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exam1_solution - MA 262 Exam I September 23 wk“ Name:...

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Unformatted text preview: MA 262 Exam I September 23 wk“ Name: PUID: — Section: — SHOW ALL YOUR WORK. NO CALCULATORS, BOOKS, OR PAPERS ARE ALLOWED. Points awarded 1. (15 points) 2. (15 points) 3. (15 points) 4. (10 points) 5. (15 points) 6. (15 points) 7. ( 15 points) Total Points: 1. (15 points) Solve the following initial value problem. You may leave your solution in implicit form. (1/2 + cos :12)d3: + (2mg + Sin y)dy = O, y(0) : 7r. L , 3,22? M:fl+wx 5 9/3 7“Ul jib amt! N: ZN‘I—rw‘y 5? 33-57227 345 .. .mx *2 I a! u ) :I L \ ‘ “:5 M + W X? W1 1' 3 2. (15 points) Find the general solution of ‘ \- mzy’ : x2 + my + y2. grim. 3/: Higafigwl Wmfiwj at we? (32w :5 gel/Ha; :5 \H’ ngvg ~ )+\J+UL 1% 3(321HUL “U710er “:5 "Thom: ~32}?— 5 mix) : Mme :5 v: EMMHHC) 3 .33 [9 :x-m (In/>4“) vr—y ‘ w»: 3. (15 points) Find the general solution of 23: MI W” £7NLJ’1: y1+1+x2y=$yz 1[X)G€, 4. (10 points) The number of a certain cell culture grows at a rate of proportional to the number present. Initially there were 100 and it has a doubling time of 4 hours, determine the time it takes for the culture to contain 3200 cells. Flat/N "(33¢ PCP/m 2) :, lhz/ )1 q/ 1,»: 3; /\J\?Q, I /hc':rt_ : ém 1-,!qu ram fl 1 ‘>< 1 -- AMA: L: 5 3L; 9] ‘Lw “2+3? \ A: [3: (j) AL‘L/)‘ *\ )(LL—M-X =3 X34 W XZL} / X+$j ~waW/5'S‘J’ 4,qu => 512/ W 93“! 6. (15 points) Solve the general solution to l y”+ Ey’ : 93:, a: > 0. 7. (15 points) A tank with a capacity of 30 gal originally contains 10 gal of water with 100 lb of salt in solution. Water containing 1 lb/gal salt is entering at a rate of 3 gal/min, and the well mixed mixture is allowed to flow out of the tank at a rate of 2 gal/min. Let A(t) be the amount of salt in the tank at time t and V(t) be the volume of solution in the tank at time t. Find the amount of salt in the tank just before the solution overflows. °‘ V01“) 3041 /-l~« =1“ C: =l IA/jq'l Y’ 33 r2 algal/“W: A 62:”;— Nw‘ «a jam/l AM) new V 7 th‘bfivmwew V(T):goja{ olv ._—-—-—: YJ-vY : --r2 '5 ' 0H; 1 3 l :2) V: /o+b E :' 77—2/0 )m orzuwalo VtT)‘=«3° OM 2A 1/»! W: CIrJ—CZ r1, : 3“ V : 3)— hfi-b‘ h (Jr/WdJ/V [law/L"- o/A 2 , 0H? + Io-f't A “‘2'. '2. j oft .r‘l't 1(_t,: Q U 3.: ol ' ‘2 _ 1 :2 Unfit) A] " 3 (hr-Hg) ~§ (10+ht)2/Ql: j3Uu+t51cft+C :(/u+‘t)5+c C =5 A; N+t+ 6WD" f9 (:2 7m A0: —:> A : {0+t+ 'OOFt); Z/leO):5D.+ 331% :lédouér} 5 ...
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This note was uploaded on 03/01/2012 for the course MA 262 taught by Professor Ber during the Fall '08 term at Purdue.

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exam1_solution - MA 262 Exam I September 23 wk“ Name:...

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