Page 174 - Numerical Analysis Using MATLAB and Excel

Using the Method of Undetermined Coefficients for the Forced Response

t –
yt() = y + y = k e + k e – 3t – 3e – 2t (5.40)
N
2
F
1
The constants k 1 and k 2 are evaluated from the given initial conditions. For this example,
0
0
y0() = 1 = k e + k e – 3e 0
1
2
or
k + k = 4 (5.41)
2
1
Also,
t –
------
y' 0() = – 1 = dy = – k e – 3k e – 3t + 6e – 2t
dt 1 2 t = 0
t = 0
or
– k – 3k = – 7 (5.42)
2
1
Simultaneous solution of (5.41) and (5.42) yields k = 2.5 and k = 1.5 . By substitution into
2
1
(5.40), we obtain
t –
yt() = y + y = 2.5e + 1.5e – 3t – 3e – 2t (5.43)
F
N
Check with MATLAB:
y=dsolve('D2y+4*Dy+3*y=3*exp(−2*t)', 'y(0)=1', 'Dy(0)=−1')

y =
(-3*exp(-2*t)*exp(t)+3/2*exp(-3*t)*exp(t)+5/2)/exp(t)
pretty(y)

-3 exp(-2 t) exp(t) + 3/2 exp(-3 t) exp(t) + 5/2
------------------------------------------------
exp(t)
The plot is shown in Figure 5.2 was produced with the MATLAB script

y=dsolve('D2y+4*Dy+3*y=3*exp(−2*t)', 'y(0)=1', 'Dy(0)=−1'); ezplot(y,[0 8])

Example 5.8

Find the total solution of the ODE
2
dy
d y + 6------ + 9y = 0 (5.44)
t d 2 dt

subject to the initial conditions y0() = – 1 and y' 0() = 1

Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−13

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