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MAP 2302 Exam 1 Solutions (08-2)

# MAP 2302 Exam 1 Solutions (08-2) - ‘ O Show all work in...

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Unformatted text preview: ‘ O Show all work in order to receive full credit. / ' E _~% Class: MAP 2302 Spring 2009 (08-2): Exam 1 6!.V¢\ W IVY: .372 \ > Date: ‘ ‘ ID . 1.: State the order, dependent variable, and independent variable of each ODE. Also determine whether each equation is linear or nonlinear. (8 points) a.) t‘ym — (t2+ 1)y"'+ (cos t) - y = 0 orders, 5 dependent var. 3 independent var. Z -~ linear rnonlinear? dzy dy b.) Ez- +y'gx' +y — smx order 1 dependent var. :2 independent var. X linear @? 2. Find all constant solutions y a c (where c is a constant), if any, of the DE y' = 6y2— 13y — 8. (4 points) ._;‘ _? -> A-"/ C:§/ I 2Q, C ” /3 ' Answerz' .6 '3 2'! 5 l '3. Can an initial value problem have more than one [email protected] or No (2 points) 4. State in words the Existence/Uniqueness Theorem for a lst-order IVP (you may use the version given in class or the version given in the book, Theorem 1.1). Also show its geometric representation. (6 points) di:;C—K,£1\}I :(Xo§:%ig Z Name: ‘ 71D: A i ' ‘ ~ » 5.; Determine whether the Existence/Uniqueness Theotem foi a lst—order IVP guarantees that the differential . equation y ’= o i y? A 25 possesses a unique "solution through the point given points. (6 points) a.)(2,,-5)YESor®: - b.)(i,0)[email protected] c.)(-3,6)YI:3 orNOYeS 2:75 035—, «ET; _,_ 2—): dx i K , . - r ‘ v 7151 4 5 \‘4 X113: 11'2"; \. <4. 11‘2\$>° i) 215: ’2 , -. . O ‘. i): :— L (1‘43) 911) *m , =5 , :1 4‘ 5 . - ~ } 6. Verify that the indicated family of functions satisﬁes the given DE. (8 points) ~ d . - x 2 -d—:-+2xy= 1; y=e_‘2Ie'2dt+c,e" ‘ 0 x 0%. a 16"; C etc»? Hﬁ'XY 5R“ F [C —_ L; a \ . X 05" an Jo ty- 2%. 3 RH)“: gm W2“? ' ”do. an i” - 2. , ' CL x. .x1 X - ( ‘XL efxlgexl . + Q ‘e‘ -2)! \e f: _, 2K 0 d>sL ~70. | 'L -xL x a}: ‘2'“ CL 9' - - _'2, 2<_& S ' I o , , : X1 X *1 1 C C,)¢l 4. i +— 1X LL - _ . + 0L1+ZX1>-Z.X‘Z€' 30 C. M_ x dx , W ,. Name; ID: A l . ' 3 . x \ 7. Given that x(t)_= _c cos I + c2 sint is a two—parameter family of solutions of the DE x+x= 0, solve the IVP: x'+x— -— 0, x(7r/4)=_ J5, x (7r/4)= 2J— (10 points) «Tic X(\T/q): Jig“ +— _.2 p . 2. C\'E'C'L L G %(W : Zc‘ +0201) V ”(x 47(1- l: (4 Cl) 2; 2. .> Wm”; - F‘s 1CL= L t _ l G)» . rs l2 '- I. L .1 ((1:7) Q0 I ( (ST - ’ *Cl X‘zsev *9 5‘” 4" “3 c. ' 7,” law“ X(T'7L(3=*E<. + «age; L f ‘ ‘ 2&3.in— ‘3—1 3V3— : “EC, ‘6 1:};ch )(()H——-cos* + _'Ui( L Answer: X(,¢.—) = ~ Casi: +- 351%; 8 Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution 18 pumped into the tank at the rate of 3 gal/min, and when the solution 18 well stirred, it is then pumped out at a solwer rate of 2 gal/min. If the concentratiOn of the solution entering is 2 lb/gal, determine a differential equation for the amount of salt A(t) in-the tank at time t. Do not solve the equation. (6 points) ‘15::33’ Gust—3909’ ' ‘ dﬁ : ct“. V‘m - WM “0991» “m=.~"\$“"‘“1" TN?" : gal/L“ I‘ .Af ”(03+ ("”6 M3}: -, V _ q“; 2\béd’q ‘ 0.9:. z 6 __8 (Ear—— AC1), (“J/((3): 3eogw£ V ”U“ " 350+L3'A)X - .‘ ' Answer: E 3.? n it_ . - \ Name: ‘ . ID: A ‘9. F 1nd the general solution of the given differential equation in explicit form. Also, give the largest interval ' over which the general solution 1s deﬁned. (15 points) "(x+1)Ex-+y=lnx, y(1)=10 . \V‘ K 1—; X‘s!) “A xix-l (:1 +_\_ 2 LL" ~ Uo+e . “4145) xi" :1? ; x+x ><+‘ J—cﬁ ‘ go {Lat x >01 ‘ X‘FL’X _ W“ \ -X +Q 41c»): @- ‘ - (x+1)‘;L= X “X ‘ \V‘C>““‘) ._ ‘ _ ”((K): E. (law, 14,):10‘3‘V25: [7({)=x*‘ ' ‘LolO :(‘WU-l‘i‘et W i ’ ~ ‘1 l—Q _ [w OLX 20 k EL [C’HAWDLE dvx -3 :x\\n‘k 9W: d1 [2L : Q dx 4!“. .Lolsc V; )s 7 J — K ltd” : X\\«K i + ll x+1 (xii-0 ‘1. c x \mx — § e” +Q ‘ W“; ' x+ L Answer: 10. . Solve the given differential equation. Leave our result in implicit form. (15 points) .‘ym%_ * [w] @9104ka V 11. Given that the differential eqhation xiix + (x2y+4y)dy = 0 is not exact: (15 points) a.)‘Shovt' that the diﬁ‘erentialequation is not exact. (2 points) _ _ Mzt'gt" «— M :0 .. ' _ ‘_ r7 1 A :) ngt-Mx 3? Ud- exact“, __ 7. t— ” = 1x 1 ' U‘xl"u’f‘-f§’€/x “L b2) Find an appropriate integrating factor that will In ' , and verify your result. (-3 points) V . ”(C K) =-_;_‘-_-‘—_V(' ,. I o r . - .‘ ,' . 1 '.-i M15-Nx ; c9 2..in ' _ | ,‘ ._ 2.; G if 1c *1th 3“.“ 0.) Use the integrating factor found m (b) to solve the equation. Leave your result in im form. (10 points) . . ' 7- 11 ~,__ :0 _- .5 ‘1' . eixaw + C L ‘1‘?“‘né‘i ' A ... ' ' ' ' ‘ ‘33- , .. _. «A = can. =3 My 74“ w. c» and ' " A/ = {‘3' (t11‘43-«5 =57 All ' 2.x" 1* .1. Q) ‘33— - ﬁg + km.) a m; t a» , ,5 we— S"? 9" 2, it wine“ Qx ‘13- :} Ci) . 9(— 2‘1 LLe‘J' 4'\~\93-—\ ‘ x 3'. 6/ 9'5;— .. x‘ie *‘W‘J—Q 77:); ,-—-—' " L ' 1 21—7” 1 ___ n ,. r a U ‘ l 1. “LL VLSC'JA =I NIB—e +‘(l x he ' ”/fjrﬁl 1.. . "L :u-u, \rxt —HS‘1{'\LJUL éwtg} [W'— - ‘13” - ' . .p Aka: 1 . - , \ : Le. W l .. a LMCﬂx: 11 ’ .. ~ 5 . a - v Name: 2 \ ID: A- l 12. Solve the given initial-value 1problem. Leave your result In implicit form. (15 points) xix—m: 3y“, ya): 13. -(Bonus.10 points) Solve the given initial-value problem. Leave your result 1n implicit form 41)) (7x =-C§S(x +y) y(0)=- y 0““ -x' - cosw “.355 494: 5c S :: co<v~+‘ . 9-541 i\$u+\ 3 . (L—LGTKOLM: K-‘rQ, (+(05m \-—Co\$V‘ ‘ u:-x+“x 'Jw = WA}:— ‘33 JL‘“:A_::A ) 1‘, - 1’ _ S i—wsft— A p. 7,. L-PQ. - [‘CU‘IN ‘S'MW ‘ . Q, , ~ f uJﬁggkawowx ._)<+C, _,’ Q. 60h; 4r Uh“- x‘\'¢’ cs «(if‘oA "c‘ﬂ'u‘yﬁJ , . 9c, .4. - \ .— C? C. (Tr/q. ’COJrC—‘Vbb ' KO 1-7 5 ~c. ’ ‘f -. ld—L —.\ .— / Answer: . ' (so (.K+~,q_co-\—CKHJ_) 5' 9‘ +61“ ...
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