Question

8. Determine the appropriate form of the particular solution for the following non-homogeneous linear differential equation with constant coefficients. (8 Puan y(4) +9y" = 5+ &'(x-3) + 4sin (3x). none of these O Ar? + Bx cos(3x) + Cx sin(3x) + De' + Exet Ar + B + C sin(3x) + D cos(3x) + Exe" A + Bre-3x + Crer + De + Exet O Ar? + Bxe- + Crex + Det + Exe! A + B sin(3x) + Cxsin(3x) + De' + Exet

Answer 1

Solution : Given (4.) 0 ૧H'S +9y"578(2-3)+4 sin(3x) (D+90%) = 54 € (-3) + 4.5in(31) --O Auxiliary Equation is mtt9m²=0 m² (n +9)=0 m (millis)-0 meo (m 311 312-0 m=0, 0, mat3i :: Complementary function (C.F) is 4 = ( + C2x) €4+ [z Cos(30) +64 sin(3x) Jeon Yc CitC₂24 C3 cos (32) + C4 sin (39) Now we have to find particular Integral (PL) (5+ é? CX-3) +4 sin (3x)] (64+90°) gps L. [5+ke"-38" + 4 sin(34) ] 04 +90² Yp (5) + (xes + I (329) + 4 sin(3x) D4+90² 04490² D4+9 8² 047903 LⓇ Yp 9 pz Now take up (5) 0490 (say) v + + + ge ypa ser Spr - C C C YA C 5.com 2 D4+9p Lean f(D) I pan C op 5 en D² (B+9) fra) it fra) #0

Ap () 1 | (ofq) Sex 는 du D² JS s dx h 5 & D² 7 19 p () q :)))) - 특 0 21 ขอ 592 . Illl It IIS)

& 4 otthon ale xe" D*+90² a ex a ser (070) 490010) a 9 = en (04+4D?+*748 +1) +9 (0420 +1) yo a ex D4 +4pP+150°+22D +10 이 Pr en 3 10 + 104 +40° +150° +220 10

X 10 46 - e* to [1 +($*+ 40 + 150+20)] Yeaed to [!-(04484 180420) *P*o£150*20). (23)] : 22 D2 + 22 (1) el 10 10 a c y ole "384(X) = 0 : DCO a 12-11 To 5 G (52-1) 10 5 c 402 c e" 50 (52-10 e C & Hope ll (-3e") D4+91² a༣ le foo I ean f(a) sen ea if fras 0 -3 D"+90² Liet dp = -3 yoles 17 -3 Le 179 ye -3 el 10

f 1 You I = s du D One )dy 어 yea of 5 coseBuj] [sin cew) (4) Sincsa) Yea (4) otto - Substituting up," Ip Mp3 & IP values in ③ we get a 5x² + (52-11) - 304 14 Sin (3x) 079" (4 sina) (0²79). 4 (D?+9) o 4 (029) Dota I sin (3x) D² +3² -X cos (3x) 213) 66) e cos(38) 6 2X COS (32) 27 18 cle? Yea () Colax e 4m=6) 2013 I sincax)= -2 D²tq² za 914 yea - - 7 E @=> 4= 52€ + + 2X COS(30) 27 T - Зе? + 10 22 COS(3) Op = 5x2 18 50 27 27 COS(39) + 8p> 572 +2e- Te-15e 27 50 18 Up = 5x²t xe? 18 26 en + 22 Cos (38) 50 27

Чр qe per te! - 19 2 + 24 cos c33 The general solution is y = 4 + Up y - G+C_X + C COS (3X) + Casin(3%) + 5x+2e" _ 13 ext 23.cos(32) 18 10 25 27 1 which is the required general Solution.