2. Koefisien → mencari dan
,
Sehingga = 0
(akar kembar)
Sehingga solusi (akar-akar persamaan indikator, teorema 2)
, X > 0
Mencari koefisien
m - 1 = s
m = s + 1
3. Koefisien
s a as s
+ - -
[(2 1) ]
2 2
1
s r sr s r
+ + + + +
as+1 = 2 2 2 1
untuk r1 = r2 = 0 , maka :
s a as s
+ - -
s s
[(2 1) ]
2
as+1 = 2 1
1
+ +
=
s a as s
+ - -
s
[(2 1) ]
2
1
+
( 1)
4. 1 0 a a a - a
+ -
[(2.1 1) ] 3
1 0
S = 1 ® a= (1 1)
4
2 2
=
+
2 1 a a a - a
+ -
[(2.2 1) ] 5
2 1
S = 2 ® a= (2 1)
9
3 2
=
+
11 5
15 5 4
5(3 )
= 9
36
36
4
1 0 1 1 0
1
1 0
a a a a a a
a a
-
=
- -
=
-
-
Solusinya adalah :
11 5 1 0 a - a
= + + + 36
+ ….
, X > 0
=