Nasab Ika
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Soal xy’’ + 3y’ + 4x3y = 0 Tentukan basis solusi dari persamaan diferensial dengan metode frobenius. c( x) b( x ) = 4x3 =3 2 x x c( x) = 4 x 5 b( x ) = 3 x ∞
y ( x) = ∑ a m x m + r m =0 ∞
y ' ( x) = ∑ (m + r )a m x
m + r −1
m=0 ∞
y ' ' ( x) = ∑ (m + r )(m + r − 1)a m x m + r −2 m =0
xy’’ + 3y’ + 4x3y = 0 ∞
∑ (m + r )(m + r − 1)am x m+ r −2
∞
∑ ( m + r ) am x
m + r −1
∞
∑a
m
x m+r
x m =0 + 3 m =0 + 4x3 m =0 =0 r-2 r-1 r-1 r x [r(r-1)a0x + (1+r)ra0x +…] + 3[ra0x + (1+r)a1x + …] + 4x3 [a0xr + a1xr+1 + … ]=0 persamaan indikator dengan xr-1 , maka a0 r(r-1)a0 + 3r a0 = 0 (r(r-1) + 3r ) a0 = 0 a0 ≠ 0 maka haruslah r(r-1) + 3r = 0 r(r-1) + 3r = 0 r2 – r +3r = 0 r2 + 2r = 0 r(r+2) = 0 r1=0 r2=-2 selisih r1 dan r2 adalah bilangan bulat, maka gunakan teorema 2 kasus 3 ( selisih kedua akarnya bilangan bulat). *sehingga solusinya : y1(x) = xr1(a0 + a1x + a2x2 +…) = x0 (a0 + a1x + a2x2 +…) = a0 + a1x + a2x2 +… y2(x) = k y1(x).ln (x) + xr2 (A0 + A1x + A2x2 + … ) , dimana r – r > 0, k ∈ R = k (a0 + a1x + a2x2 +… ).ln(x) + x -2 (A0 + A1x + A2x2 +… ) *mencari koefisien a0,a1,a2,… A0,A1,A2,… Persamaan xy’’ + 3y’ + 4x3y = 0
∞
∑ (m + r )(m + r − 1)am x m+ r −2
m =0
∞
∑ ( m + r ) am x
m + r −1
+ 3 m =0 + 4x3 misal m + r + 3 = s + r m+3=s m=s–3
Misal m + r – 1 = s + r m – 1= s m= s + 1 samakan koefisien x s + r (s + 1 + r)(s + r)as+1xs+r + 3 (s + 1 + r) as+1xs+r + 4 as-3xs+r = 0 (s + 1 + r)(s + r)as+1 + 3 (s + 1 + r) as+1 + 4 as-3 − 4a s −3 a s +1 = ( s + 1 + r )( s + r ) + 3( s + 1 + r ) Untuk r1 = 0 − 4a s −3 a s +1 = ( s + 1) s + 3( s + 1) − 4a −3 a1 = 3 − 4a − 2 a2 = 8 − 4a −1 − 4a −1 a3 = = 6+9 15 − 4a 0 − 4a 0 a4 = = 12 + 12 24 s=0 Untuk r2 = -2 A s +1 =
=
− 4 As −3 ( s + 1 − 2)( s − 2) + 3( s + 1 − 2)
− 4 As −3 ( s − 1)( s − 2) + 3( s − 1)
∞
∑a
m =0
m
x m+r =0
− 4 A−3 2 + (−3) − 4a − 2 A2 = 0 − 4a −1 A3 = = 3 − 4a 0 − 4 a 0 A4 = = 2+6 8 A1 =
y ( x) = ∑ a m x m + r m =0 ∞
y ' ( x) = ∑ (m + r )a m x
m + r −1
m=0 ∞
y ' ' ( x) = ∑ (m + r )(m + r − 1)a m x m + r −2 m =0
xy’’ + 3y’ + 4x3y = 0 ∞
∑ (m + r )(m + r − 1)am x m+ r −2
∞
∑ ( m + r ) am x
m + r −1
∞
∑a
m
x m+r
x m =0 + 3 m =0 + 4x3 m =0 =0 r-2 r-1 r-1 r x [r(r-1)a0x + (1+r)ra0x +…] + 3[ra0x + (1+r)a1x + …] + 4x3 [a0xr + a1xr+1 + … ]=0 persamaan indikator dengan xr-1 , maka a0 r(r-1)a0 + 3r a0 = 0 (r(r-1) + 3r ) a0 = 0 a0 ≠ 0 maka haruslah r(r-1) + 3r = 0 r(r-1) + 3r = 0 r2 – r +3r = 0 r2 + 2r = 0 r(r+2) = 0 r1=0 r2=-2 selisih r1 dan r2 adalah bilangan bulat, maka gunakan teorema 2 kasus 3 ( selisih kedua akarnya bilangan bulat). *sehingga solusinya : y1(x) = xr1(a0 + a1x + a2x2 +…) = x0 (a0 + a1x + a2x2 +…) = a0 + a1x + a2x2 +… y2(x) = k y1(x).ln (x) + xr2 (A0 + A1x + A2x2 + … ) , dimana r – r > 0, k ∈ R = k (a0 + a1x + a2x2 +… ).ln(x) + x -2 (A0 + A1x + A2x2 +… ) *mencari koefisien a0,a1,a2,… A0,A1,A2,… Persamaan xy’’ + 3y’ + 4x3y = 0
∞
∑ (m + r )(m + r − 1)am x m+ r −2
m =0
∞
∑ ( m + r ) am x
m + r −1
+ 3 m =0 + 4x3 misal m + r + 3 = s + r m+3=s m=s–3
Misal m + r – 1 = s + r m – 1= s m= s + 1 samakan koefisien x s + r (s + 1 + r)(s + r)as+1xs+r + 3 (s + 1 + r) as+1xs+r + 4 as-3xs+r = 0 (s + 1 + r)(s + r)as+1 + 3 (s + 1 + r) as+1 + 4 as-3 − 4a s −3 a s +1 = ( s + 1 + r )( s + r ) + 3( s + 1 + r ) Untuk r1 = 0 − 4a s −3 a s +1 = ( s + 1) s + 3( s + 1) − 4a −3 a1 = 3 − 4a − 2 a2 = 8 − 4a −1 − 4a −1 a3 = = 6+9 15 − 4a 0 − 4a 0 a4 = = 12 + 12 24 s=0 Untuk r2 = -2 A s +1 =
=
− 4 As −3 ( s + 1 − 2)( s − 2) + 3( s + 1 − 2)
− 4 As −3 ( s − 1)( s − 2) + 3( s − 1)
∞
∑a
m =0
m
x m+r =0
− 4 A−3 2 + (−3) − 4a − 2 A2 = 0 − 4a −1 A3 = = 3 − 4a 0 − 4 a 0 A4 = = 2+6 8 A1 =
