Initial Value Problems - Numerical Methods in Engineering with MATLAB - page 263

Solution Letting y 1 =

y and y 2 = y , we write the equivalent first-order equations as

y 1
y 2

y 2

y = F ( x , y ) = =

− 0 . 1 y 2 − x

which arecodedinthe following function:

_

functionF=fex7 4(x,y)

% Differential. eqs. used in Example 7.4
F = zeros(1,2);
F(1) = y(2); F(2) = -0.1*y(2) - x;

Comparing the function fex7 4 here with fex7 2 in Example 7.2 we note that it
is much simpler to input the differentialequationsfor the Runge-Kuttamethod than
for the Taylor series method. Here are the results of integration:

_

>> [x,y] = runKut4(@fex7 4,0,[0 1],2,0.25);

>> printSol(x,y,1)
x y1 y2

0.0000e+000 0.0000e+000 1.0000e+000

2.5000e-001 2.4431e-001 9.4432e-001

5.0000e-001 4.6713e-001 8.2829e-001

7.5000e-001 6.5355e-001 6.5339e-001

1.0000e+000 7.8904e-001 4.2110e-001

1.2500e+000 8.5943e-001 1.3281e-001

1.5000e+000 8.5090e-001 -2.1009e-001

1.7500e+000 7.4995e-001 -6.0625e-001

2.0000e+000 5.4345e-001 -1.0543e+000

These results are the same asobtainedbythe Taylor seriesmethod inExample 7.2.
This wasexpected,since both methods are of the sameorder.

EXAMPLE 7.5
Use the fourth-order Runge-Kuttamethod to integrate

y = 4 e − x 3 y − y (0) = 1

= = .

from x 0to10 in stepsof h 0 1. Compare the result with the analytical solution

e − x . y =