[8]
1. a) In a perfect environment, the population of Norway rat that breeds on the MIT
campus increases by a factor of e ' 2.718281828459045 . . . each year. Model this natural
growth by a di erential equation.
What is the growth rate k?
[4]
b) MIT is a
18.03 Problem Set 2
I encourage collaboration on homework in this course. However, if you do your homework
in a group, be sure it works to your advantage rather than against you. Good grades
for homework you have not thought through translate to poor grad
18.03 Problem Set 1
I encourage collaboration on homework in this course. However, if you do your homework
in a group, be sure it works to your advantage rather than against you. Good grades for
homework you have not thought through translate to poor grad
18.03 Problem Set 3: First Half
This is the rst two problems of PS3. The rest will be available on February 26.
I encourage collaboration on homework in this course. However, if you do your homework
in a group, be sure it works to your advantage rather th
[8]
[4]
[8]
1. a) For what value of k is the system represented by x + x + kx = 0 critically damped?
b) For k greater than that value, is the system overdamped or underdamped?
c) Suppose a solution of x + x + kx = 0 vanishes at t = 1, and then again for
18.03 Pactice Final Exam, 2010
1. This problem concerns the dierential equation
dy
= x2 y 2
dx
()
Let y = f (x) be the solution with f (2) = 0.
(a) Sketch the isoclines for slopes 2, 0, and 2, and sketch the direction eld along them.
(c) On the same diagr
18.03 Practice Hour Exam I (2010)
1. A certain computer chip sheds heat at a rate proportional to the dierence between
its temperature and that of its environment.
(a) Write down a dierential equation controlling the temperature of the chip, as a
function
1. A certain periodic function has Fourier series
f (t) = 1 +
cos( t) cos(2 t) cos(3 t) cos(4 t)
+
+
+
+
2
4
8
16
[4]
(a) What is the minimal period of f (t)?
[4]
(b) Is f (t) even, odd, neither, or both?
[8]
(c) Please give the Fourier series of a perio
18.03 Problem Set 4: Part II Solutions
Part I points: 13. 3, 14. 8, 15. 5, 16. 4.
5+4i
13. a) [4] z + 2z = e 3+4i)t has solution zp = e 3+4i)t /(3 + 4i) + 2) = 25+16 e3t (cos(4t) +
5
4
i sin(4t) so xp = Re(zp ) = e3t ( 41 cos(4t) + 41 sin(4t). The homogen
18.03 Hour Exam I Solutions: February 24, 2010
1. (a) x(t) = number of rats at time t; t measured in years. x = kx. So x(t) = x(0)ekt . ex(0) = x(1) =
x(0)ekt implies k = 1.
x
x
x= 1
x.
(b) x = k 1
R
1000
x
3
(c) x = 1
x a. The pest control people hope
18.03 Recitation 23 May 4 2010
Linear phase portraits
The matrices I want you to study all have the form A =
a
2
2
.
1
1. Compute the trace, determinant, characteristic polynomial, and eigenvalues, in
terms of a.
The trace is the sum of diagonals: a 1. Th
x
y
x = (2
y = (x
x y)x
1)y
1.
y=2
x=0
x = x(2 x y) = 0
x=2 y=1
y = y(x
2
J =
2 0
0
1
(0, 0)
x
(1, 1)
2
y
(2, 0)
0
1
1
1) = 0
x=y=0
(0, 0) (2, 0)
(1, 1)
2x
y
y
2
0
x
x 1
2
1
1
3/2
1
0
2
y = 0
x = 0
x(t)
x(t)
(x(t), y(t)
y(t)
t
t
x = x(2
y =
y(t)
x
y
y
+
x
u(t)e t (t2 + 1)
L[e t ] =
1
s+1
L[t2 ] =
2
s3
L[e t (t2 + 1)] = L[e t t2 ] + L[e t ]
2
1
=
+
3
(s + 1)
s+1
2
s + 2s + 3
=
.
(s + 1)3
f (t) = e
t
F (s) = L[f (t)]
t
cos(3t)
0
f (t)
L[cos(3t)] =
s
s2 +9
F (s) = L[e t cos(3t)]
= L[cos(3t)](s + 1)
s+1
=
(s +
mx + bx + kx = 0
x(0)
x(0)
m
b
k
t=0
t
(x(t), x(t)
m = 0.50 b = 1.5
k = 0.625
k = 0.625 = 5/8
k = 0.62
x(0) = x0
x(0) = x0
( x 0 , x0 )
ce
rt
x(0) = x0
x0 = 0.25
x0
t
x0
t
x0
m = 0.50
k = .625
b
mx + bx + kx = 0
b
b = 0.25
m b
k
(x0 , x0 )
x(0) = 1 x(0) =
18.03 Problem Set 9: Solutions
Part I: 34. 4; 35. 10; 36. 6.
(a) [18] A =
0.5
1
2.25 0.5
has characteristic polynomial p A () = 2 +2.5, and eigenvalues
3i/2
1
v1 = 0.
9/4 3i/2
1
1
One choice is v1 =
. The normal mode is then e(1+3i)t/2
, which has real an
x = cos(!t)
x = sin(!t)
x + !2x = 0
x = cos(!t)
x = ! sin(!t)
x = ! 2 cos(!t) =
2
sin(!t)
x = ! cos(!t)
x = ! sin(!t) = ! 2 x
!2x
x =
!
A cos(!t
2
)
x+! x=0
x = A cos(!t
!2x
)
x=
A! sin(!t
x (t) = A cos(!t
x(0) = A cos
A = 0
x (0) = 0
cos = 0
= /2, 3/2, 5
u(t)
t>a
t<
a
a
t=0
u(t
b) = (t
b)
t
a<b
a
f (t)
(
a) d = u(t
a)
f (n) (a )
f (t)
0
fs (t)
f (t)
c
a
0
fr (t)
f 0 (t) dt = f (c)
f (t) = fr (t) + fs (t)
f (t)
f (t)
(f (a+) f (a ) (t a)
c+
f (a)
a
f 0 (t)
f 0 (t) dt = f (c+)
t
p(D)x = q(t)
x(t) = 0
t<0
f
18.03 Recitation 22, April 29, 2010
Eigenvalues and Eigenvectors
1. Well solve the system of equations
x =
y =
5x 3y
6x + 4y
a) Write down the matrix of coe cients, A, so that we are solving u = Au. What
is its trace? Its determinant? Its characteristic p
18.03 Problem Set 8: Part II Solutions
Part I points: 29. 4, 30. 0, 32. 8, 33. 4.
Part II solutions:
29. (a) [8] (1) cfw_1, i, i: For large t, these functions have exponential growth rate
et . (This means that for any a < 1 < c, eat < |f (t)| < ect for la
Recitation 5 February 18 2010
p
z = 1 + 3i
zn
n = 1, 2, 3, 4
Aei
r=2
z 2 = 4ei2/3 =
z4
2n ein/3
z
a + bi
n = 0, 1, 2, 3, 4
n
= /3
z2 p
2/3
3
2 + 2 3i z
4/3
n = 1, 2, 3, 4 z 0 = 1
z k
/3
p
8 3i z n
8
k/3
2
k
k = 1, 2, 3, 4
p
ea+bi = 1+ 3i
a+bi
b
en(a+bi)
18.03 Problem Set 7: Part II Solutions
Part I points: 26. 6, 27. 10, 28. 12.
I.26. e t sin(3t) =
1
1
2i s ( 1 + 3i)
1
2i
e
1+3i)t
1
e
1
( 1
, so L[e t sin(3t)] =
1 (s + 1 + 3i) (s + 1
=
2i
(s + 1)2 + 9
3i)t
3i)
3
.
s
3i)
(s + 1)2 + 9
Z 1
Z 1
st
26. a) [1
18.03 Problem Set 1: Part II Solutions
Part I points: 0. 9, 1. 6, 2. 4, 3. 4.
0. (T 5 Feb) (a) [4] The growth rate k(t) has units years1 (so that k(t)x(t)t has the
same units as x(t). The variable t has units years, so the a added to it must have the
same
18.03 Problem Set 5, First Half
The second half will be available by Monday, March 29.
I encourage collaboration on homework in this course. However, if you do your homework
in a group, be sure it works to your advantage rather than against you. Good grad
18.03 Problem Set 7
I encourage collaboration on homework in this course. However, if you do your homework
in a group, be sure it works to your advantage rather than against you. Good grades for
homework you have not thought through will translate to poor
18.03 Problem Set 2: Part II Solutions
Part I points: 4. 4, 5. 6 , 6. 8, 7. 6.
4. (a) [2] x falls; y rises and then falls; and z rises. With = .1, = .2:
(b) [4] Startium obeys the natural decay equation, x = x, with solution x = x(0)et .
To relate to its
f (t) = 2
0<t<
f (t)
f (t) = 0
2
1
a0 =
n>0
1
an =
1
=
2
2
<t<
1
f (t)dt =
/2
2dt = 2.
/2
f (t) cos(nt)dt
/2
0dt +
/2
2 cos(nt)dt +
/2
2
/2
sin(nt)| /2
n
4
=
sin(n/2)
n
n
0dt
/2
!
=
n
n
4
n
1
4
sin(n/2) = n
4
n
f (t)
bn = 0
1
bn =
1
=
4
n
n
3
n>0
f (t) si
f (t) = (u(t)
u(t
2) sin(t)
t
f (t) = e
t
+ 3e
3t
f (t)
f (t) = cos(2t) + e
F (s)
t
f (t) = 1
sin t
f (t)
f (t)
cfw_1, i, i
F (s)
cfw_ 1 + 4i, 1
4i
cfw_ 1
w(t) = u(t)e
W (s)
t/2
sin(3t/2)
b
k
|W (s)|
f (t) = g(t)
F (s) = G(s)
cfw_0
e
2s
G(s) =
cfw_ 1, 3
g
t
x (t)
a
t
t
k x (t)
k
t
a t
t) ' x(t) + k x(t) t
x(t +
x (t)
a t,
t
dx
= kx
dt
a.
a=0
a = 0
x = kx
dx
= k dt,
x
c kt
x 0
x (t) = Cekt
x (t) = e e
C
ln |x (t)| = kt + c
x = kx
C
x (0)
T
ekT = 2
k 6= 0
x=
x(T ) = 2C
C
T =
k =0
C
c
ln 2
.
k
a
x (t) =
at +