Page 183 - Numerical Analysis Using MATLAB and Excel

Chapter 5 Differential Equations, State Variables, and State Equations

e t – 0
du e – t – 12 12e t – 3t
2
-------- = --------------------------------- = --------------- = – 6e (5.79)
dt – 2e – 4t – 2e – 4t
Now, integration of (5.78) and (5.79) and substitution into (5.75) yields

3t
∫
t
∫
t
3t
u = 6e t = 6e + k 1 u = – 6 e d = – 2e + k 2 (5.80)
d
t
1
2
3t
t –
t –
t
y = u e + u e – 3t = ( 6e + k ) 1 e + – ( 2e + k ) 2 e – 3t
2
1
(5.81)
t –
t –
= 6 + k e – 2 + k e – 3t = k e + k e – 3t + 4
1
2
1
2
We observe that the last expression in (5.81) is the same as (5.73) of part (a).
Check with MATLAB:
y=dsolve('D2y+4*Dy+3*y=12')
y =
(4*exp(t)+C1*exp(-3*t)*exp(t)+C2)/exp(t)
Example 5.13
Find the total solution of
2
d y 4y = tan 2t (5.82)
-------- +
dt 2
in terms of the constants k 1 and k 2 by any method.
Solution:

This ODE cannot be solved by the method of undetermined coefficients; therefore, we will use
the method of variation of parameters.

2
The characteristic equation is s + 4 = 0 from which s = ± j2 and thus the natural response is
y N = k e j2t + k e j – 2t (5.83)
2
1
We let
y = cos 2t and y = sin 2t (5.84)
2
1
Then, by (5.68) the solution is
y = u y + u y = u cos 2t + u sin 2t (5.85)
2 2
1
1 1
2
5−22 Numerical Analysis Using MATLAB® and Excel®, Third Edition

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