Question

3 pts Question 7 is The solution of the initial-Value Problem (IVPI 38 - 2y = 4(x - 2) v(I) = 4 W(1) = -1 4 +-9 + 1 O 1 +? - 24 0 417 +-+3 None of them o 1 +- 2c + 4 D Question 8 3 pts The power series solution of the initial Value Problem (IVP) (x+1) + xy + 2xy = 0 (0) = 2 (0) = 3 given by None of them 2125 V=2+3+ 7 6 + + + 4 = 2 +3 - Tary 6 + 213 4 2 y = 21- + 3. 20 + .. + + 3 6 2 y=30:- + + ... + 2(1- + 3 20 2

Answer 1

③ Given that (x + y yo tay' tray 20 40) 22, (o)=3 a 20 is an ordinary point of o - E en an de a solution of clearly let ya no then y nenan- nal and u 2- - 2 m-1 - E n(n-1) en an-2 using above equation becomes nena + 2nd cuan ww, En(n-1) enant 8 n(n-1) en 20-2 menants and the lo (x²+1) { n(n-1) ena" ta n21 no 222 nol noo noz ns2 nenan or, & n ( menan + m + 2) (+1) en+ 2x + nol 1120 22 +2 Cn-l nal bor, 2e2+ 6ezx + ax + acont E n (n-1) en + (+ 2)(n+ 19 ent 2 +men+2en–177"=0 Since egnation on ② is an identity on 2 2c220 6 CB + G + 2 coro and nm-gent + 2) (n+1) eut 24 hentzenyo ie. C2 20 est-$(4+260) and necu + (2+2) cm + 1) ent2 +2 eur, zo 2 en-1 at nhen W, entz in72 (n+1) (0+2)

er, y = 2 + 32 - ( 2 / 3 + 2) 23 _ x4 6 2 y = > 2+32 - Fa3 a4 This is the required solution of ① Hence option @ is the answer