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