Initial Value Problems - Numerical Methods in Engineering with MATLAB - page 283

functionF=fex7 8(x,y)

% Diff. eqs. used in Example 7.8
F = zeros(1,2);
F(1) = y(2);
F(2) = -9.80665...
+ 65.351e-3 * y(2)ˆ2 * exp(-10.53e-5 * y(1));

_

The commandsforperforming the integration and displaying the results are
shown below. Wespecifiedaper-step error tolerance of 10 − 2 in runKut5 .Consid-
ering the magnitudeof y ,this shouldbeenoughforfive decimal point accuracyinthe
solution.

_ >> [x,y] = runKut5(@fex7 8,0,[9000 0],10,0.5,1.0e-2);

>> printSol(x,y,1)

Execution of the commands resultedinthe following output:

>> x y1 y2

0.0000e+000 9.0000e+003 0.0000e+000

5.0000e-001 8.9988e+003 -4.8043e+000

1.9246e+000 8.9841e+003 -1.4632e+001

3.2080e+000 8.9627e+003 -1.8111e+001

4.5031e+000 8.9384e+003 -1.9195e+001

5.9732e+000 8.9099e+003 -1.9501e+001

7.7786e+000 8.8746e+003 -1.9549e+001

1.0000e+001 8.8312e+003 -1.9519e+001

The first integration stepwascarried out with the prescribed trial value h =

0 5 s. Apparently the errorwas well within the tolerance,sothat the stepwas accepted.

Subsequent step sizes, determined fromEq. (7.24), wereconsiderably larger.
Inspecting the output, we see that at t

.

= 10 s the object is moving with the speed

=− = . =

v y 19 52 m/s at an elevation of y 8831 m.

EXAMPLE 7.9
Integrate the moderately stiff problem

19
4

y =− 10 y y (0) y − y (0) =− 9 = 0

=

from x 0to10 with the adaptive Runge-Kuttamethod and plot the results (this

problemalso appeared in Example 7.7).