:3.1.8}$ shows that

$$T_i={h^2\over2}y''(\tilde x_i). \nonumber$$

Recalling Equation $\ref{eq:3.1.9}$, we can establish the bound

$$\label{eq:3.1.10} |T_i|\le{Mh^2\over2},\quad 1\le i\le n.$$

Although it may be difficult to determine the constant $M$, what is important is that there’s an $M$ such that Equation $\ref{eq:3.1.10}$ holds. We say that the local truncation error of Euler’s method is of order $h^2$, which we write as $O(h^2)$.

Note that the magnitude of the local truncation error in Euler’s method is determined by the second derivative $y''$ of the solution of the initial value problem. Therefore the local truncation error will be larger where $|y''|$ is large, or smaller where $|y''|$ is small.

Since the local truncation error for Euler’s method is $O(h^2)$, it is reasonable to expect that halving $h$ reduces the local truncation error by a factor of 4. This is true, but halving the step size also requires twice as many steps to approximate the solution at a given point. To analyze the overall effect of truncation error in Euler’s method, it is useful to derive an equation relating the errors

$$e_{i+1}=y(x_{i+1})-y_{i+1}\quad \text{and} \quad e_i=y(x_i)-y_i. \nonumber$$

To this end, recall that

$$\label{eq:3.1.11} y(x_{i+1})=y(x_i)+hf(x_i,y(x_i))+T_i$$

and

$$\label{eq:3.1.12} y_{i+1}=y_i+hf(x_i,y_i).$$

Subtracting Equation $\ref{eq:3.1.12}$ from Equation $\ref{eq:3.1.11}$ yields

$$\label{eq:3.1.13} e_{i+1}=e_i+h\left[f(x_i,y(x_i))-f(x_i,y_i)\right]+T_i.$$

The last term on the right is the local truncation error at the $i$th step. The other terms reflect the way errors made at previous steps affect $e_{i+1}$. Since $|T_i|\le Mh^2/2$, we see from Equation $\ref{eq:3.1.13}$ that

$$\label{eq:3.1.14} |e_{i+1}|\le |e_i|+h|f(x_i,y(x_i))-f(x_i,y_i)|+{Mh^2\over2}.$$

Since we assumed that $f_y$ is continuous and bounded, the mean value theorem implies that

$$f(x_i,y(x_i))-f(x_i,y_i)=f_y(x_i,y_i^*)(y(x_i)-y_i)=f_y(x_i,y_i^*)e_i, \nonumber$$

where $y_i^*$ is between $y_i$ and $y(x_i)$. Therefore

$$|f(x_i,y(x_i))-f(x_i,y_i)|\le R|e_i| \nonumber$$

for some constant $R$. From this and Equation $\ref{eq:3.1.14}$,

$$\label{eq:3.1.15} |e_{i+1}|\le (1+Rh)|e_i|+{Mh^2\over2},\quad 0\le i\le n-1.$$

For convenience, let $C=1+Rh$. Since $e_0=y(x_0)-y_0=0$, applying Equation $\ref{eq:3.1.15}$ repeatedly yields

$$\begin{align} |e_1| & \le {Mh^2\over2}\nonumber\\[4pt] |e_2| & \le C|e_1|+{Mh^2\over2}\le(1+C){Mh^2\over2}\nonumber\\[4pt] |e_3| & \le C|e_2|+{Mh^2\over2}\le(1+C+C^2){Mh^2\over2}\nonumber\\[4pt] & \vdots \nonumber \\[4pt]|e_n| & \le C|e_{n-1}|+{Mh^2\over2}\le(1+C+\cdots+C^{n-1}){Mh^2\over2}.\label{eq:3.1.16} \end{align}$$

Recalling the formula for the sum of a geometric series, we see that

$$1+C+\cdots+C^{n-1}={1-C^n\over 1-C}={(1+Rh)^n-1\over Rh} \nonumber$$

(since $C=1+Rh$). From this and Equation $\ref{eq:3.1.16}$,

$$\label{eq:3.1.17} |y(b)-y_n|=|e_n|\le{(1+Rh)^n-1\over R}{Mh\over2}.$$

Since Taylor’s theorem implies that

$$1+Rh<e^{rh}\nonumber$$

(verify),

$$(1+Rh)^{n}<e^{nRh}=e^{R(b-x_{0})}\quad (\text{since }nh=b-x_{0}).\nonumber$$

This and Equation $\ref{eq:3.1.17}$ imply that

$$\label{eq:3.1.18} |y(b)-y_n|\le Kh,$$

with

$$K=M{e^{R(b-x_0)}-1\over2R}.\nonumber$$

Because of Equation $\ref{eq:3.1.18}$ we say that the global truncation error of Euler’s method is of order $h$, which we write as $O(h)$.

Semilinear Equations and Variation of Parameters

An equation that can be written in the form

$$\label{eq:3.1.19} y'+p(x)y=h(x,y)$$

with $p\not\equiv0$ is said to be semilinear. (Of course, Equation $\ref{eq:3.1.19}$ is linear if $h$ is independent of $y$.) One way to apply Euler’s method to an initial value problem

$$\label{eq:3.1.20} y'+p(x)y=h(x,y),\quad y(x_0)=y_0$$

for Equation $\ref{eq:3.1.19}$ is to think of it as

$$y'=f(x,y),\quad y(x_0)=y_0, \nonumber$$

where

$$f(x,y)=-p(x)y+h(x,y). \nonumber$$

However, we can also start by applying variation of parameters to Equation $\ref{eq:3.1.20}$, as in Sections 2.1 and 2.4; thus, we write the solution of Equation $\ref{eq:3.1.20}$ as $y=uy_1$, where $y_1$ is a nontrivial solution of the complementary equation $y'+p(x)y=0$. Then $y=uy_1$ is a solution of Equation $\ref{eq:3.1.20}$ if and only if $u$ is a solution of the initial value problem

$$\label{eq:3.1.21} u'=h(x,uy_1(x))/y_1(x),\quad u(x_0)=y(x_0)/y_1(x_0).$$

We can apply Euler’s method to obtain approximate values $u_0$, $u_1$, …, $u_n$ of this initial value problem, and then take

$$y_i=u_iy_1(x_i) \nonumber$$

as approximate values of the solution of Equation $\ref{eq:3.1.20}$. we will call this procedure the Euler semilinear method.

The next two examples show that the Euler and Euler semilinear methods may yield drastically different results.

Example 3.1.2

In Example 3.1.7 we had to leave the solution of the initial value problem

$$\label{eq:3.1.22} y'-2xy=1,\quad y(0)=3$$

in the form

$$\label{eq:3.1.23} y=e^{x^2}\left(3 +\int^x_0 e^{-t^2}dt\right)$$

because it was impossible to evaluate this integral exactly in terms of elementary functions. Use step sizes $h=0.2$, $h=0.1$, and $h=0.05$ to find approximate values of the solution of Equation $\ref{eq:3.1.22}$ at $x=0$, $0.2$, $0.4$, $0.6$, …, $2.0$ by (a) Euler’s method; (b) the Euler semilinear method.

Solution a

Rewriting Equation $\ref{eq:3.1.22}$ as

$$\label{eq:3.1.24} y'=1+2xy,\quad y(0)=3$$

and applying Euler’s method with $f(x,y)=1+2xy$ yields the results shown in Table 3.1.5 . Because of the large differences between the estimates obtained for the three values of $h$, it would be clear that these results are useless even if the “exact” values were not included in the table.

Table 3.1.5 : Numerical solution of $y'-2xy=1,\ y(0)=3$, with Euler’s method.

$x$	$h=0.2$	$h=0.1$	$h=0.05$	Exact
0.0	3.000000000	3.000000000	3.000000000	3.000000000
0.2	3.200000000	3.262000000	3.294348537	3.327851973
0.4	3.656000000	3.802028800	3.881421103	3.966059348
0.6	4.440960000	4.726810214	4.888870783	5.067039535
0.8	5.706790400	6.249191282	6.570796235	6.936700945
1.0	7.732963328	8.771893026	9.419105620	10.184923955
1.2	11.026148659	13.064051391	14.405772067	16.067111677
1.4	16.518700016	20.637273893	23.522935872	27.289392347
1.6	25.969172024	34.570423758	41.033441257	50.000377775
1.8	42.789442120	61.382165543	76.491018246	98.982969504
2.0	73.797840446	115.440048291	152.363866569	211.954462214

It’s easy to see why Euler’s method yields such poor results. Recall that the constant $M$ in Equation $\ref{eq:3.1.10}$ – which plays an important role in determining the local truncation error in Euler’s method – must be an upper bound for the values of the second derivative $y''$ of the solution of the initial value problem Equation $\ref{eq:3.1.22}$ on $(0,2)$. The problem is that $y''$ assumes very large values on this interval. To see this, we differentiate Equation $\ref{eq:3.1.24}$ to obtain

$$y''(x)=2y(x)+2xy'(x)=2y(x)+2x(1+2xy(x))=2(1+2x^2)y(x)+2x, \nonumber$$

where the second equality follows again from Equation $\ref{eq:3.1.24}$. Since Equation $\ref{eq:3.1.23}$ implies that $y(x)>3e^{x^2}$ if $x>0$,

$$y''(x)>6(1+2x^2)e^{x^2}+2x,\quad x>0. \nonumber$$

For example, letting $x=2$ shows that $y''(2)>2952$.

Solution b

Since $y_1=e^{x^2}$ is a solution of the complementary equation $y'-2xy=0$, we can apply the Euler semilinear method to Equation $\ref{eq:3.1.22}$, with

$$y=ue^{x^2}\quad \text{and} \quad u'=e^{-x^2},\quad u(0)=3. \nonumber$$

The results listed in Table 3.1.6 are clearly better than those obtained by Euler’s method.

Table 3.1.6 : Numerical solution of $y'-2xy=1,\ y(0)=3$, by the Euler semilinear method.

$x$	$h=0.2$	$h=0.1$	$h=0.05$	Exact
0.0	3.000000000	3.000000000	3.000000000	3.000000000
0.2	3.330594477	3.329558853	3.328788889	3.327851973
0.4	3.980734157	3.974067628	3.970230415	3.966059348
0.6	5.106360231	5.087705244	5.077622723	5.067039535
0.8	7.021003417	6.980190891	6.958779586	6.936700945
1.0	10.350076600	10.269170824	10.227464299	10.184923955
1.2	16.381180092	16.226146390	16.147129067	16.067111677
1.4	27.890003380	27.592026085	27.441292235	27.289392347
1.6	51.183323262	50.594503863	50.298106659	50.000377775
1.8	101.424397595	100.206659076	99.595562766	98.982969504
2.0	217.301032800	214.631041938	213.293582978	211.954462214

We cannot give a general procedure for determining in advance whether Euler’s method or the semilinear Euler method will produce better results for a given semilinear initial value problem Equation $\ref{eq:3.1.19}$. As a rule of thumb, the Euler semilinear method will yield better results than Euler’s method if $|u''|$ is small on $[x_0,b]$, while Euler’s method yields better results if $|u''|$ is large on $[x_0,b]$. In many cases the results obtained by the two methods don’t differ appreciably. However, we propose the an intuitive way to decide which is the better method: Try both methods with multiple step sizes, as we did in Example [example:3.1.4}, and accept the results obtained by the method for which the approximations change less as the step size decreases.

Example 3.1.5

Applying Euler’s method with step sizes $h=0.1$, $h=0.05$, and $h=0.025$ to the initial value problem

$$\label{eq:3.1.25} y'-2y={x\over1+y^2},\quad y(1)=7$$ on $[1,2]$ yields the results in Table 3.1.7 .

Table 3.1.7 : Numerical solution of $y'-2y=x/(1+y^2),\ y(1)=7$, by Euler’s method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$	Exact
1.0	7.000000000	7.000000000	7.000000000	7.000000000
1.1	8.402000000	8.471970569	8.510493955	8.551744786
1.2	10.083936450	10.252570169	10.346014101	10.446546230
1.3	12.101892354	12.406719381	12.576720827	12.760480158
1.4	14.523152445	15.012952416	15.287872104	15.586440425
1.5	17.428443554	18.166277405	18.583079406	19.037865752
1.6	20.914624471	21.981638487	22.588266217	23.253292359
1.7	25.097914310	26.598105180	27.456479695	28.401914416
1.8	30.117766627	32.183941340	33.373738944	34.690375086
1.9	36.141518172	38.942738252	40.566143158	42.371060528
2.0	43.369967155	47.120835251	49.308511126	51.752229656

Applying the Euler semilinear method with

$$y=ue^{2x}\quad \text{and} \quad u'={xe^{-2x}\over1+u^2e^{4x}},\quad u(1)=7e^{-2}\nonumber$$

yields the results in Table 3.1.8 . Since the latter are clearly less dependent on step size than the former, we conclude that the Euler semilinear method is better than Euler’s method for Equation $\ref{eq:3.1.25}$. This conclusion is supported by comparing the approximate results obtained by the two methods with the “exact” values of the solution.

Table 3.1.8 : Numerical solution of $y'-2y=x/(1+y^2),\ y(1)=7$, by the Euler semilinear method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$	Exact
1.0	7.000000000	7.000000000	7.000000000	7.000000000
1.1	8.552262113	8.551993978	8.551867007	8.551744786
1.2	10.447568674	10.447038547	10.446787646	10.446546230
1.3	12.762019799	12.761221313	12.760843543	12.760480158
1.4	15.588535141	15.587448600	15.586934680	15.586440425
1.5	19.040580614	19.039172241	19.038506211	19.037865752
1.6	23.256721636	23.254942517	23.254101253	23.253292359
1.7	28.406184597	28.403969107	28.402921581	28.401914416
1.8	34.695649222	34.692912768	34.691618979	34.690375086
1.9	42.377544138	42.374180090	42.372589624	42.371060528
2.0	51.760178446	51.756054133	51.754104262	51.752229656

Example 3.1.6

Applying Euler’s method with step sizes $h=0.1$, $h=0.05$, and $h=0.025$ to the initial value problem

$$\label{eq:3.1.26} y'+3x^2y=1+y^2,\quad y(2)=2$$

on $[2,3]$ yields the results in Table 3.1.9 . Applying the Euler semilinear method with

$$y=ue^{-x^3}\quad \text{and} \quad u'=e^{x^3}(1+u^2e^{-2x^3}),\quad u(2)=2e^8 \nonumber$$

yields the results in Table 3.1.10 . Noting the close agreement among the three columns of Table 3.1.9 (at least for larger values of $x$) and the lack of any such agreement among the columns of Table 3.1.10 , we conclude that Euler’s method is better than the Euler semilinear method for Equation $\ref{eq:3.1.26}$. Comparing the results with the exact values supports this conclusion.

Table 3.1.9 Numerical solution of $y'+3x^2y=1+y^2,\quad y(2)=2$, by Euler’s method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$	Exact
2.0	2.000000000	2.000000000	2.000000000	2.000000000
2.1	0.100000000	0.493231250	0.609611171	0.701162906
2.2	0.068700000	0.122879586	0.180113445	0.236986800
2.3	0.069419569	0.070670890	0.083934459	0.103815729
2.4	0.059732621	0.061338956	0.063337561	0.068390786
2.5	0.056871451	0.056002363	0.056249670	0.057281091
2.6	0.050560917	0.051465256	0.051517501	0.051711676
2.7	0.048279018	0.047484716	0.047514202	0.047564141
2.8	0.042925892	0.043967002	0.043989239	0.044014438
2.9	0.042148458	0.040839683	0.040857109	0.040875333
3.0	0.035985548	0.038044692	0.038058536	0.038072838:

Table 3.1.10 : Numerical solution of $y'+3x^2y=1+y^2,\quad y(2)=2$, by the Euler semilinear method.

$x$	$h=0.1$	$h=0.05$	$h=0.025$	$h=.0125$
2.0	2.000000000	2.000000000	2.000000000	2.000000000
2.1	0.708426286	0.702568171	0.701214274	0.701162906
2.2	0.214501852	0.222599468	0.228942240	0.236986800
2.3	0.069861436	0.083620494	0.092852806	0.103815729
2.4	0.032487396	0.047079261	0.056825805	0.068390786
2.5	0.021895559	0.036030018	0.045683801	0.057281091
2.6	0.017332058	0.030750181	0.040189920	0.051711676
2.7	0.014271492	0.026931911	0.036134674	0.047564141
2.8	0.011819555	0.023720670	0.032679767	0.044014438
2.9	0.009776792	0.020925522	0.029636506	0.040875333
3.0	0.008065020	0.018472302	0.026931099	0.038072838

In the next two sections we will study other numerical methods for solving initial value problems, called the improved Euler method, the midpoint method, Heun’s method and the Runge- Kutta method. If the initial value problem is semilinear as in Equation $\ref{eq:3.1.19}$, we also have the option of using variation of parameters and then applying the given numerical method to the initial value problem Equation $\ref{eq:3.1.21}$ for $u$. By analogy with the terminology used here, we will call the resulting procedure the improved Euler semilinear method, the midpoint semilinear method, Heun’s semilinear method or the Runge- Kutta semilinear method, as the case may be.