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2
.b
2
a
2
/ˇ
D
Z
b
1
.b
a/˛
C
y.t/
dt
;
a
(5.116)
3
.b
3
a
3
/ˇ
D
Z
b
1
2
.b
2
a
2
/˛
C
1
ty.t/
dt
:
a
The linear system (
5.116
) determines ˛ and ˇ. Let us now apply the method to
the same functions as we considered for constant approximations.
Example 5.4.
Consider
y.t/
D
sin.t /
(5.117)
defined on 0
t
=2.Since
Z
=2
sin.t /
dt
D
1
0
and
Z
=2
t sin.t /
dt
D
1;
0
the linear system (
5.116
) now reads
=2
2
=8
2
=8
3
=24
˛
ˇ
D
1
1
(5.118)
and the solution is given by
0
@
1
A
0:115
0:664
˛
ˇ
D
;
8
24
96
1
2
(5.119)
24
so
p.t/
0:115
C
0:664 t:
Example 5.5.
Consider
y.t/
D
e
t
e
t
(5.120)
