Metode Frobenius
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Selesaikan persamaan berikut dengan menggunakan Metode Frobenius xy” + (1 – 2x)y’ + (x – 1)y = 0 Jawab :
xy” + (1 – 2x)y” + (x – 1)y = 0
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
Koefisien
as+1
[(2s + 1)a s − a s −1 ] 2 = s + r + 2 sr + 2 s + 2r + 1 2
untuk r1 = r2 = 0 , maka :
as+1
[(2 s + 1) a s − a s −1 ] s 2 + 2s + 1 = [(2 s + 1) a s − a s −1 ] ( s + 1) 2 =
[(2.1 + 1)a1 − a 0 ] 3a1 − a 0 = 2 4 ( 1 + 1 ) → S=1 a2 = [(2.2 + 1)a 2 − a1 ] 5a 2 − a1 = 2 9 ( 2 + 1 ) → S=2 a3 = 5(3a1 − a 0 ) − a1 15a1 − 5a 0 − 4a1 11a1 − 5a 0 4 = = 9 36 36 = Solusinya adalah :
=
+
+
11a1 − 5a0 36 +
+ …. ,X>0
=
xy” + (1 – 2x)y” + (x – 1)y = 0
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
Koefisien
as+1
[(2s + 1)a s − a s −1 ] 2 = s + r + 2 sr + 2 s + 2r + 1 2
untuk r1 = r2 = 0 , maka :
as+1
[(2 s + 1) a s − a s −1 ] s 2 + 2s + 1 = [(2 s + 1) a s − a s −1 ] ( s + 1) 2 =
[(2.1 + 1)a1 − a 0 ] 3a1 − a 0 = 2 4 ( 1 + 1 ) → S=1 a2 = [(2.2 + 1)a 2 − a1 ] 5a 2 − a1 = 2 9 ( 2 + 1 ) → S=2 a3 = 5(3a1 − a 0 ) − a1 15a1 − 5a 0 − 4a1 11a1 − 5a 0 4 = = 9 36 36 = Solusinya adalah :
=
+
+
11a1 − 5a0 36 +
+ …. ,X>0
=
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