Question

4. (18 marks) Find the solution of the following initial value problem for a nonho- mogeneous second order differential equation with constant coefficients: Y" – 3y = 3e36 - 18t2 +4, y(0) = 0, y(0) = 4

Answer 1

selly Given equation hon home geneous 2ND anden onden offenen till with constant coefficient i 3 e3t _ 18 t² + 4 0) (Initial conditions yo) 4 (3) Eqh (1) day be written as 3 day 3 e3t - 184² + 4 (4) dd² du We know that General Selution (God.) Complementary Solution Co s.) + Particular Integral (P. J.) 05 (5) c Homogeneous egh (D? - 3 D)y= & Awe leary egh of 6 - (m? 3 m) & but Dom ů mcm-3) - U m=o,3 3 CF = Geol + G en a ten where G & G are arbitrary contouts IGRE

Now finding the particular viaktions of egn.ch (Integral P.I (363t -184² +4.) D²-3D 3 out D2-3D (-18 (2) + D2 3D 3 18 D²3D 22 begin D²3D beg, 403 3 18 2 400t pat 9-9 & care fach) D23 D (30)!-] c) (1 + $ to ('potal) - 3+1 G est + & 2D-3 18 2 + D3 33 32 I 4 02 3.0 eot - (care fail + D²3D 3t (0²-3 D) 3t 3t + 6 6(1) [4? + 2.3-3 2t 3 + 2 32 (1) + 4t Leot - 38 2D 3 3+ (13-3) 3t + + 2 9 cot It (2) +44 13.0-3 [ + 2 + 7 +42(7) e at 2 +² 3 2 3 cox

Page No Dato: - Gt Gest te3t+ (a1 + 24² + aux test + gt3 + gt² in K, then we get. G.S 404) -68) Now wing intial condition in (8) bring 6 + 2 (0)3 +260)2 t 03.0 0 (0) 4 Gtge. + G 4 GG) + + GtG GE-G eqh (8) Now differentiate y' (t) 6t²4tt writ. t p3t + 23+ 3t + Get (3) (9) wing y lol A 6(0)2 + 4 (6) at e30 te 3.0 3 (0) + Ge 30 (3) 4 = otot lt tG (1) (3) 4 It 36 739 = 3 [6=1 uring in (4) Now general solution becomes yet) = gt3 + g 2² + est t+ (1) + G) eit 2 + 3 + 24 + 3t (t+1) -