Page 214 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

y = u y + u y = ( ln cos t + k ) 1 cos t + ( t + k ) 2 sin t
1 1
2 2
(
= k cos + k sin + tsin + cos t ln cos t )
t
t
t
1
2
Check with MATLAB:
y=dsolve('D2y+y=sec(t)'); f=simple(y)
f =
sin(t)*t+log(cos(t))*cos(t)+C1*sin(t)+C2*cos(t)
5. Differentiating the given integro−differential equation with respect to we obtain
t

2
dv
dv
dv 3 k -------- + k ------ + k v = 3cos 3t – 3sin 3t = 3 cos 3t – sin 3t )
(
-------- +
dt 3 3 dt 2 2 dt 1
or
2
dv
dv
dv 3 – k -------- k ------ k v + 3 cos 3t – sin 3t ) (1)
(
–
-------- =
–
dt 3 3 dt 2 2 dt 1
We let
------ =
v = x 1 dv x = x · 1 dv 2 2 x = x · 2
-------- =
2
3
dt
dt
Then,
dv 3 ·
-------- = x 3
dt 3
and by substitution into (1)
·
(
x = – k x – k x – k x + 3 cos 3t – sin 3t )
3 3
2 2
3
1 1
Thus, the state equations are
·
x = x 2
1
·
x = x 3
2
·
(
x = – k x – k x – k x + 3 cos 3t – sin 3t )
1 1
3 3
2 2
3
and in matrix form
·
x 1 0 1 0 x 1 0
·
(
x 2 = 0 0 1 ⋅ x 2 + 0 ⋅ 3 cos 3t – sin 3t )
· – k – k – k x 1
x 3 1 2 3 3
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−53
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