Higher Engineering Mathematics, Sixth Edition

462 Higher Engineering Mathematics

which is a statement calledTaylor’s series.

Ifhis the interval between two new ordinatesy 0 and

y 1 , as shown in Fig. 49.3, and if f(a)=y 0 andy 1 =

f(a+h), then Euler’s method states:

f(a+h)=f(a)+hf′(a)

i.e. y 1 =y 0 +h(y′) 0 (2)

y

y 5 f(x)

P

h

a x

Q

y 0 y 1

0 (a 1 h)

Figure 49.3

The approximation used with Euler’s method is to take

only the first two terms of Taylor’s series shown in

equation (1).

Hence ify 0 ,hand(y′) 0 are known,y 1 , which is an

approximate value for the function atQin Fig. 49.3,

can be calculated.

Euler’s method is demonstrated in the workedproblems

following.

49.3 Worked problems on Euler’s

method

Problem 1. Obtain a numerical solution of the

differential equation

dy

dx

= 3 ( 1 +x)−y

given the initial conditions thatx=1wheny=4,

for the rangex= 1 .0tox= 2 .0 with intervals of 0.2.

Draw the graph of the solution.

dy

dx

=y′= 3 ( 1 +x)−y

Withx 0 =1andy 0 = 4 ,(y′) 0 = 3 ( 1 + 1 )− 4 = 2.

By Euler’s method:

y 1 =y 0 +h(y′) 0 ,from equation( 2 )

Hence y 1 = 4 +( 0. 2 )( 2 )=4.4,sinceh= 0. 2

At pointQin Fig. 49.4,x 1 = 1. 2 ,y 1 = 4. 4

and(y′) 1 = 3 ( 1 +x 1 )−y 1

i.e.(y′) 1 = 3 ( 1 + 1. 2 )− 4. 4 =2.2

If the values ofx, y andy′found for pointQare

regarded as new starting values ofx 0 ,y 0 and(y′) 0 ,the

above process can be repeated and values found for the

pointRshown in Fig. 49.5.

Thus at pointR,

y 1 =y 0 +h(y′) 0 from equation( 2 )

= 4. 4 +( 0. 2 )( 2. 2 )=4.84

Whenx 1 = 1 .4andy 1 = 4 .84,

(y′) 1 = 3 ( 1 + 1. 4 )− 4. 84 =2.36

y

P

h

x

Q

y 0

x 051 x 15 1.2

y 1

0

4

4.4

Figure 49.4

y

P

h

x

Q

R

y 0

x 05 1.2 x 15 1.4

y 1

0 1.0

Figure 49.5