Metode Frobenius

Soal : Nomor 11:

+

Jawab: y(x) = y(x) = y(x) =

 (x-1)2

(

+



Pangkat terkecil untuk x ialah r (r-1) r(r-1) = 0 r1 = 0 , r2 = 1

sehingga solusinya:

;

sehigga persamaan indikatornya menjadi: sehinga :

y1 (x) = x0 ( = y2 (x) = x1 (

Mencari koefisien dari: a0, a1, . . . dan A0, A1 , . . . Persamaan:



 m+r = s+r m=r

m+r-1 = s+r

m+r-2 = s+r

m= s+r

m = s+2

 (s+r) (s+r-1) as – 2(s+r+1) (s+r) as+1 + (s+2+r) (s+r+1) as+2 + (s+r) as – (s+1+r) as+1 – 4 as = 0 Untuk r = 0, maka:

 s (s-1) as – 2(s+1) s as+1 + (s+2) (s+1) as+2 + s as – (s+1) as+1 – 4 as = 0



 Jika s=0

Jika s=1

Jika s=2

Sehingga solusi yang pertama: y1(x) = =( = Untuk r=1, maka :  (s+1) (s+1-1) as – 2(s+1+1) (s+1) as+1 + (s+2+1) (s+1+1) as+2 + (s+1) as – (s+1+1) as+1 – 4 as = 0

 (s+1) s as – 2(s+2) (s+1) as+1 + (s+3) (s+2) as+2 + (s+1) as – (s+2) as+1 – 4 as = 0





Jika s=0

Jika s=1

Jika s=2

Jika s=3

Sehingga solusi yang kedua: