Question

find the partial fraction expansion of

993 +6192 + 275s + 375 s4 +993 +43s2 + 75s

pay attention, the final answer should not have complex or imaginary numbers.

Answer 1

consider 1. fcs) = 95 +615²+2755.+375 - s4 +993 +4352+755 ->F(1) = 958 +6152+27554375 5(83+9524435+75) → FCS) = 950 +618? +2755+3 75 Sis? +35° +652 +185+255+75) - 953 +6187+2 75 5+375 sts9c3+3)+6565+3) +2565+3)] - 95+6132+2755+375 S (5+3) CS? +65+25) Now taking pastral fraction decomposition, we get fiss has the form f(3) = A + B + CS+D (1) Sa+46 5725 , q3 +6 1 3 +375 13tr ( S+ 3)(x1+6 S+RS)+BSCR16 TRS) s(5+3) (s?+G5+25) + Ces+D) C5 (5+3)) S (5+3)65? +65+25) The denominators are equal, so we get as3+615²+2755+375= ACS+3)(57 +65725) + B (83 +632+258) + estD) (S?+35) St3 = AC33+652t25573527185+75)+B683 +632+255) + CC53+3c52 +0.52+385) = AC 52 + 9.52 + 435 +75)+B(33+652+255) t(CS3+ (3C+D)58 +305) = (A+B+C) 33+ (QA +GB+30+D)s +(43A +258 +3D) 5+75A.

Equating coefficients of powers of s on both sides, we get AtB+C = 9. 9 A+GB+ C+D=61 - (3) 43A +253 +30= 375 — 4). 75A= 375 => A = 375 = 5 > A=5 substitute A=5 in eam (P), (3) and ell, we get 5+B+0=9 => B+C = 9-5 = 4 ~ (5) 9x5 +63 +3c+D=6] => 63 +30+D = 16 — (6) 4345 +253 +3 D=275=> 25 B+3D= 60 — (7) from ean(7), we get 25B+30=60 => 3D = 60-951 -> D = 20 - 25 B - (8) substitute this in ean (6), we get GB+30+20-25 B = 16 -> (6 - ) B+3 = 16-20 >> BB+30= -4 =-73 +96=-13 (9) multiply t in eam (s) and add with (9), we got 7 B+76=28 -7B +96=-12 166=16 =) (C=1] substitute c=1 in ean (9), we get -7B ta=-12 3-) -7B = -19~9 = -21. => B = 3 )

substitute B=3 in ean (8), we get D = 20- 25x2= 20-25 =-5. s-5 Str Now plugging the value of A, B, C and Din egna), we get f(3) = 3 + 343t spostas => 983 +615* +2755+375 = 5+3 t rastor S&S+3) (S?+65+25) Cans Intl 5