Question

5. Solve the recurrence relation an = 3an-1 + 4an-2 +10:4with ao = 5 and a1 = 32. 6. State the general solution of an = -16an-3 + 341 – 11.

Answer 1

answers Givch data, We have a ahs 3an-1 + 496-9 tlo.th do : 5 and 9,532 an - 39n-1 -4an-2 -10.45 - 0 its homo geneal reaction is an - 3an-1 Han-2=0 its characteristic equaction is, ra_ 30-4=0 (31-6) (0+1) = 0 on H, -1 an = a1-10h tea (4)h - @ let, =944 be particular soluction. p4" - 38471 - 4928 - 10.40 P[4" - 3ynas cun!] -10.45 p{um-(31)] .10125 Pl4" un ] : 10.40 ; To.ua

not possiabile, 50, - an = pth Therefore, is not a pack icular sonuctions an = Pn un lying in O, r we get Phq" - 39th-) -1 ..чP (h-a) n-3- ч үп qh - 39h-h+ 3pun-1 - үPhүrax 8үчhao- от ч"- 39 un-1 ++ 38 hr) ч" + 8чта. По-чn ?" - чеn un-1 + рец от 194" . : 10-ч" -4° 11 to ahol 4H- е: qn цо, "n in is particular saluation. an г с е" +9 (4)" } чo huh до - S

toe get, 20:41 (1) tca (4) +40 (0) 4° se cit ca citcais - ③ weget, Elo Hill an+ ,(h) 8) +,(17) 32 = -474 C2+ 160 -1 tul2 =32 -1 352-160 u © u -ci tuca = 192 - ☺ to we get sca = 192 +ş 192+55 192755 ca = 247 using in 6 55 weget, ant 2012 25 a = 5-247 55

= 275 -247 *3-247 55 Therefore, i ana () 204+ 247 (4)+ mun 55 is the required solution