Higher Engineering Mathematics

NUMERICAL METHODS FOR FIRST ORDER DIFFERENTIAL EQUATIONS 463

I

A graph of the solution of

dy

dx

+y= 2 x, with initial

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

0 0.2 0.4 0.6 0.8 1.0 x

0.5

1.0

y

Figure 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

atx=0,y=2, for the rangex=0(0.1)0.5.

Draw the graph of the solution.

(b) By an analytical method (using the inte-

grating factor method of Chapter 48), the

solution of the above differential equation is

given byy=x+ 1 +ex.

Determine the percentage error atx= 0 .3.

(a)

dy

dx

=y′=y−x.

If initiallyx 0 =0 andy 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

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

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

For line 4, wherex 0 = 0 .3:

y 1 =y 0 +h(y′) 0

= 2. 41 +(0.1)(2.21)=2.631

and (y′) 0 =y 0 −x 0

= 2. 631 − 0. 3 =2.331

For line 5, wherex 0 = 0 .4:

y 1 =y 0 +h(y′) 0

= 2. 631 +(0.1)(2.331)=2.8641

and (y′) 0 =y 0 −x 0

= 2. 8641 − 0. 4 =2.4641

For line 6, wherex 0 = 0 .5:

y 1 =y 0 +h(y′) 0

= 2. 8641 +(0.1)(2.4641)=3.11051

A graph of the solution of

dy

dx

=y−xwithx=0,

y=2 is shown in Fig. 49.8.

(b) If the solution of the differential equation

dy

dx

=y−xis given byy=x+ 1 +ex, then when

x= 0 .3,y= 0. 3 + 1 +e^0.^3 = 2 .649859.

By Euler’s method, whenx= 0 .3 (i.e. line 4 in

Table 49.3),y= 2 .631.

Percentage error

=

(

actual−estimated

actual

)

×100%

=

(

2. 649859 − 2. 631

2. 649859

)

×100%

=0.712%