Nasab2 Bagus
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Sabtu 11 april 2009 Ikfi Mubarokah Mulia Rahmayani “METODE FROBENIUS” 9. (x+2)2 y " + (x+2) y ' – y = 0
y"+
y' -
y=0
Misalkan : X m+r
y:
m
X m+r-1
y ':
m
X m+r-2
y ":
m
(x+2)2 y " + (x+2) y ' – y = 0 •
(x2 + 4 x + 4) m+r
X m+r-2 + (x+2)
m
m
X m+r-1 -
X
m
=0
•
m+r
m
X
+
m
X
m+r-2
m
X
m+r
m
X
m
X m+r-2 = 0
4
m
+
m
m+r
X
m+r-1
X
+ 2
m
X
+
4
m+r-1
-
X m+r = 0
m
•
•
+
m+r-1
X
m
+
2
m
m
0 •
Mencari r ( pangkat terkecil)
m+r
+
m
m
X
X
m+r
m+r-1
+ 4 +
4
Xm+r
X
m+r-1
+ +
m
X
m+r-2
=
4r (r-1) = 0
•
4r = 0
r–1=0
r=0
r=1
Sehingga : Y1 (x) = xr1 ( a0 + a1 + a2x2 + … ) Y2(x) = k y1(x) ln (x) + xr2 (A0 + A1 x+ A 2x + …) Mencari koefisien a0, a1, a2, … , A0, A1, A2, …
Koefisien untuk :
•
xs+r = xm+r-2
xs+r = xm+r-1
m+ r = s + r
s + r = m+ r -2
s + r = m+ r -1
m=s
s+2=m
s+1=m
[ (s + r) (s + r – 1) + (s + r) – 1 ] a s + [ 4 (s + 1+ r) (s + 1+ r - 1) + 2 (s + 1+ r) ] a s+1 + [ 4 (s + 2+ r) (s + 2 + r - 1) ] as+2 = 0
•
[ (s + r) (s + r – 1 + 1) – 1 ] as + [ (s + 1+ r) (4s + 4 + 4r – 4 + 2)] as+1 + [ 4 (s + r + 2) (s + r + 1) ] as+2 = 0
•
[ (s + r)2 – 1] as + [ (s + 1+ r) (4s + 4r + 2)] as+1 + [ 4 (s + r + 2) (s + r + 1) ] as+2 = 0
•
as+2 =
Untuk r = 0
•
as+2 =
S=0
a2 =
=
S=1
=
a3 =
=
=
=
S=2
-
+
a4 =
=
S=3
=
[
=
+
]=
a5 =
=
=
[
=
-
]=
Untuk r = 1
As+2 =
S=0
A2 =
=
=
=
S=1
A3 =
=
=
=
S=2
A4 =
=
=
=
=
S=3
A5 =
=
=
=
[
= Jadi : y1 (x) = x0 (a₀ + a1x + (
-
) x² + (
+
) x³ + (
-
) x4 + (
+
)
x5 +(
=
=
a₀
(
) x6 + …
-
1
+
-
-
…)
+
(2 + ( )² + ( )³ + ( )4 + ( )⁵ + ( )⁶ + ….) +
( )+ ( )² + ( )³ + ( )4 + ( )⁵ + …)
=
(
)–
=
(
)+
( )+ +
( (
) )
a1
+
(
x
-
(2 +
=
+
=
+
+
y 2 (x) = K ( x4 +
+
) ln (x) + x'(A₀ + A1x +(
) x² +
x³ +(
+
+ …)
)
x⁵ +…)
=K(
+
) ln (x) + A₀x +A1 (x²-
+
=K(
+
) ln (x) + A₀x + A1x² [ 1 –(
=K(
+
) ln (x) + A₀x + A1x² [
–
- ( )² – (
)⁵ - …] ]
Jadi solusinya : y = y1 (x) + y2 (x) = =
+ +
+K(
+
) ln (x) + A₀x + A1x² [
(1+ K ln (x)) + A₀x + A1x² [
]
-
]
-(
-(
y"+
y' -
y=0
Misalkan : X m+r
y:
m
X m+r-1
y ':
m
X m+r-2
y ":
m
(x+2)2 y " + (x+2) y ' – y = 0 •
(x2 + 4 x + 4) m+r
X m+r-2 + (x+2)
m
m
X m+r-1 -
X
m
=0
•
m+r
m
X
+
m
X
m+r-2
m
X
m+r
m
X
m
X m+r-2 = 0
4
m
+
m
m+r
X
m+r-1
X
+ 2
m
X
+
4
m+r-1
-
X m+r = 0
m
•
•
+
m+r-1
X
m
+
2
m
m
0 •
Mencari r ( pangkat terkecil)
m+r
+
m
m
X
X
m+r
m+r-1
+ 4 +
4
Xm+r
X
m+r-1
+ +
m
X
m+r-2
=
4r (r-1) = 0
•
4r = 0
r–1=0
r=0
r=1
Sehingga : Y1 (x) = xr1 ( a0 + a1 + a2x2 + … ) Y2(x) = k y1(x) ln (x) + xr2 (A0 + A1 x+ A 2x + …) Mencari koefisien a0, a1, a2, … , A0, A1, A2, …
Koefisien untuk :
•
xs+r = xm+r-2
xs+r = xm+r-1
m+ r = s + r
s + r = m+ r -2
s + r = m+ r -1
m=s
s+2=m
s+1=m
[ (s + r) (s + r – 1) + (s + r) – 1 ] a s + [ 4 (s + 1+ r) (s + 1+ r - 1) + 2 (s + 1+ r) ] a s+1 + [ 4 (s + 2+ r) (s + 2 + r - 1) ] as+2 = 0
•
[ (s + r) (s + r – 1 + 1) – 1 ] as + [ (s + 1+ r) (4s + 4 + 4r – 4 + 2)] as+1 + [ 4 (s + r + 2) (s + r + 1) ] as+2 = 0
•
[ (s + r)2 – 1] as + [ (s + 1+ r) (4s + 4r + 2)] as+1 + [ 4 (s + r + 2) (s + r + 1) ] as+2 = 0
•
as+2 =
Untuk r = 0
•
as+2 =
S=0
a2 =
=
S=1
=
a3 =
=
=
=
S=2
-
+
a4 =
=
S=3
=
[
=
+
]=
a5 =
=
=
[
=
-
]=
Untuk r = 1
As+2 =
S=0
A2 =
=
=
=
S=1
A3 =
=
=
=
S=2
A4 =
=
=
=
=
S=3
A5 =
=
=
=
[
= Jadi : y1 (x) = x0 (a₀ + a1x + (
-
) x² + (
+
) x³ + (
-
) x4 + (
+
)
x5 +(
=
=
a₀
(
) x6 + …
-
1
+
-
-
…)
+
(2 + ( )² + ( )³ + ( )4 + ( )⁵ + ( )⁶ + ….) +
( )+ ( )² + ( )³ + ( )4 + ( )⁵ + …)
=
(
)–
=
(
)+
( )+ +
( (
) )
a1
+
(
x
-
(2 +
=
+
=
+
+
y 2 (x) = K ( x4 +
+
) ln (x) + x'(A₀ + A1x +(
) x² +
x³ +(
+
+ …)
)
x⁵ +…)
=K(
+
) ln (x) + A₀x +A1 (x²-
+
=K(
+
) ln (x) + A₀x + A1x² [ 1 –(
=K(
+
) ln (x) + A₀x + A1x² [
–
- ( )² – (
)⁵ - …] ]
Jadi solusinya : y = y1 (x) + y2 (x) = =
+ +
+K(
+
) ln (x) + A₀x + A1x² [
(1+ K ln (x)) + A₀x + A1x² [
]
-
]
-(
-(
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