Page 163 - Numerical Analysis Using MATLAB and Excel

Chapter 5 Differential Equations, State Variables, and State Equations

We can find the value of the constant by making use of the initial condition, i.e., at t = , 0
k
v C = V 0 and (5.4) then becomes
V = 0 + k (5.5)
0
or k = V 0 , and by substitution into (5.4),

v t() = It + V 0 (5.6)
C
This example shows that when a capacitor is charged with a constant current, a linear voltage is pro-
duced across the terminals of the capacitor.

Example 5.2

Find the current i t() through an inductor whose slope at the coordinate ti,( L ) is cos t and the
L
current passes through the point π 2⁄ ,( 1 ) .
i
L
Solution:

We are given that
di
L
------- = cos t (5.7)
dt
By separating the variables we obtain
di = cos tdt (5.8)
L
and integrating both sides we obtain
i t() = sin t + k (5.9)
L
where represents the constants of integration of both sides.
k
We find the value of the constant by making use of the initial condition. For this example,
k
⁄
ω = 1 and thus at ωt = t = π 2 , i = 1 . With these values (5.9) becomes
L
π
1 = sin --- + k (5.10)
2
or k = 0 , and by substitution into (5.9),

i t() = sin t (5.11)
L

5.2 Classification

Differential equations are classified by:
1. Type - Ordinary or Partial

5−2 Numerical Analysis Using MATLAB® and Excel®, Third Edition

Copyright © Orchard Publications