Substitusi: r (r -1)a0+ 2r a0 = 0 (r 2 -1)a0+ 2r a0 = 0 (r 2 –r +2r) a0 = 0 (r 2 + r ) a0 = 0 r (r +1) = 0 r 1 = 0 r 2= -1 selisih r 1 dan r 2 bilngan bulat, maka merupakan kasus 3 dengan solusi: y1 (x) =
x
r 1
(a0 + a1 x + a2 x2+ ....)
y2 (x) = k y1 (x) ln (x) +
x
r 2
(A0 + A1 x + A2 x2+ ....)
Langkah 2; mencari koefisien koefisien a0, a1, a2 , ... dan A0, A1, A2, ... x
m + r −1
=
x
s + r
m + r – 1= s + r m=s+1
x
m + r
=
x
m+r=s+r m=s
s + r
x
m + r +1
=
x
s + r
m + r + 1 = s + r m=s–1
substitusi: (s + 1 + r -1) (s + 1 + r) a s + 1+ 2 (s + 1 + r) a s + 1 -2 (s + r) a s + as – 1 – 2 as = 0 (s + r) (s + 1 + r) a s + 1 + (2s + 2 + 2r) a s + 1 – (2s +2r) as + as – 1 – 2 as = 0 (s2 + 2 sr + s + r + r 2 + 2s + 2 + 2r) a s + 1 = (2s +2r - 2) a s + as – 1 (s2 + 2 sr + 3s + 3r + r 2 + 2) as + 1 = (2s +2r - 2) a s + as – 1
+
x
m + r
=0
as + 1 =
s = 1, maka a2 =
s = 2, maka a3 =
s = 1, maka A2 =
s = 2, maka A3 =
(2(1) − 2) a1 + a 0 2
1
=
+ 3(1) + 2
(2(2) − 2)a 2 + a1 22
=
+ 3(2) + 2
s
a0
6 2a 2
+ a1
=
12
a0 + a a0 + 3a1 1 6 = 3
2
(2 s − 4) A s
x
y1 (x) =
x
+ A −1 2 s − 2 s + 3 s (2(0) − 2) A0 + A−1 − 2 A0 + A−1 = 0 2 − 2(0) + 3(0) 0 (2(1) − 2) A1 + 2
1
(2(2) − 2) A2 22
A0
− 2(1) + 3(1)
=
+ A1
− 2(2) + 3(2)
r 1
(a0 + a1 x + a2 x2+ ....)
0
( a0 +
y1 (x) = a0 + atau,
=
12
12
s
A0
2
=
2 A2
+ A1 6
=
A0 + A A + A 1 0 1 = 2 6
2
6
∴ Jadi, solusi dari persamaan di atas adalah y1 (x) =
s
+ a −1 2 s + 3 s + 2 (2(0) − 2) a 0 + a −1 − 2a 0 + a −1 = 0 2 + 3(0) + 2 2
