Page 172 - Numerical Analysis Using MATLAB and Excel

Using the Method of Undetermined Coefficients for the Forced Response

Example 5.6

Find the total solution of the ODE

2
dy
d y + 4------ + 3y = 0 (5.31)
t d 2 dt
⁄
subject to the initial conditions y0() = 3 and y' 0() = 4 where y' = dy dt
Solution:

This is a homogeneous ODE and its total solution is just the natural response found from the char-
2
acteristic equation s + 4s + 3 = 0 whose roots are s = – 1 and s = – 3 . The total response is:
2
1
t –
yt() = y () = k e + k e – 3t (5.32)
t
2
1
N
The constants k 1 and k 2 are evaluated from the given initial conditions. For this example,
0
0
y0() = 3 = k e + k e
1
2
or
k + k = 3 (5.33)
1
2
Also,
dy t – – 3t
y' 0() = 4 = ------ = – k e – 3k e
dt 1 2 t = 0
t = 0
or
– k – 3k = 4 (5.34)
2
1
Simultaneous solution of (5.33) and (5.34) yields k = 6.5 and k = – 3.5 . By substitution into
1
2
(5.32), we obtain
t –
yt() = y () = 6.5e – 3.5e – 3t (5.35)
t
N
Check with MATLAB:
y=dsolve('D2y+4*Dy+3*y=0', 'y(0)=3', 'Dy(0)=4')

y =
(-7/2*exp(-3*t)*exp(t)+13/2)/exp(t)

pretty(y)
- 7/2 exp(-3 t) exp(t) + 13/2
-----------------------------
exp(t)
The function y = f t() , of relation (5.35), shown in Figure 5.1, was plotted with the use of the
MATLAB script

Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−11

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