Page 182 - Numerical Analysis Using MATLAB and Excel

Using the Method of Variation of Parameters for the Forced Response

the roots s = – 1 and s = – 3 . Therefore, the natural response is
2
1
t –
y N = k e + k e – 3t (5.72)
2
1
a. Using the method of undetermined coefficients we let y = k 3 (a constant). Then, by substitu-
F
tion into (5.71) we obtain k = 4 and thus the total solution is
3
t – – 3t
yt() = y + y = k e + k e + 4 (5.73)
F
1
N
2
b. With the method of variation of parameters we begin with the natural response found above as
(5.72) and we let the solutions y 1 and y 2 be represented as
t –
y = e and y = e – 3t (5.74)
1
2
Then by (5.68), the total solution is
y = u y + u y
2 2
1 1
or
t –
y = u e + u e – 3t (5.75)
1
2
Also, from (5.69),
du du
2
1
--------y + --------y = 0
dt 1 dt 2
or
du 1 – t du 2 – 3t
--------e + --------e = 0 (5.76)
dt dt
and from (5.70),
du dy du dy
2
2
1
1
⋅
⋅
-------- -------- + -------- -------- = ft()
dt dt dt dt
or
du du
-------- –( 1 e ) t – -------- –( + 2 3e – 3t ) = 12 (5.77)
dt dt
Next, we find du dt⁄ and du ⁄ dt by Cramer’s rule as follows:
2
1
– 3t
0 e
du 12 – 3e – 3t – 12e – 3t – 12e – 3t t
1
-------- = ------------------------------------------- = -------------------------------- = ------------------ = 6e (5.78)
dt e t – e – 3t – 3e – 4t + e – 4t – 2e – 4t
e – t – – 3e – 3t
and

Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−21

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