MATH 202 Exam 2 with Solution, Spring 2005

Math 202 — Spring 2005. Differential Equations, sections 1-4 Quiz 2, April 14 ~ Duration: 70 minutes GRADES: 1 (12 pts) 2 (12 pts) 3 (18 pts) 4 (12 pts) 5 (18 pts) TOTAL /72 YOUR NAME: YOUR AUB ID#: PLEASE CIRCLE YOUR SECTION: Meihihepy Section 1 Section 2 Section 3 Section 4 Recitation F 1 Recitation F 2 Recitation F 12 Recitation F 3 Ms. Zantout Ms. Zantout Ms. Zantout Dr. Yamani INSTRUCTIONS: 1. Write your NAME and AUB ID number above, and circle your SECTION. 2. Solve the problems inside the booklet. Explain your steps precisely and clearly to ensure full credit. Partial solutions will receive partial credit. Each problem is worth 12 or 18 points. . You may use the back of each page for scratchwork OR for solutions. There are three extra blank sheets at the end, for extra scratchwork or solutions. If you need to continue a solution on another page, INDICATE CLEARLY WHERE THE GRADER SHOULD CONTINUE READING. . Do as much of the exam as you can, and budget your time carefully. If you cannot do a certain integral, just leave it as an integral in your solution for partial credit on the rest of the problem. . No calculators, books, or notes allowed. Turn off and put away any cell phones or beepers. 2 2 96 9 96 9 2 2k so 2c ak 2 Ac os 2 2 2s 2 2 fe ac ic ak 2k ke se 2 2c af 2 2 2 2 og 2 2 2 2s 2 . Problems 3 and 5 ask for series solutions. For full credit on these problems, you need to give a formula for the coefficients. HOWEVER, you can get ALMOST FULL CREDIT if you give the first FOUR nonzero terms for any solution. Ex- ample: if you were solving the very simple equation y” +y = 0, your solution could just say y= co| for alnost full credit (16/18 points), but for full credit you would have to say that _ (=i (= ~ 26)! (241) 2 9 i oie os 2k 2 2 oe 2c 2s 2g 2 2 2 2 2 fe 9s 2 2b 2c 9 ko 2 2 2c 24 2c fe 9 2s 9 94 2 9 2 2 2 oa oe oo a as ok Cre co, and that coed1 = GOOD LUCK! An overview of the exam problems. “ The problems are repeated inside the booklet. — PLEASE SOLVE EACH PROBLEM ON ITS CORRESPONDING PAGE INSIDE. Remember to READ the instructions on the front of this exam regarding problems 3 and 5, which ask for series solutions. 1. (12 points) Find the general solution of the equation yl! + y' — Oy = cosx + et, (Note that y’” is the THIRD derivative of y — do not make the mistake of copying it as y’’.) 2. (12 points) Find the general solution of the equation 3 x ay” — Ixy’ + 2y = row Note: find yp using variation of parameters by DIRECTLY substituting yp = yi ui + yetlg into the differential equation. Do NOT use any formulas for variation of parameters unless you prove them first. 3. (18 points) Use a series centered at the point x = 0 (which is an ordinary point) to find the general solution of (1+2°%)y” — 6ry = 0. (Hint: simplify the recurrence as much as possible before solving for co, C1, C2) 34+ +++ Chy +++ 4. (12 points total, 6 points for each part) a) Calculate the Wronskian of the FIVE functions 2a yal, yo = 2, yg =o, ye", ys =e Use this to DEDUCE whether {y1, yo. ya. 94; Ys} are linearly dependent or independent. (Remark: don’t be intimidated by the 5 x 5 determinant. You can evaluate it by expanding by minors along any row or column. Choose a row or column that makes the calculations very easy.) b) (UNRELATED) Use the SUBSTITUTION z = e' to solve the following differential equation in terms of Bessel functions. Do NOT write down any series. You MUST carefully explain HOW you do the substitution. dy ( a 1 We + Ca 5 y=0. \ se (Recall that the Bessel differential equation is ay” + vy’ + (a — v*)y = 0, and that its general solution is y = AJ,(z)+ BJ_,(z) if v is not an integer, and y = AJ, (2)+ BY,(2) if v is an integer.) 5. (18 points) Given the following differential equation with a regular singular point at wv = 0: ary” + ay! + (2? — 9)y == 0. (This is actually a Bessel differential equation, but we want to find a SERIES solution using the method of Section 6.2.) Find ONLY ONE solution yy = 33°.) cnx"? of this differential equation, corresponding to the LARGER one of the two indicial roots r;. (Here r; > r2 and you should NOT try to find the solution yg that uses r2, since yz involves a logarithm.) 0 1. (2 points) Find the general solution of the equation m yl ty — 10y = cosa +e". ele ae) (Note that y’” is the THIRD derivative of y — do not make the mistake of copying it as y”.) 2 g } -: Se. Can cren ts oe Oe cen DE so He ex\h ara ev ahvon \s oo Wre+m-\OzO., Try rab, meet stirs aio | Aches ‘| ZP-=e : (ext) or mes me ea MAD eo Lee (52) or ‘ err dm a GS (+S) or —— AD oe met (p>, wal | . ~ . (MEm) OO, ao, yon Mam —\O = (m-2)(™ +Ur§°) _ Ur + m= 1d O2t Rin Ast ( wa ian moro” Ane roots eh wha tae St © ml So 1 ooo. we we Tht Tall fsa, bye queticke 7 OF , Coed: {. VANDA , - rots & ow. oe Tork eesti ne | \ vy RQ o%estx ~Y.. Wee ee * e es + Ce *sinry einen CE trys ce DR eS 5 LCte /\ = os™ Lae) we Wor ou geek Ye, an Ge. & near comler nahn ot cos & Qa, a Lot @ ? ve. _ \ Caiaist LCesx\ & ae o>) S&S we norm 4 wn & fh 4 Lf y on LCf }? & af (st oeniee | Sink —> ae c sic [2G Cate 68 x | Sink aging = Beay t ~\Ocurx con ow r : | | “™ ‘ore Sst Lmeasye| nus td a fOr TIES Ay So wet bonwT NEED HX \ SNe 4 \ Hl \ ‘se Se, = a esx sabsher LC4p,} = “05x \ ‘ vote ant enn nwiire na myn ayer wttemaane rennet ee . «! AG v Usps far Yoo ue wk wo oh oF Le De a me Bae YK, Equsty t Tape needs 08 exe Rokr Ry , owt an aM Nc Ge ave ' LCxer ae: Gye) (a rane) —Iore™ Since ere ae fr gt 5 2. (12 points) Find the genera) solution of the equation 4 x xy" — Qay' + 2y = Vvi-z Note: find yp using variation of parameters by DIRECTLY substituting yp = y) ui + yous into the differential equation. Do NOT use any formulas for variation of parameters unless you prove them first. Cay > Ye “Ths 1K Covey -Qler Cow doale. \* xm uh xX Sgt Dy! rdys 0 (ar y'= net i. sce « 2), ¢ . We gee Clee) YEE AR LAE SO, Carncelly xy" 4 SY \Con “velo Cle-D-LUe 4120 2 ee Le &) (-Cr +0 rete fe —= 41= xX Pare f Vom cueing Cet ey = yh = XV, + tu, , { a Ue Dye, +/*0)) 4004) ae) \ \ YF 429 tre = = The Jeet glen Ym esp at . Xx 3. (18 pomts} Use a series centered at the point + = 0 (which is an ordinary point) to find the general solution of (+ 2°)y" - Gary = 0. (Hint: simplify the recurrence as much as possible before solving for ¢g, ¢:, C2, C4, +++ + Chae + Since we em Ata s is RA ecdinony Peis we J os AD Thur aF Usual 4 = Ss nfn-1) C.X . Koo “ ul = - ~ bx 2 1 S * 3 - ° -) c. NaN) ca b S Alrar) Cr x _ 2 fe» Par Reo nro \ _ - ay “Combi we eee Nat AOA-O OAK oF Db) 6, x = 0 vy ACh-t) —G = n-n- 6 reins fen a & -Dl(ne2) oh oe ee eREIRrn Renate AWwetHhel LS RH Ls he Kaen Ahe( Som Tom with be a (s A-L2 G ceinker Mae Seond ee = SRP ete nals pairs “Ake. Se Rewer lent ~s OR we covlh reinde a Deity sums 4 baw il t, Lathal idedhes C be “3 CS wB)(30)c, 20 3 ete 2 rn va, be a Ke ~t oe one Mian Meceereny | c= | =o Cie DOV) Choe = Le,=o — Pore lucian \ Ce LC Ne arb hrmey Lor] i id 25. Recor reree » bse chs 502) Cg et) Qe nal => \Cynr ry -CD . jo} N eee ese 6=P) —™ Ca elo) me -- mad) RS) 13 tay ~~ On Cc Wi “Cee 0-2.) DAW) -- (4HYTE) | res 4, (12 points total, 6 pomts for each part) a} Calculate the Wronskian of the FIVE functions z Qs 2 wel, yo = 2, y= x", wae, yw =er Use this to DEDUCE whether {y1, y2, y3, 44: Ys} are linearly dependent or independeat. (Remark: don’t be intimidated by the 5 x 5 determinant. You can evaluate it by expanding by minors along any row or column. Choose a row or column that-qnakes the calculations very easy.) ¢ ya det 4 “My 43 An 46 - t ' ‘ Se aes Os n ~ io | a: N woe ~ us lO | o ZL €~* te ™ | igh et co ee get yee _ uel! ° a ex le c2™ Vise let Calun- 4 ; tn \\ Vs ce Ve NN ! : * byt \ \ s., 5 ~ — |, dct \ “ oe S BD € e : +, = | a Le te. ro Mw) > Abbr lores &e™ 4 of) oo. ee Ge | py » Khon c\ Oo ef be™ oO; & Whe Ni ose Xs churn Crh omn co, = V2 La( 8 ae = 2+(eer"- be™ ) = jhe] ; ex ray The Corenchin is ah derkerty tem ce Uri sb ore linearly LNDEPENBDENT b) (UNRELATED) Use the SUBSTITUTION zx = e€' to solve the following differential equation in terms of Besse] functions. Do NOT write down any series. You MUST carefully explain HOW you do the substitution. dy 2t 1 ae +(e ~ )8 (Recall that the Bessel differential equation is x7y” + zy’ + (x? —i7)y = 0, and that its general solution is y = AJ,(z)+ BJ_,(x) if v is not an integer, and y = AJ,,(x)+ BY, (x) if v is an integer.) nee MO eH et ow a = — Te b% &” & ak ey dy * ves &. (18 points) Given the following differential equation with a regular singular poini at © = 0: ary” + xy! + (a? ~ Vy = 0. (This is actually a Bessel differential equation, but we want to find e SERIES solution using the method of Section 6.2.) Find ONLY ONE solution y, = 7%, c,2"*7 of this differential equation, corresponding to the LARGER one of the two indicial roots r;. (Here r) > re and you should NOT try to find the solution yo that uses r2, since yy involves a logarithm.) = La svlb oh Whe YF 2 Cy x ) 4 S Ard Cn el _ = y> ST (neoCnee “1D cane inde Xhe DE n~e on c ar TS nec = o we ner Ad D_WreKaee-r) cx a mA (nse) CA KX + > Carr _ 2 Vea ze hz Neo =O K>o Pannen piss Comlone terms ail sam eae xnee Hine coelB cied it [medtrer-0) a (ner) -—A\ Cc, = Crerearer-a QC NNN ee oe Veneers Wie 1 vol(u-D+eur 4% . a ee Ce ae oo vo) oop an Se S (neers Dey . 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