Page 210 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

5.14 Solutions to End−of−Chapter Exercises

2
1. The characteristic equation of the homogeneous part is s + 4s + 3 = 0 from which s = – 1
1
t – – 3t
and s = – 3 . Thus y N = k e + k e . For the forced response, we refer to Table 5.1 and we
2
2
1
assume a solution of the form y = k t + k 4 and the total solution is
F
3
t –
y = k e + k e – 3t + k t + k 4
3
2
1
The first and second derivatives of are
y
t –
⁄
dy dt = – k e – 3k e – 3t + k 3
2
1
2
t –
d ydt⁄ 2 = k e + 9k e – 3t
2
1
and by substitution into the given ODE
t –
t –
t –
(
(
k e + 9k e – 3t + 4 – k e – 3k e – 3t + k ) 3 + 3k e + k e – 3t + k t + k ) 4 = t – 1
1
3
2
1
1
2
2
Equating like terms we obtain
4k + 3k t + 3k = t – 1
3
4
3
3k t = t
3
4k + 3k = – 1
3
4
⁄
⁄
and simultaneous solution of the last two yields k = 13 and k = – 79 . Therefore,
3
4
t –
--t –
y = k e + k e – 3t + 1 - 7 -
--
2
1
3
9
Check with MATLAB:
y=dsolve('D2y+4*Dy+3*y=t−1’); y=simple(y)
y =
-7/9+1/3*t+C1/exp(t)+C2/exp(t)^3
2. The characteristic equation of the homogeneous part is the same as for Exercise 1 and thus
t – – 3t
y N = k e + k e . For the forced response, we refer to Table 5.1 and we assume a solution of
1
2
t
the form y = k te t – where we multiplied e t – by to avoid the duplication with k e t – . By sub-
1
F
3
stitution of this assumed solution into the given ODE and using MATLAB to find the first and
second derivatives we obtain:
t –
y = k e + k e – 3t + k te t –
1
2
3
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−49
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