MATH 202 Exam 1 with Solution, Fall 1998

Prof. N. Haidar Sec, 1 &2 : Math 202 (1-st Semester , 1998-1999) Time : | hour QUIZ I FACULTY OF ARTS & SCIENCES , A. U. B. 23 Nov., 1998 (closed book) 4) @ UL OI. @ I. V. Find the integrating factor for (2y’ - 6 xy) dx + (3 xy - 4 x2) dy =0. Use this integrating factor to solve the DE. Given that-y, = | is a solution to y =-x ltt) + xlaxt)y- y? . Find the gencral solution of this DE, Find the second-order linear DE which is equivalent to the Riccati DE of question II . Given that y, = e* is a solution to this second-order DE , find its general solution. Solve the IVP (y? + 1) dx -y sec’x dy =0; yO)=0. ‘Solve the BVP it y” - by" + zy - 8y =0 53 y= y(1)=0,y-1) = -2 “. Y © 2 L C23 —bxy dx + Gxs-4xt) dys o “oH ayedx 3 Ob = age 6% 3 gh = 1y — Se De BH Be =H Cx Byte 7 mutt atl Att eet atl goers oat yt) dy =o HT Sa ae 2 ou (wt) it —6 ares) 4% one = a9" (ma) wie gy? (42) eel mel C E (utiy-4 (uty 1 =? G) watt De wy reurd eet gn-Te+l=S xD ) Cu -tae t= > pnty-wUA-S=O 4 Cu-de—t =O - . = guard => Get) \ Entb—4-8 = 0 Mt ca z [Ht = fewer? PIO zy fx - 9 f xx 4 yb 2 x3 aut Tul 2 2y' He - OO AF i ee CF) 9 _ byt , pe RB furbe = 3% 4x3 y'2 _ go (ett) + x! (2xt#l) y-4” 5 J,= | Riccok dD. E- Paya — tet), OG) = x Gxt), Re)=-1 Cc . n22 € | L Lyrae) - & Jaber Le ee = ae “ORF a v 2 CD 2. c+ TM fener) FZ OMDST Sb Oe Pe) Qc R / Sew we Re) = Wl Qk 4 = a QO) _ S?P@) 7) tt j.- 5, |S Qe 4) 2 4 Ont Le ats é Hw _ ec _ § Paid a eC = Cc - C * 1% (x de 2& ee @& % xe ly = c tae eo ) “em © CAS cer, wen ai § S Dey (ste) dx ~ Ysectx dy=o 3 Je=0 dx y secre = go | ty im Sect te - {ean (v dy = fcotuce = L Vile caredey = %F ae sec xX , \ aed 2 O.. Wwr+t \ poe 2s) omy 7 (s +1) ($+ ca = - 4 bist eC. L qo ca 2, a Cre = toe = LSE ; ™ Llgei)= yx + & Se rw -C \e/J Ye@e)= © othoa $b &lilte = C=O “7 __ > {\ yet Vor V 6) Bue YT. fu694I24—-S9=0 , JO@)= GG)s0, ¥iI)=-2 CE mi bm P42 e= O (1) C-2)(C m* Une) = ma © — Hn + ¢ (u-2@)\Cum-2)% =o 3 2 _— . mt \ mi bm +IZ mM "FH (m-2z)F = oO mit2 me amt + itM -F —~ymepgum Ce My e mM, omy HZ Cc.) — A MeTst 4m-8 4am-9F . ; > a ae ij,=¢ e. ,=2 xe, Ix = xe fies. : LK G= tet +ane**4 6; xt | Poe) e Ca) Je)=0 = CC; +0+0 =0 mGeo de Y= Crxe “+0; xt e*™ J) = G.e*+ @; e*2 0 => Cr +Cs=0 2 Gis Gy L ux j= exe - exe ~ ~ La+2