1. (20 pts) (a). Find the solution of the following initial value problem.
y"
+ y'
- 6y
=0
y(O)
y" - 4y'
= 2,
+ 4y = O.
y' (0)
= 9.
2. (20 pts) (a). Newton's law of cooling asserts that the rate at which an object
cools is proportional to the difference between the object's temperature (T) and the
temperature of the surrounding medium (A). It therefore satisfies the ODE
= -k(T(t)
T'(t)
Solve for T(t) in terms of A, To
the proportionality constant.
= the
- A).
temperature of the body at time 0, and k
=
(b) A murder victim is discovered at midnight and the temperature of the body
is recorded at 31°C. One hour later, the temperature of the body is 29°C. Assume
that the surrounding air temperature remains constant at 21°C. Calculate the vic-
.
approximately 37°C).
----~_.-----
L~t t::;O
ef ~~.
----_ _-~
----.---
b-t
..
l\1
lif)
=-
2.1
~t
~Wt~I\\~
1f
T (i,)-"3,
- 7
it=-
7t.e
-+
(of)
t
_-------_
..
_----~~
..
_--_.--._.
__
...
..
_----
<0
,ll)::
-/
..
,IOU .
J,,-
A ~ 2I
fttl'-f-hJAr,
It-
i
it
74-n
10::: ~ I .
l"
The "normal" temperature of a living human being is
_
tim's time of death. (Note.
=-
I-
f'oJM'\
ent/tilt
v/d,,'''1') tl\l'I1~
.
C
/;v e.
-H
e..
3/-L\)
2;
=-
t
(0.
2../
~H
2-1-1-(~1-11)e
({~ ) -t
~
= -
eh1=/en-f:
(~-2)
~J
-=
70
.
+ (0· (1
ro) ~{
&-cI,
Z
-)
3 7-- L/
~f
t
-tc
_I<
i2htl
~t
((0)
J... -t I--
Iu-un
/0
s .
(6 -- 1-
h~fr<l
VHIJ"fIJ'-i.
(~/ofYl1)
1
(c). Consider the equation
= f(at + by + c)
y'
where a, b, and c are constants. Show that the substitution x
= a + bf(x).
the equation to the separable equation x'
general solution of the equation y'
~
7'-'-
-
'w~
#
=-
C\.t
o..+hfcClt4bci+C)
+ bd
tC ) =
0-+
d'::: ('jf-t
71-
t\
~
r;
ef~~
~
=-
J .eA.£~(
x'L.
tI.
ott
S~(+x~ ::::Jr tlt
X -:::: f&V\
::
iv
e'f"A+ ~~
(+x,..
C;-ll\.(~
tt.
~ll.t.t' v<>.W-
X ( ::. (J' + I ::: (J-i t )')..
tI
;)
:::CI.+bfrK).
((~~{i: .
Lbt
7 A~)
-B-.-~
f(At+ b»+c)
b~=
f(
Let
+L
b~
tU.
A ~
lA I
elf -
4f-=~t
Use this method to find the
= (y + t?
~t +bd
'X-=-
= at + by + c changes
(t + C)
/)..
.
d~ tt\A(tfC)-t.
d:: 'X~l,
sol ~-r+t\.o'YJ
= t+ C
C•...
rCfM)(
~
~'=- (j-f
T y"
.
_
1
3. (15 pts) FACT: The function y
differential equation
=
(x2
+
1)e-X2
(a). Using the above fact, find the general solution
equation
is a solution to the ordinary
y(x)
to the ordinary differential
value
y(O)
I
<:
1(0)::: (u
=-
=
1.
+ I) e-0 { J(J
( 1)
f
C)
C
~('f}:o l"f~+\){x'L(i(}(X"tl)
t C)
_
1
4. (15 pts) Suppose that the functions u and v are solutions to the linear, homogeneous
equation
y"
+ p(y)y' + q(t)y = 0
in the interval (a, (3). Prove that the Wronskian of u and v is either identically equal
to zero on (a, (3), or it is never equal to zero there.
f
r
I tu - I tr-) ,
_
5. (20 pts) (a). Show that the following equation is exact and solve it.
(2x
+ y)dx +
{)
-{)Y (2x
(x - 6y)dy
=0
{)
+
y)
= 1 = -{) (x
.7:
- 6y)
implies that the equation is exact. The solution is
1
= 0 has
(b). Suppose that (x+y)dx+2xdy
of x alone (i.e f1 = f1(x)). Find the integrating
an integrating
factor that is a function
factor and use it to solve the differential
equation.
--- "-'--------------~._------Suppose
/lex)
---
-----
..
--- - --------------------.-------..
..
is an integrating factor. We need
Jt(x)(x
+ y)d'r + ji,(x)
(2x)dy
=0
to be exact. So we have
Le.
We have solution for this ODE,
Now the equation becomes
(y'x
+ yj
y'x)dx
+ 2y'xdy = 0
Therefore we can get the solution
-----
..
~
6. (10 pts) The undamped system
2
+ kx = 0
-x"
5
is observed to have period
7f
x(O)
= 2,
x'(O)
= Vo
/2 and amplitude 2. Find k and vo.
----- -.------------------------27r k -- Wo2 -- 16 an d th erelOre
1:
k -- 5"'
32
Th e equa t'Ion
Wo 7r12
- 4', so 275
becomes x" + 16x = 0, with general solution x(t) = Cl cos4t + C2 sin4t. But
we know that x(O) = 2 = amplitude, hence x(t) = 2cos4t and therefore
Vo
= x/CO) = O.
1