Higher Engineering Mathematics

NUMERICAL METHODS FOR FIRST ORDER DIFFERENTIAL EQUATIONS 469

I

4. Obtain a numerical solution of the differential
   equation

1

x

dy

dx

+ 2 y= 1

using the Euler-Cauchy method in the range

x=0(0.2)1.0, given the initial conditions that

x=0 wheny= 1.

[see Table 49.14]

Table 49.14

xy y′

01 0

0.2 0.99 −0.196

0.4 0.958336 −0.3666688

0.6 0.875468851 −0.450562623

0.8 0.784755575 −0.45560892

1.0 0.700467925

49.5 The Runge-Kutta method

The Runge-Kutta method for solving first order dif-
ferential equations is widely used and provides a
high degree of accuracy. Again, as with the two
previous methods, the Runge-Kutta method is a
step-by-step process where results are tabulated for
a range of values ofx. Although several intermediate
calculations are needed at each stage, the method is
fairly straightforward.
The 7 step procedure for the Runge-Kutta
method, without proof, is as follows:

To solve the differential equation

dy

dx

=f(x,y)given

the initial conditiony=y 0 atx=x 0 for a range of
values ofx=x 0 (h)xn:

1. Identify x 0 , y 0 and h, and values of x 1 ,x 2 ,
   x 3 ,....


2. Evaluatek 1 =f(xn,yn) starting withn= 0


3. Evaluatek 2 =f

(

xn+

h

2

,yn+

h

2

k 1

)

4. Evaluatek 3 =f

(

xn+

h

2

,yn+

h

2

k 2

)

5. Evaluatek 4 =f(xn+h,yn+hk 3 )
   6. Use the values determined from steps 2 to 5 to
   evaluate:

yn+ 1 =yn+

h

6

{k 1 + 2 k 2 + 2 k 3 +k 4 }

7. Repeat steps 2 to 6 forn=1, 2, 3,...

Thus, step 1 is given, and steps 2 to 5 are intermediate

steps leading to step 6. It is usually most convenient

to construct a table of values.

The Runge-Kutta method is demonstrated in the

following worked problems.

Problem 7. Use the Runge-Kutta method to

solve the differential equation:

dy

dx

=y−x

in the range 0(0.1)0.5, given the initial condi-

tions that atx=0,y= 2.

Using the above procedure:

1.x 0 =0,y 0 =2 and sinceh=0.1, and the range

is fromx=0tox=0.5, thenx 1 = 0 .1, x 2 = 0 .2,

x 3 = 0 .3,x 4 = 0 .4, andx 5 = 0. 5

Letn=0 to determiney 1 :

2.k 1 =f(x 0 ,y 0 )=f(0, 2);

since

dy

dx

=y−x, f(0, 2)= 2 − 0 = 2

3.k 2 =f

(

x 0 +

h

2

,y 0 +

h

2

k 1

)

=f

(

0 +

0. 1

2

,2+

0. 1

2

(2)

)

=f(0.05, 2.1)= 2. 1 − 0. 05 =2.05

4.k 3 =f

(

x 0 +

h

2

,y 0 +

h

2

k 2

)

=f

(

0 +

0. 1

2

,2+

0. 1

2

(2.05)

)

=f(0.05, 2.1025)

= 2. 1025 − 0. 05 =2.0525

5.k 4 =f(x 0 +h,y 0 +hk 3 )

=f(0+ 0 .1, 2+ 0 .1(2.0525))

=f(0.1, 2.20525)

= 2. 20525 − 0. 1 =2.10525