MATH 2280-001
EXAM 1
Instructions. There are 5 problems, each worth the same nuniber of points. No
calculators or electronic devices ale allowed on the exam. You are allowed one side
of an 8.5 by Ii inch sheet of notes.
1. Consider t he ODE:
=
(IX
a. Find a general
b. Find
solution.
a solution satisring
unique (find
another
the initial
condition.
y(O)
=
0. Show that this is
not
sohition).
c. Explain why the Existence-Uniqueness Theorem does not hold.
I
3r
Zi
5e/aAA
LI
a
)C
I
I
—
—
3
c
9
sz
7
C.
or
OD.€I
_/
-f
15•
IHh14jk-9
1
4
4Lf
;Ii
3
?fl
red/€
(1
Co,O)
1’2D
2. (‘onsidei lift’
iiiflt’it’iil iai
ahltOllnhift)tlH
=
a. I’i11(i tiit (‘(1111111>1i111)1 501111
his.
b. D’l (‘riiiinc’ which
xl HI) i(,
I )11cS Hi’(’
(‘(jll)Il 101)
—
—
H 11(1
Wlli(11
3.r
011(’S
arc
ii ustal )1(’
c. Skel (‘11 tiic’ xkctcii I hc i’cp jill ri I 111 sol ul bus, as wcll as tim solutions that satisfy
tim aol 11)1 condil iOlls :r(O) = —1, .r(O) = 1/2, r(O) = 2 auud .r(O) = 4.
f)
12ti4:)
4
EL;flJ
h
DItt h
y x
—
—
Q/’2
X
3
9x
=
—
x(x t3) 3
—
ç°
c)
II;(d1114
egj/;/, reø
Q?;k
C/Qd4flhil
S-oIti’iS,
‘flç,i2(’
0
/
)1;
\fç)
ne
12.
I1,
x<3
M. pos. g
4
7
0
4
1•
/
y
c.z-)
—I
3. Consider a brine tank which holds 15,000 gallons of coiithxuously mixed liquid.
Let :c(I ) I o’ the amount of salt, in the tank at time t (in hours). The inflow and out
low rates are both 150 gallons per hour, and the concentration of salt flowing iii is
1 1)01111(1 CI II) gallons of vat er.
a. Front thus
iirloriiial
ion explant how .r satisfies the difierential equation
± 0.01:r
b. If there is tin salt in the tank
at
time t
=
=
15
0, find x(t)
c. What is the limit ilig amount of salt as f approaches infinity?
A
=
—
It
hi-.
O9in
4
=
1 tih
frj
(Q
1iPwrdL
o
1O
â
5
.
3
/A50
/D
0 0/
e
‘i
‘
+
ooie°
-
r
ie
I_ Se00)h/0tA
(e°°’c)’z
,
ODI
.
4
00)
-ô
C
1
d
o)
(/5()
i)
O’
.j’ V
—
1
.J
I’5
1
Jo
_
_
’é
/
/
e°°”
4a. Find the general solution of t11(’ (liflerelitial
y” + 2j/’ + 2?)’
I). Solve the
equation
(I
=
e(1110t ion
+ 2/ + 2y
wit ii the mitial couditions i(O)
1 and AU)
=
=
()
—1.
r
Ck”4h cikL
3
t2r tr-c3
r(rz2
1—
t-tQ
d
z
::-
-1
I)
WI
ic/LShn
—
°
i
F
3
e
fx
-(e
2
c,f
e
c
iirA
,
5/ni
AJr4
r
7)
1
(eV Itk-’ Is
J,<
,
-_fx
‘)
4d1)1
4ce
—7
—
—
Cié
5
—
-x
C Si 4X 4
i
C,OD
t
=
—/
J/i/A
)
e
6*
L
—
C
C;
Cl
-i
-
)1
Ii
c:i)
C—,
N
)
7
cJ
-4-
I’
cj