Numerical methods for first order differential equations 465
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, when x= 0 .3 (i.e. line 4 in
Table 49.3),y= 2 .631.
y
x
3.0
2.5
2.0
0 0.1 0.2 0.3 0.4 0.5
Figure 49.8
Percentage error
=
(
actual−estimated
actual
)
×100%
=
(
2. 649859 − 2. 631
2. 649859
)
×100%
=0.712%
Euler’s method of numerical solution of differential
equations is simple, but approximate. The method is
most useful when the intervalhis small.
Now try the following exercise
Exercise 184 Further problems on Euler’s
method
- Use Euler’s method to obtain a numer-
ical solution of the differential equation
dy
dx
= 3 −
y
x
, with the initial conditions that
x=1wheny=2, for the rangex= 1 .0to
x= 1 .5 with intervals of 0.1. Draw the graph
of the solution in this range. [see Table 49.4]
Table 49.4
x y
1.0 2
1.1 2.1
1.2 2.209091
1.3 2.325000
1.4 2.446154
1.5 2.571429
- Obtain a numerical solution of the differen-
tial equation
1
x
dy
dx
+ 2 y=1, given the initial
conditions thatx=0wheny=1, in the range
x= 0 ( 0. 2 ) 1. 0 [see Table 49.5]
- (a) The differential equation
dy
dx
+ 1 =−
y
x
has the initial conditions thaty=1at
x=2. Produce a numerical solution of
the differential equation in the range
x= 2. 0 ( 0. 1 ) 2. 5