Tugas Nasab

(1+x)x2y’’ – (1+2x)xy’+(1+2x)y = 0 Tentukan solusi dengan metode frobenius! Jawab : (x2+x3)y’’ – (x+2x2)y’ + (1+2x)y = 0 ● (x2+x3)

xm+r-2 – (x+2x2)

xm+r-1 + (1+2x)

xm+r =0 xm+r +

● =0 - 2

xm+r+1 +

xm+r+1 (am xm+r ) + 2

am xm+r+1 =0

xm+r+1 +

●

xm+r = 0 ●

m+r+1

+

xm+r = 0 * mencari r : → m=0 [(r-1)r – r+1] a0= 0 [r2 – r – r+1] a0 = 0 , a0 ≠ 0 haruslah : (r2 – r- r +1) = 0 (r2 – 2r + 1) = 0 →(r-1)2 = (r-1) (r-1) r1=1 atau r2=1 maka r1 = r2 * mencari koefisien ; s+r Xm+r+1 dan xs+r maka ; m +1+r = s+r m+1 = s

s+r = m+r m=s

m = s-1 →[(s-1)+r-1)((s-1)+r)-2((s-1)+r)+2)as-1]+[(s+r-1)(s+r)-(s+r)-(s+r)+1)as=0 →[(s+r-1)(s+r-1)-2(s+r-1)+2]as-1+[(s+r-1)(s+r)-(s+r)+1]as=0

xm+r

→[(s2+sr-s+rs+r2-r-s-r+1)-2s-2r+2+2)as-1]+[(s2+sr+rs+r2-s-r)-s-r+1]as=0 →[s2+r2-2sr-4s-4r+5]as-1 +[s2+r2+2sr-2s-2r+1]as=0 as= untuk r=1 s = 1,2,3,4,… s1=1 =s2= 2 a2= a2= s3 = 3