Math 33B: Dierential Equations
Practice problems for nal: Section B
June 2013
Instructor: Aliki M
1. Consider the initial value problem given by,
dy
= y 2 , with y (0) = y0 .
dx
(a) Show that there exists a unique solution for every initial condition.
(b)

Solution
Section 1.5 Mixing Problems
Exercise
Consider two tanks, label tank A and tank B for reference. Tank A contains 100 gal of solution in
which is dissolved 20 lb of salt. Tank B contains 200 gal of solution which is dissolved 40 lb of salt.
Pure wa

9/28/16: Lecture 3
dx
t
=
x
dt t+ 1
Example
This is a separable equation
dx
=f ( t ) g( x ) where
dt
f ( t )=
Note that
t
, g ( x )=x
t+1
g ( 0 )=0 and so we have the constant solution
x ( t )=0 for all t ( ,) .
x 0 we separate variables
dx
t
=
dt
x t+ 1

11/23/16: Lecture 24
Phase Plane Portraits
'
x =Ax
(
x : R R 2 , A=
a11 a12
a21 a22
The Eigenvalues solve
)
with constant real entries
2T+ D=0
T T 2 4 D
2
T + T 24 D
2 =
2
1 =
Case 3
2
T 4 D=0
1= 2=
We have
T
.
2
Case 3a
The eigenspace is generated by 2 l

10/7/16: Lecture 7
Example
Show the equation
(cos ( x ) sin ( x )x y 2) dx + y (1x 2 )dy =0
is exact on
R=( , ) ( , ) .
Solve this equation subject to the initial condition
y ( 0 )=2 .
2
P ( x , y )=cos ( x ) sin ( x ) x y
Q ( x , y )= y ( 1x 2 )
P( x , y

9/23/16: Lecture 1
First-Order Differential Equations
Definition
A first-order differential equation takes the form
dx
=f (t , x )
dt
t is the independent variable
x=x (t) is a function of t
The equation says that the rate of change of the function
x as t

HOMEWORK 2
Section 2.4 in the book: Exercises 4, 16, 22, 28, 40.
Section 2.5 in the book: Exercises 8, 12.
Problem 1. (a) Let a R be a constant and let b1 : R R and b2 : R R be
two continuous functions such that b1 (t) b2 (t) for all t 0. Consider the l

11/16/16: Lecture 21
A=
(
)
a11 a12
, T =a11 + a22 , D=det A=a11 a22a12 a 21
a 21 a22
x ' =Ax
2
T+ D=0
Eigenvalues solve the characteristic equation
1,2 =
Case 3
Case 3a
T T 2 4 D
2
T 2 4 D=0 =
T
2
with multiplicity
2 .
is generated by 2 linearly indepe

HOMEWORK 8
Problem 1. For each of the following matrices, perform the following tasks:
a) compute the eigenvalues and eigenvectors and write down the general solution;
b) sketch the phase plane portrait.
6 3
7 0
2 4
1. A =
2. A =
3. A =
4 2
0 7
1 2
3 2
2

11/21/16: Lecture 23
Phase Plane Portraits
'
x =Ax
x: RR
A=
(
2
a11 a12
a 21 a22
)
real entries
Eigenvalues solve
2
T+ D=0
T T 2 4 D
2
T + T 24 D
2 =
2
1 =
Case 1
Case 1a
Case 1b
Case 1c
Case 1d
2
T 4 D> 0
1 <0< 2
The origin is a saddle pointunstable equ

11/7/16: Lecture 18
Linear Systems with Constant Coefficients
Consider a second-order constant coefficient equation
'
'
x + p x +qx=f (t)
We can write this as a first-order equation for a vector.
Indeed, let
v =x '
Compute
() ( ) ( ) (
)(
d x
v
x'
v
v
0
0

11/18/16: Lecture 22
Phase Plane Portraits
(
a11 a12
2
'
x =Ax with x : R R , A=
a21 a22
Consider
Definition
An equilibrium point is a vector
)
with real entries
v 0 R2 such that A v 0=0 . The solution x ( t )=v 0 for all
t R is called an equilibrium solu

9/26/16: Lecture 2
dx 1 3
= x with x ( 0 )=1
dt 2
Example
Rewrite this equation as follows:
dx
dt
1= 3 when x 0
x
2
Integrating with respect to
t ,
'
2 x ( t ) 1
t+C= 1 dt= 3 = 2 for some C (, ) constant
x ( t ) x (t )
x 2 ( t )=
Note that no choice of
Th

10/24/16: Lecture 13
Linear Second-Order Equations With Constant Coefficients
Consider the following equation
'
'
x + p x +qx=0
with
p , q ( , ) constant.
Seek solutions of the form
x ( t )=e t
'
'
x (t ) + p x ( t ) + qx ( t )=0
2 e t + p e t + q e t =0

Math 33B: Differential Equations, Lecture 1, Spring 2017
Class CCLE page: https:/ccle.ucla.edu/course/view/17S-MATH33B-1
Instructor: Eric M. Radke, Ph.D. ([email protected]). Instructors Office: Math Sciences 5226.
Textbook: Polking, Boggess, and Arnold

11/9/16: Lecture 19
Consider
'
x =Ax
x: RR
where
n
,
A
is an
n n matrix with constant real coefficients
x ( t )=e t v is a solution if is an eigenvalue of A
We saw that
and
v is an eigenvector with
.
eigenvalue
Finding Eigenvalues and Eigenvectors
to be

10/12/16: Lecture 9
Autonomous Equations and Stability
Definition
An autonomous equation is an equation of the form
dx
=f (x ) .
dt
Without finding explicit solutions, we are interested in answering qualitative questions about the solutions, such
as:
Are

10/21/16: Lecture 12
Theorem
I be an open interval on (, ) containing t 0 and assume
continuous. Let 1 , 2 be two solutions to
Let
p , q :I (, ) are
x ' + p ( t ) x' +q ( t ) x=0.
Then all solutions are of the form
c 1 1 +c 2 2 for some c 1 , c 2 ( , ) if

10/5/16: Lecture 6
Exact Equations
Consider a first-order differential equation of the form
P ( x , y )+ Q ( x , y )
where
x is the independent variable,
dy
=0(1)
dx
y= y ( x ) is the unknown function of x , and
P ( x , y ) , Q(x , y) are two given functi

10/28/16: Lecture 15
Theorem
I be an open interval and let
Let
p , q , f : I ( , ) be continuous. Assume
x p : I ( , ) is a particular solution to the inhomogeneous equation
x ' + p ( t ) x' +q ( t ) x=f (t)
Assume also that
1 , 2 :I ( , ) form a fundame

11/1/16: Lecture 17
Variation of Parameters
I ( , ) be an open interval and let
Let
p , q , f : I ( , ) be continuous. We consider
x ' ' + p ( t ) x' +q ( t ) x=f ( t ) .
(Important: The coefficient of
x' '
is
1 !)
cfw_1 , 2 for the associated homogeneou

10/3/16: Lecture 5
Linear First-Order Differential Equations
Further
Remarks
dx
=f ( t ) x (1)
dt
Solutions:
x ( t )= A e F (t ) where A ( , ) constant and F' ( t ) =f (t )
From the formula, its clear that solutions to the homogeneous equation differ by a

10/10/16: Lecture 8
Existence and Uniqueness of Solutions
Example 1
t
dx
=4 x +t 4
dt
The general solution for this equation is
cfw_
t 4 ln|t |+C t 4 , t 0 for some constant C ( , )
.
0 , t=0
The interval of existence is
I =( , ) .
t
This means that the i

10/26/16: Lecture 14
Consider
'
'
x + p x +qx=0
Characteristic equation
2
+ p+q=0
Characteristic roots
1,2 =
Case 1
p p24 q
2
2
p 4 q> 0 1 , 2 R and 1 2
Fundamental set of solutions
1 ( t )=e
1 t
2 ( t )=e
and
2 t
General solution
1t
2 t
=c1 e +c 2 e

10/12/16: Lecture 10
Review
Exercise
Consider the differential equation
x ( , ) there exists a unique solution to the equation satisfying the
(a) Explain why for any
initial condition
dx
=x sin x
dt
x ( 0 )=0 , at least on some time interval containing 0

9/30/16: Lecture 4
Linear Homogeneous Equations:
dx
=f ( t ) x
dt
We found the solutions
x ( t )=0 for all t ( , ) . The interval of existence is (, ) .
x ( t )= A e F (t ) with A 0 constant and F' ( t ) =f ( t ) .
We wrote this in one compact formula
Rec

HOMEWORK 3
Section 2.6 in the book: Exercises 10, 12, 26, 28, 30.
Section 2.7 in the book: Exercises 8, 10, 30, 32.
Section 2.9 in the book: Exercises 10, 18, 22.
Problem 1. Consider the differential equation
dx
= |x|1/2 (1 x).
dt
(a) Explain why for a

11/14/16: Lecture 20
A=
(
a11 a12
a 21 a22
)
x : R R2
'
x =Ax
Consider
Eigenvalues are solutions of the characteristic equation
det ( Id A)=0 2T+D=0
T =a11 + a22
D=det A=a11 a22 a12 a21
1,2 =
Case 1
T T 2 4 D
2
T 2 4 D> 0
1 , 2 are distinct real eigenvalu

10/19/16: Lecture 11
Second-Order Differential Equations
Definition
A second-order differential equation takes the form
2
d x
x = 2 =f ( t , x , x ' )
dt
'
An initial condition for this problem is of the form
cfw_
x ( t 0 )=x 0
x ' ( t 0 ) =x 1
A solution

10/31/16: Lecture 16
Remark
Consider the constant coefficient homogeneous equation
x ' + p x ' +qx=0 (1)
The characteristic equation is
2
+ p+q=0
( 1 ) ( 2 ) =0
where the characteristic roots are
1,2 =
p p24 q
2
(1) as follows:
We can rewrite equation
(

12/2/16: Lecture 27
Recall
e tA =
l0
Proposition
1 l l
t A
l!
A
Let
n n matrix and let v R n . Assume
be an
[
e tA v= Id +tA + +
A 2 v=0 , then e tA v=v+ tAv
if
A
Let
and
e
Then
A+ B
]
1
t k1 A k1 v
( k1 ) !
Av =0 , then e tA v=v
In particular, if
Proposi