Page 275 - Applied Numerical Methods Using MATLAB

264 ORDINARY DIFFERENTIAL EQUATIONS
clue to the basic concept of numerical solution for a differential equation simply
and clearly. Let’s consider a first-order differential equation:

y (t) + ay(t) = r with y(0) = y 0 (6.1.1)

It has the following form of analytical solution:
r −at r

y(t) = y 0 − e + (6.1.2)
a a
which can be obtained by using a conventional method or the Laplace trans-
form technique [K-1, Chapter 5]. However, such a nice analytical solution does
not exist for every differential equation; even if it exists, it is not easy to
find even by using a computer equipped with the capability of symbolic com-
putation. That is why we should study the numerical solutions to differential
equations.
Then, how do we translate the differential equation into a form that can eas-
ily be handled by computer? First of all, we have to replace the derivative

y (t) = dy/dt in the differential equation by a numerical derivative (introduced in
Chapter 5), where the step-size h is determined based on the accuracy require-
ments and the computation time constraints. Euler’s method approximates the
derivative in Eq. (6.1.1) with Eq. (5.1.2) as
y(t + h) − y(t)
+ ay(t) = r
h
y(t + h) = (1 − ah)y(t) + hr with y(0) = y 0 (6.1.3)

and solves this difference equation step-by-step with increasing t by h each time
from t = 0.

y(h) = (1 − ah)y(0) + hr = (1 − ah)y 0 + hr
2
y(2h) = (1 − ah)y(h) + hr = (1 − ah) y 0 + (1 − ah)hr + hr (6.1.4)
3 2 m
y(3h) = (1 − ah)y(2h) + hr = (1 − ah) y 0 + (1 − ah) hr
m=0
. ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. ... .. ..
This is a numeric sequence {y(kh)}, which we call a numerical solution of
Eq. (6.1.1).
To be specific, let the parameters and the initial value of Eq. (6.1.1) be a = 1,
r = 1, and y 0 = 0. Then, the analytical solution (6.1.2) becomes

y(t) = 1 − e −at (6.1.5)