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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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