AMath 350
Assignment #7
Fall 2014
Solutions
1. ~ 0 = A~ + F (t);
x
x ~
A=
2
1
1
2
~
, and F (t) =
8te
6
t
.
Solution:
We rst need to nd the solution to the corresponding homogeneous equation:
A
I=
2
1
1
2
so the characteristic equation is ( + 2)2 1 = 0, o
y = 3x cos2 (y )
,
22
y
sec2 y dy =
tan y =
3x2
+C ,
2
y = tan1
y = 8x3 ey ,
dy
= 3xdx
cos2 y
3xdx
3x2
+C
2
y.
y (1) = 0
ey dy =
8x3 dx
ey = 2x4 + C
y = ln 2x4 + C .
1 = 2 + C
C = 1
y = ln 2x4
AMath 350
Assignment #6
Fall 2011
Due Monday, October 31st
1.
a) Show that the boundary value problem
y + ky = 0,
y (0) = 0,
y (0) = 1
has a solution for every value of k .
b) Find the eigenvalues and eigenfunctions for the boundary value problem
y + 2ky
AMath 350
Assignment #1 (Review)
Winter 2015
Solutions
C
u = y, dv = sin (y) , du = 1, v =
Z
Z
y sin (y) dy = y cos (y) + cos (y) dy
=
u=t
cos (y)
y cos (y) + sin (y) + C.
2 du = dt
Z
4
Z
1
dt =
4t + t2
Z
t
2
du =
u = ln (2x)
ln (2x)
dx =
x
2
sin (3) d =
AMath 350
Assignment #2
Winter 2015
SOLUTIONS
1. (Are the given equations separable, linear, neither, or both?)
a) Not separable, but linear.
b) Separable, but not linear.
c) Both.
d) Neither.
e) Both.
f) Separable, but not linear.
2.
a) Find the solution
AMath 350
Assignment #3
Winter 2015
Solutions
1. (Find all solutions to the following equations:)
a) x2
dy
= xy
dx
y2
dy
y y2
=
, and we
dx
x x2
y
can see that it is a so-called homogeneous equation. So, we let u = ,
x
and then, in the usual way, we can w
Dierential Equations
for Business and Economics
Course Notes for AMATH 350
Sue Ann Campbell
Department of Applied Mathematics
University of Waterloo
Winter 2013 Edition
Edited by David Harmsworth
c S.A. Campbell 2010
Contents
1 Introduction
1
2 First Orde
AMath 350
Assignment #5
Winter 2015
Solutions
1. Solve y 0 + 3y = 5e
3x
+ 6xe3x :
By inspection, the complementary function is yh = Ce
3x
.
For a particular solution, the 5e 3x term would normally require a term of the
form Ae 3x , but since this matches
AMath 350
Assignment #6
Winter 2015
Solutions
1.
a)
ax2
d2 y
dy
+ bx + cy = 0
2
dx
dx
Let x = ez (which, of course, means z = ln x, if x > 0). Then
dy
dy dz
1 dy
=
=
.
dx
dz dx
x dz
We also need the second derivative:
d2 y
d dy
d 1 dy
=
=
dx2
dx dx
dx x
Theorems on Ordinary Dierential Equations
(Proofs as Presented in Class)
1. Existence and Uniqueness Theorem for IVPs Involving 1stOrder Equations
(see course notes - proof not discussed)
2. Existence and Uniqueness Theorem for IVPs Involving Linear Equat
AMath 350
Assignment #2
Fall 2015
due Friday, September 25th
Hand in #2, 4 and 5.
1. Are the following equations separable, linear, neither, or both?
a)
b)
c)
d)
e)
f)
2.
dy
= y + ex
dx
dy
= ex sin(y)
dx
dy
= 5y 2
dx
dy
= x y2
dx
dy
x2
=y
dx
dy
y
=x
dx
a)
AMath 350
Assignment #7
Winter 2015
Solutions
1. ~ 0 = A~ + F (t);
x
x ~
A=
2
1
1
2
~
, and F (t) =
8te
6
t
.
Solution:
We rst need to nd the solution to the corresponding homogeneous equation:
A
I=
2
1
1
2
so the characteristic equation is ( + 2)2 1 = 0,
AMath 350
Assignment #8
Winter 2015
Solutions
1. Following the hint, we have
Z 1
Z
(x x2 ) dx =
e
0
1
0
=e
Now, letting u = x
1
2
2
e (x x) dx =
1
4
Z
1
e (x
Z
1
0
h
x
e (
1 2
2
)
1
4
i
dx
1 2
2
) dx.
0
, this becomes
e
1
4
Z
1
2
e
u2
du
e
u2
du.
1
2
and
AMath 350
Assignment #3
Fall 2015
due Monday, October 5th
Hand in #1-4.
I strongly recommend trying #5 as well, but #6 and #7 are primarily for your own
interest.
1. Find the general solutions to the following equations, and sketch a representative set of
AMath 350
Assignment #1 (Review)
Fall 2015
Not for Submission
1. Calculate the following indenite integrals:
Z
a)
y sin (y) dy
Z
1
b)
dt
4 4t + t2
Z
ln (2x)
c)
dx
x
Z
d)
sin2 (3) d
Z
1
e)
du
9 u2
Z
1
p
f)
dx
1 x2
Z
g)
tan () d
2. Determine if the followin
University of Waterloo
AMath 350
Midterm Examination
Winter 2015
Saturday February 28th
9:30am - 11:00am or 10:30am-12:00pm (90 minutes)
Name (print):
I.D. Number:
Signature:
Closed book.
Math-Faculty approved calculators permitted.
MARKS
Question
Marks A
AMath 350
Assignment #7
Winter 2014
Solutions
1. ~ 0 = A~ + F (t);
x
x ~
A=
2
1
1
2
~
, and F (t) =
8te
6
t
.
Solution:
We rst need to nd the solution to the corresponding homogeneous equation:
A
I=
2
1
1
2
so the characteristic equation is ( + 2)2 1 = 0,
AMath 350
Assignment #8
Winter 2014
Solutions
1. Following the hint, we have
Z 1
Z
(x x2 ) dx =
e
0
1
0
=e
Now, letting u = x
1
2
2
e (x x) dx =
1
4
Z
1
e (x
Z
1
0
h
x
e (
1 2
2
)
1
4
i
dx
1 2
2
) dx.
0
, this becomes
e
1
4
Z
1
2
e
u2
du
e
u2
du.
1
2
and
AMath 350
Assignment #9
Winter 2014
Due Wednesday, April 2nd
Hand in #4, #5, and #6ab.
1. Use the denition or Theorem 5.31 of the Course Notes to determine if the
Fourier Transform of each of the following functions exists.
a) f (x) = x
2
b) f (x) = xe x
AMath 350
Assignment #6
Winter 2014
Solutions
1.
a)
ax2
d2 y
dy
+ bx + cy = 0
2
dx
dx
Let x = ez (which, of course, means z = ln x, if x > 0). Then
dy
dy dz
1 dy
=
=
.
dx
dz dx
x dz
We also need the second derivative:
d2 y
d dy
d 1 dy
=
=
dx2
dx dx
dx x
Must-Know Formulas for Calculus
In Math 117 and Math 119 we have tried to encourage you to concentrate on understanding the material,
rather than memorizing formulas. However, there are a few things that you simply have to KNOW, by heart.
Trying to write
AMath 350
Assignment #4
Fall 2014
Solutions
1.
a) y 00 7y 0 + 12y = 0
Solution: The characteristic equation is m2 7m + 12 = 0, which gives
(m 3)(m 4) = 0 =) m = 3, 4. Therefore the general solution is
y = c1 e3x + c2 e4x .
b) y 00 8y 0 + 16y = 0
Solution:
AMath 350
Assignment #1 (Review)
Winter 2014
Solutions
1. In all solutions, C is an arbitrary constant.
a) Use integration by parts: with u = y, dv = sin (y) , du = 1, v =
we nd
Z
Z
y sin (y) dy = y cos (y) + cos (y) dy
=
b) Let u = t
cos (y),
y cos (y) +
AMath 350
Assignment #6
Fall 2014
Solutions
1.
a)
ax2
d2 y
dy
+ bx + cy = 0
2
dx
dx
Let x = ez (which, of course, means z = ln x, if x > 0). Then
dy
dy dz
1 dy
=
=
.
dx
dz dx
x dz
We also need the second derivative:
d2 y
d dy
d 1 dy
=
=
dx2
dx dx
dx x d
AMath 350
Assignment #5
Fall 2014
Solutions
1. Solve y 0 + 3y = 5e
3x
+ 6xe3x :
By inspection, the complementary function is yh = Ce
3x
.
For a particular solution, the 5e 3x term would normally require a term of the
form Ae 3x , but since this matches th
University of Waterloo
AMath 350
Midterm Examination
Winter 2014
Friday February 28th , 4:00-5:30pm or 5:00-6:30pm (90 minutes)
Name (print):
I.D. Number:
Signature:
Closed book.
MARKS
Question
Marks Available
1
7
2
8
3
16
4
9
5
10
6
10
Total
60
Marks Awa
AMath 350 Course Outline Fall 2015
Differential Equations for Business and Economics
Lectures: Mondays, Wednesdays, and Fridays, 12:30-1:20, MC 1085.
Text:
Course Notes, by Sue-Ann Campbell, available from MC 2018.
Instructor:
David Harmsworth
dlharmsw@uw
AMath 350
Possible Final Exam Content
Methods and Concepts
Youre responsible for everything weve discussed, except for the method for solving
PDEs known as separation of variables:
ODEs:
1st-order Equations
separable equations
linear equations (integr
7. Consider a commodity with constant supply S = 50 and demand curve
whose derivative is given by
dD
= 2D 5p + 10
dt
AMath 350
Assignment #7
Fall 2015
where D is the demand and p is the price. The rate of increase of the
th
Wednesday,
November
price is eq