Nasab Dt2
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Selesaikanlah dengan metode Frobenius y"+ xy'+ (1 − 2 x − 2 ) y = 0 Penyelesaian : ∞
∑ (m + r )(m + r − 1) a
m=0
m
x
m+ r − 2
+ x ∑ (m + r )am x
∑ (m + r )(m + r − 1) am x m+ r − 2 +
m=0 ∞
m=0
m + r −1
m= 0
∞
∑ [ (m + r )(m + r − 1) − 2] a
∞
∞
∑ (m + r )am x m + r + m= 0
∞
+ (1 − 2 x ) ∑ a m x m+ r = 0 −2
m=0
∞
∞
m= 0
m= 0
∑ a m x m + r − 2∑ a m x m + r − 2 = 0
∞
m
x m+ r − 2 + ∑ [ (m + r ) + 1] a m x m+ r = 0....................(*) m=0
* Mencari r yaitu dengan memilih koefisien terkecil dari (*), pilih ( x r − 2 ) Substitusi m = 0 ke [ (m + r )(m + r − 1) − 2] a m
[ r (r − 1) − 2] a0 = 0
a 0 ≠ 0 sehingga r (r − 1) − 2 = 0 r2 − r − 2 = 0 (r − 2)(r + 1) = 0 r1 = 2 r2 = −1 Kasus 1 karena selisih kedua akarnya bilangan bulat Solusi : y1 ( x) = x r1 (a0 + a1 x + a 2 x 2 + ...) = x 2 (a 0 + a1 x + a 2 x 2 + ...) y 2 ( x) = k y1 ( x) ln( x) + x r2 ( A0 + A1 x + A2 x 2 + ...)
[
]
= k x 2 (a 0 + a1 x + a 2 x 2 + ...) ln x + x −1 ( A0 + A1 x + A2 x 2 + ...)
* Mencari koefisien a0 , a1 , a2 ,..., A0 , A1 , A2 ,... * Mencari koefisien x s +r x s +r = x m +r −2 x s +r = x m +r s +r = m+r −2 s+r =m+r m = s+2 m=s * Substitusi ke (*) [ ( s + r + 2)(s + r + 2 −1) − 2] as +2 + [ s + r +1] as = 0 s + r +1 as ( s + r + 2)( s + r + 1) − 2 untuk r1 = 2 a s +2 = −
0 + 2 +1 3 a0 = − a 0 (0 + 2 + 2)(0 + 2 + 1) − 2 10 2 Jika s = 1 , a3 = − a1 9 5 5 3 3 Jika s = 2 , a4 = − a2 = − × − a0 = a0 28 28 10 56 3 Jika s = 3 , a5 = − a3 20 7 7 3 1 Jika s = 4 , a6 = − a 4 = − × a0 = − a0 54 54 56 144 * Cari a1 dengan m = 1 disubstitusikan ke [ (m + r )(m + r −1) − 2] am Jika s = 0 ,
a2 = −
[ r (r +1) − 2] a1 = 0 (r 2 + r − 2)a1 = 0
(r − 1)(r + 2)a1 = 0 r1 = 2
r2 = −1
sehingga a1 = 0 → Diperoleh : y1 ( x) = x 2 (a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6 + ...) 3 3 1 = x 2 a0 + 0 x − a 0 x 2 + 0 x 3 + a0 x 4 + 0 x 5 − a0 x 6 + ... 10 56 144 3 3 4 1 6 = x 2 a0 1 − x 2 + x − x + ... 56 144 10 3 3 6 1 8 = a0 x 2 − x 4 + x − x + ... 10 56 144
* Untuk r = − 1 s+ r +1 As ( s + r + 2)( s + r + 1) − 2 s + (− 1) + 1 As + 2 = − As ( s + (− 1) + 2)( s + (− 1) + 1) − 2 s As + 2 = − As ( s − 1)( s + 2) Jika s = 0 , A2 = 0 × A0 = 0 As + 2 = −
1 1 Jika s = 2 , A4 = − A2 = − × 0 = 0 2 2 3 Jika s = 3 , A5 = − A3 10 2 2 Jika s = 4 , A6 = − A4 = − × 0 = 0 9 9 5 5 3 3 Jika s = 5 , A7 = − A5 = − × − A3 = A3 28 28 10 56 * Cari A1 dengan m = 1 disubstitusikan ke [ (m + r )(m + r − 1) − 2] Am
[ r (r + 1) − 2] A1 = 0 (r 2 + r − 2) A1 = 0
(r − 1)(r + 2) A1 = 0 r1 = 2 r2 = − 1 sehingga A1 = 0 → Diperoleh : y 2 ( x) = k y1 ( x) ln( x) + x r2 ( A0 + A1 x + A2 x 2 + ...) 3 3 1 8 = k a0 x 2 − x 4 + x 6 − x + ... ln x + x −1 ( A0 + A1 x + A2 x 2 + ...) 10 56 144 = k a0 x 2 = k a0 x 2
3 4 3 6 1 8 3 3 x + x − x + ... ln x + x −1 A0 + A3 x 3 − A3 x 5 + A3 x 7 + ... 10 56 144 10 56 3 3 1 8 3 3 − x4 + x6 − x + ... ln x + x −1 A0 + A3 x 3 − x 5 + x 7 + ... 10 56 144 10 56 −
∴ Solusi : 3 3 1 8 y1 ( x) = a 0 x 2 − x 4 + x 6 − x + ... 10 56 144 3 3 1 8 3 3 y 2 ( x) = k a 0 x 2 − x 4 + x 6 − x + ... ln x + x −1 A0 + A3 x 3 − x 5 + x 7 + ... 10 56 144 10 56
∑ (m + r )(m + r − 1) a
m=0
m
x
m+ r − 2
+ x ∑ (m + r )am x
∑ (m + r )(m + r − 1) am x m+ r − 2 +
m=0 ∞
m=0
m + r −1
m= 0
∞
∑ [ (m + r )(m + r − 1) − 2] a
∞
∞
∑ (m + r )am x m + r + m= 0
∞
+ (1 − 2 x ) ∑ a m x m+ r = 0 −2
m=0
∞
∞
m= 0
m= 0
∑ a m x m + r − 2∑ a m x m + r − 2 = 0
∞
m
x m+ r − 2 + ∑ [ (m + r ) + 1] a m x m+ r = 0....................(*) m=0
* Mencari r yaitu dengan memilih koefisien terkecil dari (*), pilih ( x r − 2 ) Substitusi m = 0 ke [ (m + r )(m + r − 1) − 2] a m
[ r (r − 1) − 2] a0 = 0
a 0 ≠ 0 sehingga r (r − 1) − 2 = 0 r2 − r − 2 = 0 (r − 2)(r + 1) = 0 r1 = 2 r2 = −1 Kasus 1 karena selisih kedua akarnya bilangan bulat Solusi : y1 ( x) = x r1 (a0 + a1 x + a 2 x 2 + ...) = x 2 (a 0 + a1 x + a 2 x 2 + ...) y 2 ( x) = k y1 ( x) ln( x) + x r2 ( A0 + A1 x + A2 x 2 + ...)
[
]
= k x 2 (a 0 + a1 x + a 2 x 2 + ...) ln x + x −1 ( A0 + A1 x + A2 x 2 + ...)
* Mencari koefisien a0 , a1 , a2 ,..., A0 , A1 , A2 ,... * Mencari koefisien x s +r x s +r = x m +r −2 x s +r = x m +r s +r = m+r −2 s+r =m+r m = s+2 m=s * Substitusi ke (*) [ ( s + r + 2)(s + r + 2 −1) − 2] as +2 + [ s + r +1] as = 0 s + r +1 as ( s + r + 2)( s + r + 1) − 2 untuk r1 = 2 a s +2 = −
0 + 2 +1 3 a0 = − a 0 (0 + 2 + 2)(0 + 2 + 1) − 2 10 2 Jika s = 1 , a3 = − a1 9 5 5 3 3 Jika s = 2 , a4 = − a2 = − × − a0 = a0 28 28 10 56 3 Jika s = 3 , a5 = − a3 20 7 7 3 1 Jika s = 4 , a6 = − a 4 = − × a0 = − a0 54 54 56 144 * Cari a1 dengan m = 1 disubstitusikan ke [ (m + r )(m + r −1) − 2] am Jika s = 0 ,
a2 = −
[ r (r +1) − 2] a1 = 0 (r 2 + r − 2)a1 = 0
(r − 1)(r + 2)a1 = 0 r1 = 2
r2 = −1
sehingga a1 = 0 → Diperoleh : y1 ( x) = x 2 (a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6 + ...) 3 3 1 = x 2 a0 + 0 x − a 0 x 2 + 0 x 3 + a0 x 4 + 0 x 5 − a0 x 6 + ... 10 56 144 3 3 4 1 6 = x 2 a0 1 − x 2 + x − x + ... 56 144 10 3 3 6 1 8 = a0 x 2 − x 4 + x − x + ... 10 56 144
* Untuk r = − 1 s+ r +1 As ( s + r + 2)( s + r + 1) − 2 s + (− 1) + 1 As + 2 = − As ( s + (− 1) + 2)( s + (− 1) + 1) − 2 s As + 2 = − As ( s − 1)( s + 2) Jika s = 0 , A2 = 0 × A0 = 0 As + 2 = −
1 1 Jika s = 2 , A4 = − A2 = − × 0 = 0 2 2 3 Jika s = 3 , A5 = − A3 10 2 2 Jika s = 4 , A6 = − A4 = − × 0 = 0 9 9 5 5 3 3 Jika s = 5 , A7 = − A5 = − × − A3 = A3 28 28 10 56 * Cari A1 dengan m = 1 disubstitusikan ke [ (m + r )(m + r − 1) − 2] Am
[ r (r + 1) − 2] A1 = 0 (r 2 + r − 2) A1 = 0
(r − 1)(r + 2) A1 = 0 r1 = 2 r2 = − 1 sehingga A1 = 0 → Diperoleh : y 2 ( x) = k y1 ( x) ln( x) + x r2 ( A0 + A1 x + A2 x 2 + ...) 3 3 1 8 = k a0 x 2 − x 4 + x 6 − x + ... ln x + x −1 ( A0 + A1 x + A2 x 2 + ...) 10 56 144 = k a0 x 2 = k a0 x 2
3 4 3 6 1 8 3 3 x + x − x + ... ln x + x −1 A0 + A3 x 3 − A3 x 5 + A3 x 7 + ... 10 56 144 10 56 3 3 1 8 3 3 − x4 + x6 − x + ... ln x + x −1 A0 + A3 x 3 − x 5 + x 7 + ... 10 56 144 10 56 −
∴ Solusi : 3 3 1 8 y1 ( x) = a 0 x 2 − x 4 + x 6 − x + ... 10 56 144 3 3 1 8 3 3 y 2 ( x) = k a 0 x 2 − x 4 + x 6 − x + ... ln x + x −1 A0 + A3 x 3 − x 5 + x 7 + ... 10 56 144 10 56
