Higher Engineering Mathematics, Sixth Edition

464 Higher Engineering Mathematics

For line 2,wherex 0 = 0 .2andh= 0 .2:

y 1 =y 0 +h(y′), from equation( 2 )

= 1 +( 0. 2 )(− 1 )=0.8

and (y′) 0 = 2 x 0 −y 0 = 2 ( 0. 2 )− 0. 8 =−0.4

For line 3,wherex 0 = 0 .4:

y 1 =y 0 +h(y′) 0

= 0. 8 +( 0. 2 )(− 0. 4 )=0.72

and (y′) 0 = 2 x 0 −y 0 = 2 ( 0. 4 )− 0. 72 =0.08

For line 4,wherex 0 = 0 .6:

y 1 =y 0 +h(y′) 0

= 0. 72 +( 0. 2 )( 0. 08 )=0.736

and (y′) 0 = 2 x 0 −y 0 = 2 ( 0. 6 )− 0. 736 =0.464

For line 5,wherex 0 = 0 .8:

y 1 =y 0 +h(y′) 0

= 0. 736 +( 0. 2 )( 0. 464 )=0.8288

and(y′) 0 = 2 x 0 −y 0 = 2 ( 0. 8 )− 0. 8288 =0.7712

For line 6,wherex 0 = 1 .0:

y 1 =y 0 +h(y′) 0

= 0. 8288 +( 0. 2 )( 0. 7712 )=0.98304

As the range is 0 to 1.0,(y′) 0 in line 6 is not needed.

y

x

1.0

0.5

0 0.2 0.4 0.6 0.8 1.0

Figure 49.7

A graph of the solution of

dy

dx

+y= 2 x, with initial

conditionsx=0andy=1 is shown in Fig. 49.7.

Problem 3.

(a) Obtain a numerical solution, using

Euler’s method, of the differential equation

dy

dx

=y−x, with the initial conditions that at

x=0,y=2, for the rangex= 0 ( 0. 1 ) 0 .5. Draw

the graph of the solution.

(b) By an analytical method (using the integrating

factor method of Chapter 48), the solution of

the above differential equation is given by

y=x+ 1 +ex.

Determine the percentage error atx= 0. 3

(a)

dy

dx

=y′=y−x.

If initiallyx 0 =0andy 0 =2,

then(y′) 0 =y 0 −x 0 = 2 − 0 = 2.

Hence line 1 of Table 49.3 is completed.

For line 2,wherex 0 = 0 .1:

y 1 =y 0 +h(y′) 0 , from equation( 2 ),

= 2 +( 0. 1 )( 2 )=2.2

and(y′) 0 =y 0 −x 0

= 2. 2 − 0. 1 =2.1

For line 3,wherex 0 = 0 .2:

y 1 =y 0 +h(y′) 0

= 2. 2 +( 0. 1 )( 2. 1 )=2.41

and(y′) 0 =y 0 −x 0 = 2. 41 − 0. 2 =2.21

Table 49.3

x 0 y 0 (y′) 0

1. 0 2 2


2. 0.1 2.2 2.1


3. 0.2 2.41 2.21


4. 0.3 2.631 2.331


5. 0.4 2.8641 2.4641


6. 0.5 3.11051