Page 177 - Numerical Analysis Using MATLAB and Excel

Chapter 5 Differential Equations, State Variables, and State Equations

4Ae – 2t – 10Ae – 2t + 6Ae – 2t = 3e – 2t (5.52)
but the sum of the three terms on the left side of (5.52) is zero whereas the right side can never be
zero unless we let t → ∞ and this produces a meaningless result.

The problem here is that the right side of the given ODE of (5.49) has the same form as one of the
– 2t
terms of the natural response of (5.50), namely the term k e .
1
To work around this problem, we assume that the forced response has the form

y = Ate – 2t (5.53)
F
that is, we multiply (5.51) by in order to eliminate the duplication of terms in the total response.
t
Then, by substitution of (5.53) into (5.49) and equating like terms, we find that A = 3 . There-
fore, the total response is

yt() = y + y = k e – 2t + k e – 3t + 3te – 2t (5.54)
N
F
2
1
Check with MATLAB:
y=dsolve('D2y+5*Dy+6*y=3*exp(−2*t)')
y =
-3*exp(-2*t)+3*t*exp(-2*t)+C1*exp(-3*t)+C2*exp(-2*t)
We observe that the first and last terms of the displayed expression above have the same form and

thus they can be combined to form a single term C3*exp(-2*t).

Example 5.10
Find the total solution of the ODE

2
d y + 5------ + 6y = 4cos 5t (5.55)
dy
2 dt
t d
Solution:
No initial conditions are given; therefore, we will express solution in terms of the constants k 1

and k 2 . We observe that the left side of (5.55) is the same of that of Example 5.9. Therefore, the
natural response is the same, that is, it has the form

y N = k e – 2t + k e – 3t (5.56)
1
2
Next, to find the forced response and we assume a solution of the form

5−16 Numerical Analysis Using MATLAB® and Excel®, Third Edition

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