Page 178 - Numerical Analysis Using MATLAB and Excel

Using the Method of Undetermined Coefficients for the Forced Response

y = Acos 5t (5.57)
F
We can find out whether our assumption is correct by substitution of the assumed solution of
(5.57) into the given ODE of (5.56). Then,

– 25Acos 5t – 25Asin 5t + 6Acos 5t = – 19Acos 5t – 25Asin 5t = 4cos 5t

⁄
but this relation is invalid since by equating cosine and sine terms, we find that A = – 4 19 and
also A = 0 . This inconsistency is a result of our failure to recognize that the derivatives of Acos 5t
produce new terms of the form Bsin 5t and these terms must be included in the forced response.
Accordingly, we let
y = k sin 5t + k cos 5t (5.58)
3
4
F
and by substitution into (5.55) we obtain
– 25k sin 5t – 25k cos 5t + 25k cos 5t – 25k sin 5t + 6k sin 5t + 6k cos 5t = 4cos 5t
3
4
4
3
3
4
Collecting like terms and equating sine and cosine terms, we obtain the following set of equations
19k + 25k = 0
4
3
25k 19– k = 4 (5.59)
4
3
We use MATLAB to solve (5.59)
format rat; [k3 k4]=solve(19*x+25*y, 25*x−19*y−4)
k3 =
50/493
k4 =
-38/493
Therefore, the total solution is

50
yt() = y + y t() = k e – 2t + k e – 3t + --------- sin 5t + – 38 5t (5.60)
--------- cos
2
F
N
1
493
493
Check with MATLAB:
y=dsolve('D2y+5*Dy+6*y=4*cos(5*t)'); y=simple(y)
y =
-38/493*cos(5*t)+50/493*sin(5*t)+C1*exp(-3*t)+C2*exp(-2*t)

In most engineering problems the right side of the non−homogeneous ODE consists of elementary

k
n
functions such as (constant), x n where is a positive integer, e kx , cos kx , sin kx , and linear
combinations of these. Table 5.1 summarizes the forms of the forced response for a second order
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−17

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