Question

QUESTION 10 Note that yı(t) Vt and y(t) =t-1 are solutions of the linear homogeneous differential equation 2t²Y" + 3ty' – y=0. Use variation of parameters to find the general solution of the nonhomogeneous differential equation 2t’y" + 3ty' - y = 4t? + 4t. 8 2 OA Civt + Cat-1 + 74 35 oCivt+Cet-1 + 4 9 t2 + 2t OC Civt + Cat-1 + 4 9 t2 + 2 5 2 OD. 8 Civt + Cat-1 + t2 + 35 t 5 4 OL Civt + Cat-1 + t3 + 2 t2

Answer 1

TQ. (19). Answers Given Y(t) = Vt and Y f) - t are solutions of the linear homo. differential een at ylltoty-y=0 ☺ it's complementry function Tos ementary ginen y (t) GY, (t) te. GCt) y (H) = GFt + et So now we need to find out Perticular solution of non-homogenous linery diff. a 2 t gl +3 tyl G = 4+² +44

here, wen we will use method of variation of parameters so let y t = AYt + By lt) OP Whire A 4B are functions of t. and 'Altia Wly, y Y(t • Rit) at and B- ( yt Relat wly, Where Alts - Rinis part of diffe. f. Wly, y) = Wronskien of j 4% 9. y y se Wly, th) t ſt 100y=17 4y = x DA2 L 2A 1 - - t? 2

Wly = - + 2 2 - 3t دارم 2 how -- AO - (AX (48 + 4t dt Ret) - € +41 7 &yed = t' A2- (44 + etdt th( 41² 441) at 2 3 2 [of th d t + 4 t olt 1 + 4 xt - 2 DO 4 4 4 3 st Nia + Sanota =xt ht int-) 2 Ih st 2 ttt t 3 L 7 Fh * t 7 + Here 3 5

we can not use integral constants. since we already use it in complem functiona u complementy Y XR dE Thow W VAX G + ² + 47) dt Et 2 It = (4+² +4 t) at TE +ť dt BU - - 2 1 yt = AY + By 16 5 th 152 to tata +t 4 3 5 3 L 5 3 + & J = 3 1 2 ttt 13t4 3t H 8 3 35. 20

1 - др. 8 3 X3 ++ -35 20 y (t) 8t + 2t 35 5 Hence gennal folution of gren homogeneous differential equation is y (t) = = y(t) tyt GE+ GA + 8 + + 2 t 35 so option is true. 7