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Midterm2Soln

Midterm2Soln - Name 391d F753 ID number ’ MATH 2271...

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Unformatted text preview: . Name: . . 391d F753 ........... ID number: .............................. April 20, 2009 ’ MATH 2271 ' Differential Equations for Scientists and Engineers Winter 2009 ' Second Midterm Exarn Time: 50 minutes 1. Consider the ﬁrst order differential equation y’- — \/ 2 + t2. {Lt-1 (3) Sketch the corresponding direction ﬁeld in the region 0 < t _<_ 1, 0 < y_ < 1. Kg 1 (b) Given that y( (0) = 1/2, use two steps of Euler’s method (with step ‘ size h: 1/2) to estimate y(1). LCJ” (7': 21(0)=)’ _ . :Y—(gﬂt): 83 rt 30:) a 31 = g, + WNW? 4 1:41 2. (a) Show that the functions 6‘ sint and 6‘ cost comprise a fundamen- _ ‘ tal set of solutions to the equation 3/” — 2y’ + 23; = 0. Nola {Wt ‘Llwil £06 ZCtSln—l: laml ﬂal=cté€\$f Ue solutlib‘ns “t9 {1 6m: @9459“ ~ 1‘ I ‘ I’m; (4% +25%).e’v f”—Z¥’+2¥ : Zod/et’lﬁr‘ﬁvﬁlk © 4(a) = Zwe't ej/ +29}?! e1= Oi g’a) = [Exit mg? 9"- 23 923.: vzswe'L—ZMHHM ~ tha/(Cf/ ,— ~Zsfn‘t 6+ . New! w: chlille 14c Mam/dam Wt”) 41 1%: 77M i=0. . _ ® WMJM): 466(2) 1') : :1 f0" . . faw/ 1/: Mar/V m/Ww/dé sinu- ﬁt; ‘ ‘ W 2&0 Mil ﬁtﬂ’c’é: (Medial) 60166ch {a ‘lLe gm“ team/485k. Eéj (b) Find the general solution to the equation y” — 2y’ + 2y = sin t. W? Salt 4 ‘Paf'flow-(W whim N2 “(L #N% ' 3f : Acbﬁ‘c +2€h§t ”7": ~/ler\d;‘+:l?cv<‘l/ Ur”: whoa—84kt . . ” - 23/ + Z 1, : ﬂat/Em; fZKAshfﬂ‘géﬁf 5f E1 1+2 Mum/+3975” : (A ~ZB)C067§ + (5 +20%” = SW; @ 92 :32: é> 3%; Aée‘gré‘ﬁh‘s‘ﬁl 3. Let yl and y2 denote solutions to the equation 3/” + e‘y’ + t4'g = 0. E5 I (a) If yl and y1 + 312 are linearly independent, what can you conclude about the possible values of Wronskian W(y1,y2) of y1 and y2? Explain and justify your reasoning. ( If + 1 Ne {Medal Mdcymaawl Hum 41M glﬂdﬁz OM wit “to 4a 'tluK K 50 Mt . ® \ ”:4 ”131* “(1‘3 ' O ' the“ (cg—0931* «2(3 :5 »=> 0 > ‘>) :0 :0! 51) at x ~x<>d M aiwj'W’“ @ M < AltdnaﬁwJ am am Slow 79ml W[?1)(719;1):W((7ujq/>~> I 5‘ (b) Given that 311(0) = y’1(0) = y’2(0) = 1 and 112(0) = 2, express the ' solution to the initial value problem yll + etyl+ t4y = 0, y(0) = 1, . yl(0) = 0 in terms of y1 and y2. We lulou/ llxe awwa/ \$olw¥th I’M“ H“ 40"“ y: CIJMCZW‘ Q) o 4. Find the general solution to the equation 3/ ’+ y= sect. ' [Ac haw/m anew/L9 ezwb‘f‘wa ”+3 .2 0 cm [9: larva-2K as . LDZ + 1) £5 _: 0 .I . gm“ ('0 4 1): CD + 2:X D a”) rtlgc L15: ‘ gauci splay! M 3 I 81 mi * (z 6617- We use Vwc'atlb’vl p4 ”Md/’0’? “(f0 éoIVL ‘er Omaha?! iaprw‘tif/ 4/5 I/{0W<. . 0 Lo?! 0L <[af a Vco<~L/\/n\——\> (4 watt , \leh‘t + . ( Leg/Vt + “609+ ._ 096“! Cm"? '75::(7), [afbéwtl ~ (U?) ' (Maw cast) 'S'Vtgmf bit. So! u = ’t MA V 2 ’X'Lm't 1% ; +JV‘ law“ give “in: TU‘L’iCWlM 4°LW+ :0“: d? : ’thet'l'J’“ Icodlagt End of exam ‘TLC johbm/ c;a((~JV‘O'\5 5? :1f GHC) SW6 “‘(Cﬁﬁl‘KiDMt ...
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