Higher Engineering Mathematics, Sixth Edition

Numerical methods for first order differential equations 463

This step by step Euler’s method can be continued
and it is easiest to list the results in a table, as shown
in Table 49.1. The results for lines 1 to 3 have been
produced above.

Table 49.1

x 0 y 0 (y′) 0

1. 1 4 2


2. 1.2 4.4 2.2


3. 1.4 4.84 2.36


4. 1.6 5.312 2.488


5. 1.8 5.8096 2.5904


6. 2.0 6.32768

For line 4,wherex 0 = 1 .6:

y 1 =y 0 +h(y′) 0

= 4. 84 +( 0. 2 )( 2. 36 )=5.312

and (y′) 0 = 3 ( 1 + 1. 6 )− 5. 312 =2.488

For line 5,wherex 0 = 1 .8:

y 1 =y 0 +h(y′) 0

= 5. 312 +( 0. 2 )( 2. 488 )=5.8096

and (y′) 0 = 3 ( 1 + 1. 8 )− 5. 8096 =2.5904

For line 6,wherex 0 = 2 .0:

y 1 =y 0 +h(y′) 0

= 5. 8096 +( 0. 2 )( 2. 5904 )

=6.32768

(As the range is 1.0 to 2.0 there is no need to calculate
(y′) 0 in line 6). The particular solution is given by the
value ofyagainstx.

A graph of the solution of

dy

dx

= 3 ( 1 +x)−ywith initial

conditionsx=1andy=4 is shown in Fig. 49.6.
In practice it is probably best to plot the graph as
each calculation is made, which checks that there is a
smooth progression and that no calculation errors have
occurred.

y

x

6.0

5.0

4.0

1.0 1.2 1.4 1.6 1.8 2.0

Figure 49.6

Problem 2. Use Euler’s method to obtain a

numerical solution of the differential equation

dy

dx

+y= 2 x, given the initial conditions that at

x=0,y=1, for the rangex= 0 ( 0. 2 ) 1 .0. Draw the

graph of the solution in this range.

x= 0 ( 0. 2 ) 1 .0 means thatxranges from 0 to 1.0 in equal

intervals of 0.2 (i.e.h= 0 .2 in Euler’s method).

dy

dx

+y= 2 x,

hence

dy

dx

= 2 x−y, i.e.y′= 2 x−y

If initiallyx 0 =0andy 0 =1, then

(y′) 0 = 2 ( 0 )− 1 =− 1.

Hence line 1 in Table 49.2 can be completed with

x= 0 ,y=1andy′( 0 )=−1.

Table 49.2

x 0 y 0 (y′) 0

1. 0 1 − 1


2. 0.2 0.8 −0.4


3. 0.4 0.72 0.08


4. 0.6 0.736 0.464


5. 0.8 0.8288 0.7712


6. 1.0 0.98304