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Winter 2000 - Bender's Class - Final Exam

Winter 2000 - Bender's Class - Final Exam - RPM Pl MAW 21D...

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Unformatted text preview: RPM Pl MAW 21D F‘lllal F‘mem F1711’T‘lle-dihy Melii £11,270“ . :‘leaae , Icu: name, l) anbexg end section number (or time; on your olue Cook . The first page rt yoiii hliie (ink (177th s lies) my mohair n'rfés No other ;apez is allowed - saloulatoes are NOT ellowed - You must show your work to receive oredit. i 0 p35) Shame (1 7 1521/4 (1 7 W! + (1 + My 7 Us (a) Deterrn no the singu a: ;Om’usv W hi ones are resu M7 to) Suppose we obtain a power series abcu: o = L tor gm Does the series onverge et a , 27 You must give a soviet-r reason tor your eusvsi to iete v ned t , lo the szandard notation. l/(l e a)i_’l Z and 4(a) = l/(l + 110 e a)? (a) bi :e pi’ ex eot at a: = il‘ these are the singular p Regularity requires that (a we ) and (a e 920 ) have oowe: series at rig Wleii rig )Hls-yhliHl lieveiime .seiies) . point s isgiihi When 10 : 71) pm does not have a power series so this point is uegula: (or you can rev “not :ogularm (b) It converges: The power series for p o 4' short so 7 o converge :‘or M < 1' By a theororn in tho ooole so does the pow , serios tor tlat so (m m) Compute 17:2 Top e-e ti ansiri in o: my given int V'l’r‘ltult) 7 1+22‘i ml 7 1 and a/l’O) 7 2 Nat: mat you me not asked to find 3/(0, We oave (s? 1171/3 + 2/(s e ,i, Tous '/s+LI/(d71)+d7£ fi+1 (10 pts) A Lyli,ll,lllt.<tl tent. is 10 iieh equ has a tirtlilur has-e of diameter 5E ieet A hole ir, the bottom ~t the tank allows water to 1934 out aeoordiop to Torrioe 1?: law: alt/av ofht who h = Mr] is tho depth :f tho wato: in toot and t is time ih days Water is typed into the tack at a constazt late so that“ it were We: to Jole, an erhpoy tank woule fill in 5 cays Write down a diEe:ehti;1 equation tor ha) when Ute» tort its slit empty itl hoeaodw rsbe,gphmpedln , med not 37le Ma sesame, , he lutu attornt the :e.t the. the toll]; Lets u sure to es 1;in how you got the equation You : level would rise 100 ' at n 5 days, it is sing at the rate cf :oteet per ea This s the rato sowhioo water is Eowmg in Si is is flowlng out at tho rats cf EtW teet per days :he dlflexentlal equation s m 7 2J 7 SH? i‘be in tial oood tion is m: n ,s (60 was) Fiutl the petrtlttleu sulutleus Lu the fullluwlhg dlfleeehtlal equatlulls (4)19’71'7255 9(1):: lb) rt” 1:) y’ (Id) r’y’ e. (u) The e,lielirn is liner Thus s y 2* + c, m l w A Lemar/91x the equation is oomogeneous The ntegre. he Tutti, r} su( 212x2:1i+0andsoc ihj) The equation is separah e, so lo = as 7 21mm + G Since ale) ; 1 e are positive and n2 7 3 + C’ Hence 0 7 Ir] 7 3' One ran aponeub see to get a iiiiei r in MHW'I A teroativelyi the equation is meat, (o) By undeterm ned ions, yaria: on of parameters, or observe e arsrtoristio equatior e 1 e u and so the general solution is e e as +ooe7= et, Usihg the initial i-ondil iis 3+ ‘ n and in e l : 1 Thu The equation oao a n h solved by Leo algebra and paetial a oaeoioulor solution Too hornosonoous equation 74” e = o 395 (d) The eqoat on is hLLuL sue, on etLll 5/ : rhr + r Lu oluteiri L'U’ + e : 1 e r 7 r? Thus as operate va:iahi1es aod ntegrate to get l’l o‘r‘ e In: a Thus (1 we) elna l ct Ran: the initial oondition. c=1, You sen solve tor 1/ it you wish: 3/ = z e s/lniezy 5 (nu pte) one soluoioo to ”e" + og/ e o = U is 14:) En; an i dependent soluiion toi a > 0 zeductlor, 0: etc 5 We set e2 7 «mi 7 ea, einoe o; 7 o + w on: 1/" o 271 + at" 0 e :35; tee/2 7 1/2 7 L215u'+ x‘u": e :Lu+ ow) 7 ea e m" + (2 Separaeing \ iahles: Tons a partiotlar so ition is Inti’ 71m + 1/? EhrpOhentiatihg Integeating a m 72W, This gi‘yesy_ m raj/m as ah iodepeodeot solu oot (sinoe we eon nuultiol v hv a constant ahv solution oteoo torrn es 7 may” l as: is atoootahle it or ,1 Lt 3, m pts) Find the {1me seiies so ution ahout so : 0 to, :be d fieieutial eeuation (1 eeX‘iy’Wilg/JrSymo) 9(0‘i7 1, y’lfD)7 7; You my use the ’ollo‘mug test to help yoL oheoh your o atlonsg m yon may not e it to fall Hi? s mm The enswei is a 'llywt’lmlal oriethe lrm degiee a VJh set p E rigs the eoefiheien: of r“ to obtain the reours oo iirereritiate t Ln , slihstitlite LlLu the 2L1L<tuhlu> enr luluk at 02+ 21w +1)aei+h 7 am e Lise 7 402+ haw + doe 7 J Thi .\ (77.2 e n 7 else 7 4m +1ioeei oo=,. oi: 3‘ and awe: ”gym“ With a : 0, a2 : 3' With so : tag : ,1, With n : 2, a4 a Wito a : 3 a4 : Us From now on me) = D sinoo it eoponds only on :hc peovious W0 vahos as and saw, 7' (w stay (a) :‘md the general solttion to 111231” + tool is = u :or 1 > to : u‘in egene:a souion odtu +9123 7y7 z” ort>, L dth llt t‘Q”“’ wt U 7' (at This is an Eu er equation. so y o where 71:7 3i 717 0 Thus r o ,1 and i = l/fl and so the sonorol soluu on is o = ei/s l will (:b) Ary rnethod tha: produoes a partiru a: solution is acceptable Tois noludes using Lneeteernioed oerioienes even though there is no reas the: method should give a soliiioi sin e ii iileleiii iiieil loom sills is rii iiirsimii ioemi eiili liiieu equation‘ The ' to note that, in solv ng Euler‘s oquatior, the heh 1/; 5’12 7 912 Hehoe C 71 side is a tunotior of ’7 times a V so wo want to end Lp With a? we tor a part - olutio ‘i’beu we get or? oce2 o 5 alwr] so the geneisl solo is 7/ : :1/i-+ V7 + Alternatively, we oah use the torrnula tor variation cf parenneters (p 175) Note that we rnu divide toe river, equat on hr 212 so that the coefl‘lclent cf t” is one Thus 9/2 After sone oaleuletions Wlti‘e, = 3/2(23/2:I and Y 7 e2, AthLLlale/ely) you do use the lrieh i rreutirrerl [in oinyertiug eu Etllu, equation to a constant coefficie equation: Set In: : it Toe given e uation becomw 2 at? l was t T is see he solvod in Va: ors wave Tho generel solution is e e as“ + ...
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