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Section 1.1: Some Basic Mathematical Models; Direction Fields
A direction field (or a slope field) is a graph consisting of short line segments that indicate the
slopes of the tangent lines to the family of curves of a differential equation at a discrete number
of points.
An equilibrium solution is a solution to a differential equation that does not change (i.e., is a
constant). Equilibrium solutions are found by setting y = 0 (because the derivative of a constant
is zero) and solving the resulting equation.
A stable equilibrium solution is a solution that other solutions tend to converge toward.
Direction field lines point toward stable equilibriums.
An unstable equilibrium solution is a solution that other solutions tend to diverge away from.
Direction field lines point away from unstable equilibriums.
More comments on direction fields:




dy
 f t, y  .
dt
f is sometimes called the rate function (think of the exponential growth differential
dy
 ky , where k is the growth rate (or rate costant)).
equation
dt
Direction fields are nice because they do not require solving the differential equation.
Use a computer to do the work!
Direction fields are useful in studying equations of the form
Section 1.2: Solutions of Some Differential Equations
dy
b
 ay  b have a general solution y    ce at
Equations are of the form
dt
a
Notes:
 As we solve these equations by performing a single integration, a single constant of
integration will always be generated.
 To get a concrete value for the constant of integration, a desired point that the solution
should contain must be given.
 The given point on the curve is called an initial condition.
 A differential equation combined with an initial condition is called an initial value
problem (IVP).The solution obtained with the arbitrary constant of integration (due to no
specified initial condition) is called the general solution.
 All the graphs of the solutions for all possible values of the constant of integration are
called the integral curves.
 The solution to the equations in this section will be found by manipulating the equation
so that one side is an exact differential and then integrating.
 This will be analogous to the technique of “separation of variables” you may have
learned from section 7.2 of Smith & Minton’s 3rd edition of Calculus: Early
Transcendental Functions.
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Section 1.3: Classification of Differential Equations
Ordinary vs. Partial Differential Equations
Ordinary Differential Equations
 Only ordinary derivatives appear in the
equation
Example:
d 2Q  t 
dQ  t  1
L
R
 Q t   E t 
2
dt
dt
C
(RLC series circuit with a drive voltage)
Partial Differential Equations
 Only partial derivatives appear in the
equation
Example:
 2u  x, t  u  x, t 
2

x 2
t
(heat conduction equation)
Systems of Differential Equations
 …consist of multiple differential equations involving multiple variables.
 Used when one quantity depends on other quantities…
dx
 ax   xy
dt
 Example: Lotka-Volterra (predator-prey) equations:
dy
 cy   xy
dt
Order of a Differential Equation
 …is the order of the highest derivative that appears (analogy: the degree of a polynomial
is the largest exponent)
 General form for an nth order differential equation: F t , u  t  , u  t  , , u  n  t   0 for



some function F.
We’ll deal with differential equations that can be solved explicitly for the highest
derivative: y  n  t   f t , y  t  , y  t  , , y  n1  t  .


Linear vs. Nonlinear Differential Equations
 F t , y  t  , y  t  , , y  n  t   0 is linear if F is a linear function of y  t  , y  t  ,



, y  n t  ;
note: F does not have to be linear in t…
The general form of a linear differential equation of order n is:
an  t  y  n  t   an1  t  y  n1  t    a1  t  y  t   a0 t  y t   g t 


Nonlinear differential equations are not of the above form…
Many times, nonlinear equations can approximated as being linear over small regions of
the solution. This process is known as linearization.
Solutions of Differential Equations
 …are functions  such that all necessary derivatives exist over an interval of interest of
n
n 1
the independent variable and that satisfy     t   f t ,   t  ,    t  , ,     t  .



To verify a given solution, substitute it and its derivatives into the given differential
equation to show that the equation is satisfied.
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Section 2.1: Linear Equations; Method of Integrating Factors
dy
 p t  y  g t 
Form of a linear first-order differential equation:
dt
Formula for the solution of a first-order linear differential equations with constant
coefficients:
t
dy
 ay  g  t  , then   t   eat , and y  t   e at  eas g  s  ds  ce at
If
dt
t0
Formula for the solution of a general first-order linear differential equation:
t

dy
1 
 p  t  y  g  t  , then   t   exp   p   dt  , and y  t  
If

s
g
s
ds

c








dt
  t   t0

Section 2.2: Separable Equations
dy
 f  x, y 
dx
M  x
Separable first-order differential equation: f  x, y   
N  y
General form of a first-order differential equation:
Another form for a separable first-order differential equation: M  x  dx  N  y  dy  0
Solution:
M  x
dy

dx
N  y
 N  y  dy   M  x  dx
dy
y
 f  v  , where v  can be converted to a separable equation.
dx
x
Recall the following homework problem:
Note: equations of the form
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Section 2.3: Modeling with First Order Equations
Three Key Steps in Mathematical Modeling
 Construct the model (use the steps at the end of section 1.1).
 Analyze the model by solving it or making appropriate simplifications to solve it.
 Compare the results from the model with experiment and observation.
Section 2.4: Differences Between Linear and Nonlinear Equations
Linear Fist-Order Differential Equations
dy
 p t  y  g t 
dt
Solution:
First, compute   t   exp   p  t  dt 
Then answer is explicitly
1 
y t  
  t  g  t  dt  c 
 t  
Theorem 2.4.1: Existence and Uniqueness of
the Solution of a Linear First-Order
Differential Equation
If the functions p and g are continuous in an
open interval I containing the point t = t0, then
there exists a unique function y = (t) in I that
satisfies the initial value problem
y  p  t  y  g  t  and y  t0   y0 .
Nonlinear Fist-Order Differential Equations
dy
 f t, y 
dt
… we have only looked at one kind, which is
separable equations:
dy
 F  y  G t 
dt
Solution:
dy
First, separate variables:
 G  t  dt
F  y
Then integrate for an implicit equation
dy
 F  y    G t  dt
Theorem 2.4.2: Existence and Uniqueness of
the Solution of a Nonlinear First-Order
Differential Equation
f
If the functions f and
are continuous in the
y
rectangle containing the point (t0, y0), then
there exists a unique solution y = (t) in that
rectangle that satisfies the initial value problem
y  f  t , y  and y  t0   y0 .
The proof is rooted in the fact that we can write
the solution in terms of operations on
The proof is beyond the scope of this class. It
continuous functions.
should be noted that if the conditions of the
hypothesis are not satisfied, then it is possible
to get multiple solutions for one initial
condition
Note that if the uniqueness criteria are met, then graphs of two solutions can not intersect,
because if they did then that implies two solutions (i.e., contradicting uniqueness) for a given
initial condition.
Interval of Definition:
Interval of Definition:
…is determined by any discontinuities in p and …cannot be determined by looking at f.
g
… must look at the solution itself
General Solutions:
General Solutions:
…can always be written using the constant of
…many times cannot be summarized with the
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integration c.
Implicit vs. Explicit Solutions:
…the solution is stated explicitly as an
integral.
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constant of integration; example: equilibrium
solutions.
Implicit vs. Explicit Solutions:
…the solutions obtained are usually implicit.
Section 2.5: Autonomous Equations and Population Dynamics
In this section, we look at a restricted form of a nonlinear separable equation:
dy
 f  y .
dt
When f is independent of t, the equation is called autonomous (i.e., t does not drive the solution;
the solution only depends on itself…).
Logistic Growth Model
 Accounts for the fact that the growth rate depends on population  instead of
dy
 ry , use

dt
dy
 h  y  y for some function h.

dt
 We like h to model the fact that as populations get larger, the growth rate gets
smaller due to “using up” the environement.  h  y   r  ay is a simple model
that gets used extensively.
dy
  r  ay  y …called the logistic equation, and is usually written the form:

dt
r
dy
y

 r 1   y where K  is the environmental carrying capacity (or

a
dt
 K
saturation level), and r is the intrinsic growth rate.
dy
 0 , we are also solving
 When finding equilibrium solutions by solving
dt
f  y   0 ; zeros of f are called critical points.




A phase line graph shows the y-axis and how solutions either converge toward or
diverge from equilibrium solutions.
dy
For stable equilibrium solutions,
points toward the equilibrium.
dt
dy
For unstable equilibrium solutions,
points away from the equilibrium.
dt
dy
y

  r 1   y , T is called the threshold level for the
For he equation
dt
 T
population.
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Section 2.6: Exact Equations and Integrating Factors
In this section, we examine the equation M  x, y   N  x, y 
If once again the solution comes from   x, y   c , then
dy
0.
dx
d
d
  x, y    c
dx
dx
 dx  dy

0
x dx y dx
dy
 x  x, y    y  x, y   0
dx
Thus:
Theorem 2.6.1: Exact Differential Equations and Their Solution
Let the functions M , N , M y , and N x be continuous in some rectangular region R. Then the
equation M  x, y   N  x, y  y  0 is an exact differential equation in the region R if and only if
M y  x, y   N x  x, y  at each point of R. That is, there exists a function  satisfying
 x  x, y   M  x, y  and  y  x, y   N  x, y  if and only if M y  x, y   N x  x, y  .
The method of solving an exact equation is identical to finding the potential for a conservative
vector field…
Integrating Factors…
Sometimes a differential equation that is not exact can be converted to an exact equation by
multiplying the equation by an integrating factor   x, y  :
dy
0
dx
dy
  x, y  M  x, y     x , y  N  x , y   0
dx
For this equation to be exact, it must be true that
  M  y    N x
M  x  N  y
Which leads to the partial differential equation:
 y M    x N   N x
 y M  x N   M y  N x    0
If  can be found, then an exact equation is formed, whose solution can be obtained using the
prior technique. Unfortunately, the equation to find  is usually harder to solve than the original
equation (in general).
However, we can look for simplifications, like what happens if the original equation is
such that  depends only on x or y.
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Differential Equation where the integrating
factor  depends on x only (i.e.,     x  ):
  M  y    N x
d
N   Nx
dx
d  M y  Nx


dx
N
M y 
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Differential Equation where the integrating
factor  depends on y only (i.e.,     y  ):
  M  y    N x
d
M  M y   Nx
dy
d  Nx  M y


dy
M
Criterion for     x  only:
Criterion for     y  only:
M y  Nx
Nx  M y
is a function of x only.
N
Note that the equation for  is both linear and
separable now.
is a function of y only.
M
Note that the equation for  is both linear and
separable now.
Section 2.7: Numerical Approximations: Euler’s Method
y   f  x, y 
y 
y
x
y
 f  x, y 
x
y  f  x, y  x
yi 1  yi  f  xi , yi  x
yi 1  yi  f  xi , yi  x
Or, letting h = x, yi 1  yi  hyi
To perform Euler’s method quickly on your calculator:
 Store the initial conditions in variables X and Y.
 Enter the update formula for Y then for X separated by a colon.
 Repeatedly Use 2ND ENTER until you are at the final x-value.
 If you want to see the intermediate values of Y, you’ll have to type Y every time…
 Example: for y  x  e y , y(1) = -2, h = 0.25
1X
-2  Y
Y + (X + e^(-Y))*0.25  Y:X+0.25X
Y
2ND ENTER 2ND ENTER …
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Section 3.1: Homogeneous Equations with Constant Coefficients
Big idea: ay  t   by  t   cy t   0 has solution y  t   ert or y  t   tert for some values r.
Here are four fundamental second-order differential equations and their solutions:
y  t   0
y  t   c1t  c0
y  t   k 2 y  t 
y  t    2 y  t 
OR
OR
y  t   k 2 y  t   0
y  t    2 y  t   0
y  t   c1e kt  c2 e  kt
y  t   c1eit  c2 e  it
y  t   k1 cosh  kt   k2 sinh  kt 
y  t   c1 cos t   c2 sin t 
To solve a second-order differential equation with constant coefficients:
ay  t   by t   cy t   0
Assume a solution of the form y  t   ert , substitute it and its derivatives y  t   rert and
y  t   r 2ert into the equation.
ar 2 e rt  brert  ce rt  0
 ar
2
 br  c  e rt  0
Then find the values r1 and r2 that satisfy the resulting characteristic equation:
 ar 2  br  c   0
The general solution is a linear combination of the two solutions:
y  t   c1er1t  c2er2t
Section 3.2: Solutions of Linear Homogeneous Equations; the Wronskian
Big Idea: Any two solutions we find of a homogeneous linear second-order differential equation
are all that is needed if the Wronskian is non-zero on the interval of the existence of the solution.
Theorem 3.2.1: Existence and Uniqueness Theorem
(for Linear Second-Order Differential Equations)
For the initial value problem
y  p  t  y  q  t  y  g t  , y t0   y0 , y t0   y0
where p, q, and g are continuous on some open interval I containing the point t0, then there is
exactly one solution y    t  of this problem, and the solution exists throughout the interval I.
Theorem 3.2.2: Principle of Superposition
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If y1 and y2 are solutions of the differential equation
L  y   y  p  t  y  q  t  y  0 ,
then the linear combination c1y1 + c2y2 is a solution for any values of the constants c1 and c2.
The Wronskian: W 
y1  t0 
y1  t0 
y2  t 0 
 y1  t0  y2  t0   y2  t0  y1  t0 
y2  t0 
Theorem 3.2.3:
If y1 and y2 are two solutions of the differential equation
L  y   y  p  t  y  q  t  y  0
and the initial conditions y  t0   y0 , y  t0   y0 must be satisfied by y, then it is always possible
to choose the constants c1 and c2 so that
y  c1 y1  t   c2 y2  t 
satisfies this initial value problem if and only if the Wronskian W  y1 y2  y2 y1 is not zero at t0.
Theorem 3.2.4:
If y1 and y2 are two solutions of the differential equation
L  y   y  p  t  y  q  t  y  0
then the family of solutions
y  c1 y1  t   c2 y2  t 
with arbitrary constants c1 and c2 includes every solution of the differential equation if and only
if there is a point t0 where the Wronskian W  y1 y2  y2 y1 is not zero.
Notes:
 y  c1 y1  t   c2 y2  t  is called the general solution of L  y   y  p  t  y  q  t  y  0 .


y1 and y2 are said to form a fundamental set of solutions.
To find the general solution, and thus all solutions of L  y   y  p  t  y  q  t  y  0 , we
need only find two solutions of the given equation whose Wronskian is non-zero.
Theorem 3.2.5:
If the differential equation
L  y   y  p  t  y  q  t  y  0
has coefficients p and q that are continuous on some open interval I, and if t0  I, and if y1 is a
solution that satisfies the initial conditions
y1  t0   1, y1  t0   0
and if y2 is a solution that satisfies the initial conditions
y2  t0   0, y2  t0   1
then y1 and y2 form a fundamental set of solutions of the equation.
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Theorem 3.2.6: (Abel’s Theorem)
If y1 and y2 are two solutions of the differential equation
L  y   y  p  t  y  q  t  y  0
With the coefficients p and q continuous on an open interval I, then the Wronskian W  y1 , y2  t 
is given by:
W  y1 , y2  t   c exp    p  t  dt 
where c is a certain constant that depends on y1 and y2, but not on t. Further, W  y1 , y2  t  either
is zero for all t  I (if c = 0), or else is never zero in I (if c  0).
Section 3.3: Complex Roots of the Characteristic Equation
Big Idea: When there are complex roots to the characteristic equation, each of the fundamental
solutions will have a sinusoidal factor.
Euler’s Formula:
eit  cos  t   i sin  t 
Notes:
 The complex solutions from the quadratic formula always come in conjugate pairs.
  i t
  i t
 If those complex roots are r     i , then the general solution y  t   c1e   c2e 
can always be written as y  t   d1et cos  t   d2et sin  t  .
Additional topics from homework:
 A Euler equation is of the form t 2 y   ty   y  0 . It can be solved by making the
substitution x  ln  t  , which results in the constant coefficient equation

d2y
dy
   1   y  0
2
dx
dx
A general linear homogeneous equation y  p  t  y  q t  y  0 can be solved using the
substitution x  u  t    q  t dt if and only if
q  t   2 p  t  q  t 
 q t 
3
2
 constant .
ODE
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Section 3.4: Repeated Roots; Reduction of Order
Big Idea: When there are repeated roots to the characteristic equation, the second of the
fundamental solutions is formed by multiplying the first solution by a factor of the independent
variable.
Summary of Sections 3.1 – 3.4:
 To solve a second-order linear homogeneous differential equation with constant
coefficients:
ay  by  cy  0


Assume a solution of the form y  t   ert , which leads to a corresponding characteristic
equation:
ar 2  br  c  0
Solve the characteristic equation to find its two roots r1 and r2:
 If the roots are real and unequal, then the general solution is
y  t   c1er1t  c2er2t

If the roots are complex conjugates    i , then the general solution is
y  t   c1et cos  t   c2et sin  t 

If the roots are the same, then the general solution is
y  t   c1er1t  c2ter1t

If we have a more complicated second-order initial value problem like:
y  p  t  y  q  t  y  0, y  t0   y0 , y t0   y0 , and we know that y1 and y2 are solutions
of the differential equation, then we can be sure that they form a fundamental set of
y  t  y2  t 0 
0
solutions if the Wronskian is nonzero: W  1 0
y1  t0  y2  t0 

Reduction of Order (end of 3.4): If we know a solution y1 of the more general secondorder differential equation: y  p  t  y  q t  y  0 , then we can derive the second
solution y2 by assuming y2  t   v  t  y1  t  and substitute y2 into the original equation,


resulting in y1v  2 y1  py1 v  0 , which is a first order equation for v.
ODE
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Section 3.5: Nonhomogeneous Equations; Method of Undetermined Coefficients
Big idea: One way to solve ay  by  cy  g  t  is to “guess” a particular solution that has the
same form as g(t), then work out the value(s) of the coefficient(s) that make that guess work.
Theorem 3.5.1: The Difference of Nonhomogeneous Solutions Is the Homogeneous Solution
If Y1 and Y2 are two solutions of the nonhomogeneous equation
L  y   y  p  t  y  q t  y  g t  , then their difference Y1  Y2 is a solution of the corresponding
homogeneous solution L  y   y  p  t  y  q  t  y  0 . If, in addition, y1 and y2 are a
fundamental set of solutions of the homogeneous equation, then
Y1  Y2  c1 y1  t   c2 y2 t  ,
for certain constants c1 and c2.
Theorem 3.5.2: General Solution of a Nonhomogeneous Equation
The general solution of the nonhomogeneous equation L  y   y  p  t  y  q  t  y  g  t  , can
be written in the form y    t   c1 y1  t   c2 y2 t   Y t  m where , y1 and y2 are a fundamental
set of solutions of the corresponding homogeneous equation, c1 and c2, are arbitrary constants,
and Y is some specific solution of the nonhomogeneous equation.
Method of Undetermined Coefficients:
 When g(t) is an exponential, sinusoid, or polynomial, make a guess for the particular
solution that is the same exponential, sinusoids of the same frequency, or polynomial of
the same degree, except with arbitrary coefficients that you determine by substituting
your guess into the nonhomogeneous equation.
 This technique works because derivatives of these types of functions result in the same
type of function.
 If the form for the particular solution replicates any terms of the homogeneous solution,
then factors of t must be applied until there is no replication.
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Forms for the Particular Solution of ay  by  cy  g  t  for Assorted g(t):
g t   Pn t  (i.e., a polynomial of degree n)
g t   A cost   B sin t 
g t   A  e kt
g  t   ekt  Pn  t 
g  t   ekt  A cos t   B sin t  
y p  t   Rn  t  (i.e., a polynomial of degree n,
but with different coefficients than Pn)
Exception #1: r = 0 is a root of the
homogeneous equation 
y p  t   t  Rn t 
Exception #2: r = 0 is a double root of the
homogeneous equation 
y p  t   t 2  Rn  t 
y p t   C1 cost   C 2 sin t 
Exception: r = ±i are roots of the
homogeneous equation 
y p  t   t C1 cos t   C2 sin t  
y p t   C1e kt
Exception #1: r = k is a root of the
homogeneous equation 
y p t   C1tekt
Exception #2: r = k is a double root of the
homogeneous equation 
y p t   C1t 2 e kt
y p  t   Rn  t  ekt (i.e., the product of a
polynomial of degree n, and an exponential).
Exception #1: r = k is a root of the
homogeneous equation 
y p  t   tRn  t  ekt
Exception #2: r = k is a double root of the
homogeneous equation 
y p  t   t 2 Rn  t  ekt
y p  t   ekt  C1 cos t   C2 sin t  
Exception #1: r = k ± i are roots of the
homogeneous equation 
y p  t   tekt  C1 cos t   C2 sin t  
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g  t   Pn  t  cos t   Qn t  sin t 
y p  t   Rn  t  cos t   Sn  t  sin t  (i.e., the
sum of the products of distinct polynomials of
degree n with sinusoids).
Exception: r = ±i are roots of the
homogeneous equation 
y p  t   t  Rn t  cos t   Sn t  sin t  
g  t   ekt Pn t  cos t   ekt Qn t  sin t 
y p  t   ekt Rn  t  cos t   ekt Sn t  sin t 
Exception #1: r = k ± i are roots of the
homogeneous equation 
y p  t   t  e kt Rn  t  cos t   e kt S n  t  sin t  
Section 3.6: Variation of Parameters
Big Idea: The method of variation of parameters works by assuming a solution that is of the
form of a sum of the products of arbitrary functions and the fundamental homogeneous solutions.
The basic idea behind the method of variation of parameters, due to Lagrange, is to find two
fundamental solutions y1 and y2 of a corresponding homogeneous equation, then assume that the
solution of the nonhomogeneous equation is:
y  t   u1 t  y1 t   u2 t  y2 t 
with the additional criterion that u1  t  y1  t   u2  t  y2  t   0 .
Theorem 3.6.1: Solution of a Nonhomogeneous Second-Order Equation
If the functions p, q, and g are continuous on an open interval I, and if the functions y1 and y2 are
a fundamental set of solutions of the homogeneous equation
y  p  t  y  q t  y  0 corresponding to the nonhomogeneous equation
y  p  t  y  q t  y  g t  , then a particular solution is:
t
y2  s  g  s 
y s g s
Y  t    y1  t  
ds  y2  t   1
ds ,
W  y1 , y2  s 
W  y1 , y2  s 
t0
t0
t
where t0 is any conveniently chosen point in I. The general solution is
y  t   c1 y1 t   c2 y2 t   Y t  .
ODE
FINAL EXAM Review
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Section 3.7: Mechanical and Electrical Vibrations
Big Idea: Second order differential equations with constant coefficients provide very good
models for oscillating systems, like a mass hanging from a spring or a series RLC circuit.
Mu   u  ku  0
Recall: A cos t   B sin t   R cos t    , where R  A2  B 2 and
 B   0 if A  0
  tan 1    
.
 A   if A  0
Resonant (angular) frequency: 0 

2
1 2
Period: T  
f

k
M
Frequency: f 
Quasifrequency of the damped system:   0 1 
2
4MK
,
ODE
FINAL EXAM Review
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Section 3.8: Forced Vibrations
Big Idea: A damped spring-mass (or circuit) system described in section 3.7 that is driven by a
sinusoidal force will eventually settle into a steady oscillation with the same frequency as the
driving force. The amplitude of this steady state solution increases as the drive frequency gets
closer to the natural frequency of the system, and as the damping decreases. An undamped
spring-mass system that is driven by a sinusoidal force results in an oscillation that is the average
of the two frequencies and modulated by a sinusoidal envelope that is the difference of the
frequencies.
Forced Vibrations with Damping
In this section, we will restrict our discussion to the case where the forcing function is a sinusoid.
Thus, we can make some general statements about the solution:
The equation of motion with damping will be given by:
mu   u  ku  F0 cos t 
Its solution will be of the form:
u  t   c1u1  t   c2u2  t   A cos t   B sin t 
homogeneous solution uh  t 
"transient solution"
particular solution u p  t 
"steady state solution"
Notes:
 The homogeneous solution uh  t   0 as t   , which is why it is called the “transient
solution.”
 The constants c1 and c2 of the transient solution are used to satisfy given initial
conditions.
 The particular solution u p  t  is all that remains after the transient solution dies away,
and is a steady oscillation at the same frequency of the driving function. That is why it is
called the “steady state solution,” or the “forced response.”
 The coefficients A and B must be determined by substitution into the differential
equation.
 If we replace u p  t   U  t   A cos t   B sin t  with u p  t   U  t   R cos t    ,

m 0 2   2 
2
F0

then R  , cos   
, sin   
,   m 2 0 2   2    2 2 , and



k
0 2  . (See scanned notes at end for derivation)
m
Note that as   0 , cos    1 and sin    0    0 .

Note that when   0 ,  



Note that as   ,    (mass is out of phase with drive).

2
The amplitude of the steady state solution can be written as a function of all the
parameters of the system:
ODE
FINAL EXAM Review
R

F0
F0

2

m 2 0 2   2    2 2
F0
2
 2 
2
m 0 1  2    20 2 2
0
 0 
F0
2

Page 18 of 56
4
2
k2  2 
k 2
m 2 1  2    2
m  0 
m 0 2
2


F0
2
 2   2 2
k 1  2  
2
 0  mk 0
F0 / k
2
 2 
2
1  2    2
0
 0 
, 
2
mk
 k 
1
R  
2
 F0 
 2 
2
1




2 
0 2
 0 





 k 
Notice that R   is dimensionless (but proportional to the amplitude of the motion),
 F0 
F
since 0 is the distance a force of F0 would stretch a spring with spring constant k.
k
  mass 2 


  2    time  
2
 1
Notice that  
is dimensionless…   
mk
 mk   mass mass 

time2 

 k 
F
Note that as   0 , R    1  R  0 .
k
 F0 
Note that as   , R  0 (i.e., the drive is so fast that the system cannot respond to it
and so it remains stationary).
The frequency that generates the largest amplitude response is:
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d   k 
 R    0
d    F0  
1    2  2 
 
1




2
1


2



2



2
2
2
2   0  0 
0 2 
d    2 
2  
 1 
   2   
3
d    0 2 
0
2 2

 
   2 2



 1  2    2 
  0 
0 



    2 
 2   2 1  2      0
  0    0 

 2 
  0,  2 1  2     0
 0 
2 
1 2 
0
2



2
max
 0 2  1  
2

2
max


2
max

k 2
 0 
m 2mk
2
 0 
2
2
2m 2
Plugging this value of the frequency into the amplitude formula gives us:
F0
Rmax 
0 1 



2
2
4mk
 1 , then the maximum value of R occurs for   0 .
4mk
Resonance is the name for the phenomenon when the amplitude grows very large
because the damping is relatively small and the drive frequency is close to the undriven
frequency of oscillation of the system.
If
ODE
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Page 20 of 56
Forced Vibrations Without Damping
The equation of motion of an undamped forced oscillator is:
mu  ku  F0 cos t 
When   0 (non-resonant case), the solution is of the form:
u  t   c1 cos 0t   c2 sin 0t  
F0
k
cos t  , 0 
2
2
m
m 0   
When   0 (resonant case), the solution is of the form:
F
u  t   c1 cos 0t   c2 sin 0t   0 t sin 0t 
2m
For the non-resonant case with initial condition u  0  0, u  0  0 , c1  
and u  t  
F0
 cos t   cos 0t   , which can be written as
m 0 2   2 

 0    t    0    t 
2 F0
u t   
sin

  sin 
.
2
2
2
2
 m 0    
  

F0
, c2  0 ,
m  0 2   2 
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Section 4.1: General Theory of nth Order Linear Equations
Usual form of an nth order linear differential equation:
dny
d n 1 y
dy
P0  t  n  P1  t  n 1   Pn 1  t   Pn  t  y  G  t 
dt
dt
dt
Assumptions:
 The functions P0  t  ,
, Pn  t  , G  t  are continuous and real-valued on some interval
I :   t   , and P0  t   0 for any t  I .
Linear differential operator form:
dny
d n 1 y
L  y   n  p1  t  n1 
dt
dt
 pn 1  t 
dy
 pn  t  y  g  t 
dt
Notes:
 Solving an nth order equation ostensibly requires n integrations
 This implies n constants of integration
 Also implies n initial conditions to completely specify an IVP:
n 1
n 1
o y  t0   y0 , y  t0   y0 , y    t0   y0 
Theorem 4.1.1: Existence and Uniqueness Theorem
(for nth-Order Linear Differential Equations)
If the functions where p1 , , pn , and g are continuous on the open interval I, then exists exactly
one solution y    t  of the differential equation
dny
d n1 y
dy

p
t
  pn 1  t   pn  t  y  g  t  that also satisfied the initial conditions
1 
n
n 1
dt
dt
dt
 n 1
y  t0   y0 , y  t0   y0 , y
t0   y0 n1 . This solution exists throughout the interval I.
The Homogeneous Equation:
L  y   y  n  p1  t  y  n1   pn1  t  y  pn  t  y  0
Notes:
 If the functions y1 , y2 ,
, yn are solutions, then so is a linear combination of them,
y  t   c1 y1  t   c2 y2 t  

 cn yn t 
To satisfy the initial conditions, we get n equations in n unknowns:
c1 y1  t0   c2 y2  t0    cn yn  t0   y0
c1 y1  t0   c2 y2  t0  
c1 y1 n 1  t0   c2 y2 n 1  t0  
 cn yn  t0   y0
 cn yn n 1  t0   y0 n 1
ODE

FINAL EXAM Review
This system will have a solution for c1 , c2 , , cn provided the determinant of the matrix of
coefficients is not zero (i.e., Cramer’s Rule again). In other words, the Wronskian is
nonzero, just like for second-order equations.
y1
y2
yn
W  y1 , y2 ,

y1
yn  
y2
0
, pn are continuous on the open interval I, if the functions
, yn are solutions of y n  p1  t  y  n1 
W  y1 , y2 ,
 pn1  t  y  pn  t  y  0 , and if
yn   0 for at least one point in I, then every solution of the differential equation can
be written as a linear combination of y1 , y2 ,
Notes:
 y  t   c1 y1  t   c2 y2 t  
y n  p1  t  y  n1 

yn
y1 n 1 y2 n 1
yn n 1
Note that a slightly modified form of Abel’s Theorem still applies:
W  y1 , y2 , yn  t   c exp    p1  t  dt 
Theorem 4.1.2:
If the functions where p1 ,
y1 , y2 ,
Page 22 of 56
y1 , y2 ,
, yn .
 cn yn t  is called the general solution of
 pn1  t  y  pn  t  y  0 .
, yn are said to form a fundamental set of solutions.
Linear Dependence and Independence:
 The functions f1 , f 2 , , f n are said to be linearly dependent on an interval I if there exist
constants k1 , k2 ,

, kn , NOT ALL ZERO, such that k1 f1  t   k2 f 2 t  
FOR ALL t  I .
The functions f1 , f 2 ,
dependent there.
kn f n  t   0
, f n are said to be linearly independent on I if they are not linearly
ODE
FINAL EXAM Review
Theorem 4.1.3:
If y1 , y2 , , yn is a fundamental set of solutions to y n  p1  t  y  n1 
Page 23 of 56
 pn1  t  y  pn  t  y  0
on an interval I, then y1 , y2 , , yn are linearly independent on I. Conversely, if y1 , y2 , , yn are
linearly independent solutions of the equation, then they form a fundamental set of solutions on
I.
The Nonhomogeneous Equation:
L  y   y n  p1  t  y  n1   pn1  t  y  pn  t  y  g  t 
Notes:
 If Y1(t) and Y2(t) are solutions of the nonhomogeneous equation, then
L Y1  Y2   t   L Y1  t   L Y2  t   g t   g t   0



I.e., the difference of any two solutions of the nonhomogeneous equation is a solution of
the homogeneous equation.
So, the general solution of the nonhomogeneous equation is:
y  t   c1 y1  t   c2 y2  t    cn yn t   Y t  , where Y(t) is a particular solution of the
nonhomogeneous equation.
We will see that the methods of undetermined coefficients and reduction of order can be
extended from second-order equations to nth-order equations.
Section 4.2: Homogeneous Equations with Constant Coefficients
For equations of the form
L  y   a0 y  n  a1 y  n1   an1 y  an y  0 , “it is natural to anticipate that”
y = ert is a solution for correct values of r. Under this anticipation,
L ert   ert  a0 r n  a1r n 1   an 1r  an   e rt Z  r   0 , where the polynomial
Z  r   a0r n  a1r n1 
 an1r  an
is called the characteristic polynomial, and Z  r   0 is called the characteristic equation.
Recall that a polynomial of degree n has n zeros, and thus the characteristic polynomial can be
written as:
Z  r   a0  r  r1  r  r2   r  rn 
Practice: All roots are real and unequal…
 y  t   c1er1t  c2er2t  cnernt
Practice: Some roots are complex…
 Recall that if a polynomial has real coefficients, then it can be factored into linear and
irreducible quadratic factors.
 The linear irreducible quadratic factors will factor into complex conjugate roots    i ,
which will correspond to solutions of et cos  t  , et sin  t  .
ODE
FINAL EXAM Review
Page 24 of 56
Practice: Some roots are repeated…
 If a root r1 is repeated s times, then that repeated root will generate s solutions:
er1t , ter1t , t 2er1t , , t s 1er1t
 The same applies if the repeated roots are complex.
Section 4.3: The Method of Undetermined Coefficients
If the nonhomogeneous term g(t) of the linear nth order differential equation with constant
coefficients
L  y   a0 y  n  a1 y  n1   an1 y  an y  g  t 
is of an “appropriate form” (i.e., a sinusoidal, polynomial, or exponential function), then the
method of undetermined coefficients can be used to find the particular solution.
Recall that if any term of the proposed particular solution replicates a term of the homogeneous
solution, then the entire particular solution must be multiplied by a sufficient number of factors
of t to eliminate the replication.
Section 4.4: The Method of Variation of Parameters
Big Idea: The method of variation of parameters for determining a particular solution of the
nonhomogeneous nth order linear differential equation
L  y   y n  p1  t  y  n1   pn1  t  y  pn  t  y  g  t  is a direct extension of the method for
second-order differential equations in section 3.6.
The basic idea is that once the homogeneous solution is determined,
yh  t   c1 y1  t   c2 y2 t    cn yn t 
Then the particular solution is of the form
y p  t   u1  t  y1 t   u2 t  y2 t    un t  yn t 
with the (n – 1) additional criteria
u1  t  y1  t   u2  t  y2  t    un  t  yn  t   0,
u1  t  y1  t   u2  t  y2  t  
 un  t  yn  t   0,
u1  t  y1  t   u2  t  y2  t  
 un  t  yn  t   0,
u1  t  y1 n  2  t   u2  t  y2 n  2  t  
This results in the system of equations:
 un  t  yn  n  2  t   0.
ODE
FINAL EXAM Review
y1  t  u1  t   y2  t  u2  t  
Page 25 of 56
 yn  t  un  t   0,
y1  t  u1  t   y2  t  u2  t  
 yn  t  un  t   0,
y1  t  u1  t   y2  t  u2  t  
 yn  t  un  t   0,
y1 n  2  t  u1  t   y2 n  2  t  u2  t  
 yn  n  2  t  un  t   0,
y1 n 1  t  u1  t   y2 n 1  t  u2  t  
 yn  n 1  t  un  t   g  t  .
The solution for any one function um(t) is:
g  t  Wm  t 
um  t  
, where W(t) is the Wronskian W  t   W  y1 , y2 , , yn  t  , and Wm(t) is the
W t 
determinant obtained from W by replacing the mth column with the column [0, 0, …, 1].
n
t
m 1
t0
Thus, y p  t    ym  t  
g  s Wm  s 
ds .
W s
ODE
FINAL EXAM Review
Page 26 of 56
Section 5.1: Review of Power Series

1. Definition of convergence of a power series: A power series
a x  x 
n 0
m
n
n
0
is said to
converge at a point x if lim  an  x  x0  exists for that x.
m 
n
n 0

2. Definition of absolute convergence of a power series: A power series
a x  x 
n 0

said to converge absolutely at a point x if
 a x  x 
n 0
n
0
n
n
0
n
is
converges.
a. Absolute convergence implies convergence…
3. Ratio test: If, for a fixed value of x, lim
n 
an 1  x  x0 
an  x  x0 
n 1
n
 x  x0 lim
n 
an 1
 x  x0 L , then
an
the power series converges absolutely at that value of x if x  x0 L  1, and diverges if
x  x0 L  1. If x  x0 L  1, then the test is inconclusive.

4. If
a x  x 
n 0
n
n
0
converges at x = x1, it converges absolutely for x  x0  x1  x0 , and if
it diverges at x = x1, it diverges for x  x0  x1  x0 .
5. Radius and interval of convergence: The radius of convergence  is a nonnegative

number such that
a x  x 
n 0
n
0
n
converges absolutely for x  x0   and diverges for
x  x0   .
a. Series that converge only when x = x0 are said to have  = 0.
b. Series that converge for all x are said to have  = .
c. If  > 0, then the interval of convergence of the series is x  x0   .
ODE
FINAL EXAM Review

Given that

Page 27 of 56
 an  x  x0  and  bn  x  x0  converge for x  x0   …
n
n 0
n
n 0

6. Sum of series:


 an  x  x0    bn  x  x0     an  bn  x  x0 
n
n 0
n
n 0
n
n 0
7. Product and Quotient of series:



n
n
a
x

x


0
 n
   bn  x  x0  
 n 0
  n 0

2
3
2
3
  a0  a1  x  x0   a2  x  x0   a3  x  x0   ... b0  b1  x  x0   b2  x  x0   b3  x  x0   ...




 a0 b0  b1  x  x0   b2  x  x0   b3  x  x0   ...
2
3


 a1  x  x0  b0  b1  x  x0   b2  x  x0   b3  x  x0   ...
2

a  x  x  b  b  x  x   b  x  x 
3


 b  x  x   ...
 a2  x  x0  b0  b1  x  x0   b2  x  x0   b3  x  x0   ...
2
3
3
0
0
1
0
2
0
2
2
3
3
3
0
...
 a0b0   a0b1  a1b0  x  x0    a0b2  a1b1  a2b0  x  x0    a0b3  a1b2  a2b1  a3b0  x  x0   ...
2

 n

n
    ak bn  k   x  x0 
n 0  k 0

To do a quotient, can write as a multiplication and equate terms:

a x  x 
n 0

n
b  x  x 
n 0
n
n
0
n




n
n
n
n
  d n  x  x0     d n  x  x0     bn  x  x0     an  x  x0 
n 0
 n 0
  n 0
 n 0
0
OR can do long division…
8. Derivatives of a series:

d 
n
n 1
a
x

x

nan  x  x0 




n
0


dx  n 0
 n 1
2


d 
n
n2
a
x

x

n  n  1 an  x  x0 



0
2  n

dx  n 0
 n2
9. Taylor series:
f  n   x0 
n
 x  x0  is the Taylor series for a function f(x) about the point x = x0.

n!
n 0

3
ODE
FINAL EXAM Review
Page 28 of 56
10. Equality of series: every corresponding term is equal…
11. Analytic functions: have a convergent Taylor series with non-zero radius of convergence
about some point x = x0.
12. Shift of index of Summation…
Section 5.2: Series Solutions Near an Ordinary Point, Part I
d2y
dy
Will consider homogeneous equations of the form P  x  2  Q  x   R  x  y  0 where the
dx
dx
polynomial coefficients are polynomials, like:
The Bessel equation: x 2 y  xy   x 2   2  y  0
The Legendre equation: 1  x 2  y  2 xy     1 y  0
Ordinary point: a point x0 such that P(x0)  0.  y  p  x  y  q  x  y  0
Singular point: a point x0 such that P(x0) = 0.  p(x) or q(x) becomes unbounded (See sections
5.4 – 5.7)
To solve a differential equation near an ordinary point using a power series technique:


Assume a power series form of the solution y  x    an x n which converges in the
n 0
interval x  x0  





Plug the power series into the equation and equate like terms
Obtain a recurrence relation for the coefficients of higher-order terms in terms of earlier
coefficients.
If possible, try to find the general term, which is an explicit function for the coefficients
in terms of the index of summation.
Write the final answer in terms of those coefficients.
Check the radius of convergence.
ODE
FINAL EXAM Review
Page 29 of 56
Section 5.3: Series Solutions Near an Ordinary Point, Part II
To justify the statement that a solution of P  x  y  Q  x  y  R  x  y  0  y  p  x  y  q  x  y

m
can be written as y    x    an  x  x0  , we must be able to compute m!am      x0  for
n
n 0
any order derivative m based only on the information given by the IVP.
Note:
So it seems that p and q need to at least be infinitely differentiable at x0, but in addition they also
need to be analytic at x0. In other words, they must have Taylor series expansions that converge
in some interval about x0.
Theorem 5.3.1 (due to Fuchs)
If x0 is an ordinary point of the differential equation P  x  y  Q  x  y  R  x  y  0 (i.e.,
pQ
P
and q  R

P
are analytic at x0), then its general solution is
y   an  x  x0   a0 y1  x   a1 y2  x  ,
n
n 0
where a0 and a1 are arbitrary, and y1 and y2 are two power series solutions that are analytic at x0.
The solutions y1 and y2 form a fundamental set of solutions. Also, the radius of convergence of
y1 and y2 is at least as large as the minimum of the radii of convergence of p and q.
Note:
From the theory of complex-valued rational expressions, it turns out that radius of convergence
of a power series of a rational expression about a point x0 is the distance from x0 to the nearest
zero of the (complex-valued) denominator.
ODE
FINAL EXAM Review
Page 30 of 56
Section 5.4: Euler Equations; Regular Singular Points
P  x  y  Q  x  y  R  x  y  0
Singular points are x = x0 such that P(x0) = 0.
Euler equations have singular points at x = 0:
x 2 y   xy   y  0
To solve Euler equations, make an assumption for the solution of y = xr.
When the roots are real and distinct, then the two fundamental solutions are obtained.
When the roots are repeated, multiply the first fundamental solution by ln(x) to obtain the second
fundamental solution.
When the roots are complex, r    i  , the fundamental solutions are
y1  x cos   ln  x   and y2  x sin   ln  x  
In summary, the solutions of x 2 y   xy   y  0 are:
 r1
r2
c1 x  c2 x , given r1 and r2 are distinct real roots of r  r  1   r    0

y  x    c1  c2 ln  x   x r1 , given r1 is a real double root of r  r  1   r    0


 c1 cos   ln  x    c2 sin   ln  x   x , given   i  are complex roots of r  r  1   r    0


For negative x values, one can make the substitution    x , and show that, for any x value, the
solutions are:
r
r

c1 x  c2 x
r

y  x    c1  c2 ln  x   x


 c1 cos   ln  x    c2 sin   ln  x   x

1
2
1


We do not have a general theory of how to handle any possible singularity of
P  x  y  Q  x  y  R  x  y  0 . So, for the next few sections, we will restrict the discussion to
the power series solution of this type of equation near regular singular points.
A regular singular point at x = x0 is a singular point with the additional restrictions that
Q  x
2 R  x
lim  x  x0 
is finite , and lim  x  x0 
is finite .
x  x0
x  x0
P  x
P  x
ODE
FINAL EXAM Review
Page 31 of 56
Section 5.5: Series Solutions Near a Regular Singular Point, Part I
Big Idea: According to Frobenius, it is valid to assume a series solution of the form
y   x  x0 
r

a x  x 
n 0
n
n
0
to a second-order linear differential equation near a regular singular
point x = x0.
Recall:
If x = x0 is a regular singular point of the second order linear equation
P  x  y  Q  x  y  R  x  y  0 ,
then lim x
x  x0
Q  x
R  x
 lim xp  x   finite and lim x 2
 lim x 2 q  x   finite .
x  x0
P  x  x  x0
P  x  x  x0


This means xp  x    pn  x  x0  and x 2 q  x    qn  x  x0  are convergent for some radius
n
n 0
n
n 0
about x0.
Thus, we can write the original equation as: x 2 y   xp  x   xy   x 2 q  x   y  0
If all the coefficients pn and qn are zero except for p0 and q0, then the equation reduces to
x 2 y  p0 xy  q0 y  0 , which is an Euler equation. In fact, this is called the corresponding
Euler equation when all the coefficients pn and qn are not zero, and the roots of this
corresponding Euler equation play a role in the solution called “the exponents at the singularity.”
Since the equation we are trying to solve looks like an equation with “Euler coefficients” times
power series, we will look for a solution that is of the form of an “Euler solution” times a power
series:
y   x  x0 
r


a x  x   a x  x 
n 0
n
n
0
n 0
n
r n
0
As part of the solution, we must determine:
 The values of r that make this a valid solution.
 The recurrence relation for the coefficients an.
 The radius of convergence.
The theory behind a solution of this form is due to Frobenius.
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To solve a linear second order equation near a regular singular point using the method of
Frobenius:
 Identify singular points and verify they are regular.


Assume a solution of y   an  x  x0 
r n
and its derivatives:
n 0

y   an  r  n  x  x0 
n 0
r  n 1

, y   an  r  n  r  n  1 x  x0 
r  n2
n 0

Substitute the assumed solution and its derivatives into the given equation.
o You may have to re-write coefficients in terms of  x  x0  …

Shift indices so that all series solutions have  x  x0 

“Spend” any terms needed so that all series start at the same index value.
o This should result in a term that looks like the characteristic equation of the
corresponding Euler equation.
o This is called the indicial equation.
o The roots of the indicial equation are called the exponents of the singularity.
o They are the same as the roots of the corresponding Euler equation.
r n
Set the coefficient of  x  x0  to zero to get the recurrence relation.



r n
in the general term.
Use the recurrence relation with each exponent to get the general term for each of the two
solutions for each exponent.
o If the exponents of the singularity are equal or differ by an integer, then it is only
valid to get the series solution for the larger root. (What to do for the second
solution will be covered in 5.6).
Compute the radius of convergence for each solution.
ODE
FINAL EXAM Review
Page 33 of 56
5.6: Series Solutions Near a Regular Singular Point, Part II
To solve a linear second order equation near a regular singular point using the method of
Frobenius REGARDLESS OF THE EXPONENTS OF THE INDICIAL EQUATION:
 Identify singular points and verify they are regular.
 Find the exponents of the singularity by solving the corresponding Euler equation
Q  x
R  x
x 2 y  p0 xy  q0 y  0 , where p0  lim x
and q0  lim x 2
, or using the indicial
x 0
x 0
P  x
P  x
equation.


Assume a solution of y  x r1  an  x  x0  .
n
n 0


Substitute the assumed solution and its derivatives into the given equation.
r n
Shift indices so that all series solutions have  x  x0  in the general term.


“Spend” any terms needed so that all series start at the same index value.
o This should result in a term with a factor that looks like the characteristic equation
of the corresponding Euler equation; this is the indicial equation.
r n
Set the coefficients of  x  x0  to zero to get the recurrence relation.

Get the second solution using the theorem below…
ODE
FINAL EXAM Review
Page 34 of 56
Theorem 5.6.1: General Solution of a Second Order Equation with Real Exponents of the
Singularity near a Regular Singular Point
Consider the differential equation x 2 y  x  xp  x   y   x 2 q  x   y  0 , where x = 0 is a regular
singular point. Then xp  x  and x 2 q  x  are analytic at x = 0 with convergent power series


expansions xp  x    pn x n and x 2 q  x    qn x n for x   where   0 is the minimum of
n 0
n 0
the radii of convergence for xp  x  and x q  x  . Let r1 and r2 be the roots of the indicial equation
2
F  r   r  r  1  p0r  q0  0 , with r1  r2 if r1 and r2 are real. Then in either the interval
   x  0 or 0  x   , there exists a solution of the form


r 
y1  x   x 1 1   an  r1  x n 
 n 1

where the an  r1  are given by the recurrence relation
n 1
F  r  n  an   ak  r  k  pn k  qn k   0
k 0
with a0 = 1 and r = r1. There are three cases for the second solution:

If r1  r2 is not zero or a positive integer, then in either the interval    x  0 or 0  x   ,
there exists a second solution of the form


r 
y2  x   x 2 1   an  r2  x n  .
 n 1

The an  r2  are determined by the same recurrence relation as the an  r1  , with with a0 = 1 and
r = r2. These power series solutions converge for at least x   .


If r1 = r2 then the second solution is of the form


r 
y2  x   y1  x  ln x  x 1   bn  r1  x n  .
 n 1

If r1  r2 = N, a positive integer, then the second solution is of the form


r 
y2  x   ay1  x  ln x  x 2 1   cn  r2  x n  .
 n 1

The coefficients an , bn  r1  , cn  r2  and the constant a can be determined by substituting the form
of the series solution for y2 into the original differential equation. The constant a may turn out to
be zero. Each of these series converge at least for x   and defines an analytic function near x
= 0.
In all three cases, the two solutions form a fundamental set of solutions.
ODE
FINAL EXAM Review
Page 35 of 56
Section 5.7: Bessel’s Equation
Bessel’s Equation: x 2 y  xy   x 2  2  y  0





x = 0 is a regular singular point
The roots of the indicial equation are .
The value of  is called the “order” of the equation.
The first solution for a given value of  is called the “Bessel function of the first kind of
order ,” and is denoted by J(x).
The second solution for a given value of  is called the “Bessel function of the second
kind of order ,” and is denoted by Y(x).
Bessel Equation of Order Zero (i.e.,  = 0): x 2 y  xy  x 2 y  0
 The roots of the indicial equation are r1 = r2 = 0.
m
m



1 x 2 m 
1 x 2 m


 y1  x   a0 1   2 m
 J 0  x   1   2m
2 
2
m 1 2
 m!
 m1 2  m ! 

J0(x)  1 as x  0.


 2 2

J0  x  
 cos  x   as x   .
4
x 

1


y2  x   J 0  x  ln  x   
m 1
o
Hm  1
1 1
 
2 3
 1
2
2m

m 1
Hm
 m!
2
x2m
m
1
1
  ; i.e., a partial sum of the harmonic series up to m.
m k 1 k
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FINAL EXAM Review
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
The traditional Bessel function of the second kind of order zero is a linear combination of
y2  x  and J 0  x  :

Y0  x  

 y  x      ln  2   J 0  x  
 2
m 1

1 H m 2 m 

2
Y0  x      ln  2   J 0  x    2 m
x 
2
 
m 1 2
 m!

o  is the “Euler-Mascheroni” constant:   lim  H n  ln  n    0.577 215 665
2
n
2

Y0  x  


 2 2

Y0  x   
 sin  x   as x   .
4
x 


ln  x  as x  0 .
1
ODE
FINAL EXAM Review
Page 37 of 56
1

Bessel Equation of Order One-Half (i.e.,  = ½ ): x 2 y  xy   x 2   y  0
4

 The roots of the indicial equation are r1 = + ½ , r2 = - ½ .
m
m


1 
1 x 2 m 
1 x 2 m1


1
1
2
2
 y1  x   x 1  
 x 2 sin  x 
x 
m  0  2m  1 !
 m1  2m  1! 
1



 2  2
By convention, J 1  x   
 sin  x  , x > 0.
2
x 
n

   1n x 2 n
1 x 2 n 1 
cos  x 
sin  x 

y2  x   x  a0 
 a1 
 a1 1
  a0
1
n 1  2n  1 ! 
 n 1  2n !
x 2
x 2

The Bessel function of the second kind of order one-half is:
1
2
1

 2 2
J 1  x  
 cos  x  , x > 0.
2
x 
ODE
FINAL EXAM Review
Page 38 of 56
Bessel Equation of Order One (i.e.,  = 1): x 2 y  xy   x 2  1 y  0

The roots of the indicial equation are r1 = 1, r2 = -1.
2m

 1  x 
x   1 x
J1  x    2 m

 
2 m 0 2  m  1 !m ! m 0  m  1!m !  2 
series can be written as powers of (x/2)…
m






(Note: J1(x) = ½ y1(x) so that the
3 
 2 2

J1  x   
 cos  x   as x   .
4 
x 

m

1  H m  H m1  2 m 

1
y2  x    J1  x  ln  x   1  
x , x>0
x  m1 22 m m ! m  1!

The traditional Bessel function of the second kind of order one is:
2
Y1  x     y2  x      ln  2   J1  x  

2
Y1  x    as x  0 .
x
1

2 m 1
J1(x)  0 as x  0.
1

m
3 
 2  2

Y1  x   
 sin  x   as x   .
4 
x 

ODE
FINAL EXAM Review
Page 39 of 56
Bessel Equations of Higher Positive Integer Order (i.e.,  = positive integer,  > 1):
x 2 y  xy    x 2  2  y  0

The roots of the indicial equation are r1 = , r2 = -.

 1  x 2 m
J  x   
 
m0  m  !m !  2 

J(x)  0 as x  0.


 2  1   as x   .
 2  2
J  x   

 cos  x 
4
x 


m

1



Weisstein, Eric W. "Bessel Function of the First Kind." From MathWorld--A Wolfram
Web Resource.
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.htmlhttp://mathworld.wolfr
am.com/BesselFunctionoftheFirstKind.html
J  x  cos    J   x 
Y  x   
sin  
Y  x   
  1!  2 
  as x  0 .
 x


 2  1   as x  
 2  2
Y  x   
.

 sin  x 
4
x 


Weisstein, Eric W. "Bessel Function of the First Kind." From MathWorld--A Wolfram
Web Resource.
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.htmlhttp://mathworld.wolfr
am.com/BesselFunctionoftheSecondKind.html
1


ODE
FINAL EXAM Review
Page 40 of 56
Section 6.1: Definition of the Laplace Transform
Review of Improper Integrals (Calc 2):

If f is continuous on the interval [a, ), then the improper integral
 f  t  dt
is defined as:
a

A
 f  t  dt  lim  f  t  dt .
A
a
a
If the limit exists as A  , then the integral is said to converge. Otherwise, the integral
diverges.
Piecewise Continuous Functions:
A function f is piecewise continuous on an interval   t   if the interval can be partitioned by
n points   t0  t1  ...  tn   such that
 f is continuous on each subinterval Ii = ti 1  t  ti
 the limit of f as the endpoints are approached from within each limit are finite.
Upshot: f is continuous everywhere except at a finite number of jump discontinuitues.
To integrate a piecewise continuous function:

t1
t2
 f  t  dt   f t  dt   f t  dt 

t1


 f t  dt
tn1
It can be difficult to tell if an improper integral with any given piecewise converges when the
antiderivative can’t be written in terms of elementary functions…
ODE
FINAL EXAM Review
Page 41 of 56
Theorem 6.1.1: Integral Comparison Test for Convergence and Divergence
If f is piecewise continuous for t  a , and if f  t   g  t  when t  M for some positive constant
M, and if


M
a
 g  t  dt converges, then  f  t  dt
t  M and if


M
a
 g  t  dt diverges, , then
also converges. However, if f  t   g  t   0 for
 f  t  dt
also diverges.
Integral Transforms:

An integral transform, in general, is a relation of the form F  s    K  s, t  f  t  dt , where K is a
a
given function, called the kernel of the transformation. Alpha and beta need not be finite
numbers. Note that an integral transform transforms a function f into another function F, called
the “transform of f.”
The Laplace Transform:
…has a kernel of e  st .

…is defined by
L  f  t    F  s    e st f  t  dt
0
…we will use the Laplace transform to:
 Transform an initial value problem for an unknown function f in the t domain to an
unknown function F in the s domain.
 Solve the resulting algebraic problem to find F.
 Recover f from F by “inverting the transform.”
Theorem 6.1.2: Criteria for a Laplace Transform to Exist
Suppose that:
1. f is piecewise continuous on the interval 0  t  A for any positive A.
2. f  t   Ke at when t  M for some real, positive constants K and M, and a real a.

Then the Laplace transform L  f  t    F  s    e  st f  t  dt exists for s  a .
0
Functions that satisfy the hypotheses of theorem 6.1.2 are described as being “of exponential
2
order” as t   . f  t   et is not of exponential order.
ODE
FINAL EXAM Review
Page 42 of 56
Section 6.2: Solution of Initial Value Problems
Theorem 6.2.1: Laplace Transform of the Derivative of a Function
Suppose that f is continuous and that f  is piecewise continuous on any interval 0  t  A and that
there exist constants K, a, and M such that f  t   Ke at for t  M . Then L  f   t   exists for
s > a and is given by:
L  f   t    sL  f  t    f  0  .
Laplace Transform of the second derivative of a function:
L  f   t    s 2 L  f  t    sf  0   f   0 
Theorem 6.2.2: Laplace Transform of the nth Derivative of a Function
…. Then L  f  n   t   exists for s > a and is given by:
L  f  n  t   s n L  f  t   s n1 f  0  
sf  n 2  0   f  n 1  0  .
 To solve an initial value problem with a Laplace transform:
 Transform the differential equation.
 Solve the resulting algebraic equation for L  f  t   .

Look up each term in a table of Laplace transforms (page 317 or the table on the next
page); the actual solution is the inverse of each term of L  f  t   . Note that some
algebra will most likely need to be done to get the transform to look like sums of the
transforms in the table below.
ODE
FINAL EXAM Review
f  t   L  F  s  
1
1
e at
t n , n = positive integer
t p , p > -1
sin  at 
cos  at 
sinh  at 
cosh  at 
eat sin  bt 
eat cos  bt 
t n e at , n = positive integer
uc  t 
Page 43 of 56
Laplace Transform Table
F  s   L  f  t  
1
,s>0
s
1
,s>0
sa
n!
,s>0
s n 1
  p  1
,s>0
s p 1
a
,s>0
2
s  a2
s
,s>0
2
s  a2
a
, s > |a|
2
s  a2
s
, s > |a|
2
s  a2
b
,s>a
2
 s  a   b2
sa
s  a
2
 b2
n!
s  a
n 1
,s>a
,s>a
uc  t  f  t  c 
e  cs
,s>0
s
ecs F  s 
ect f  t 
F  s  c
f  ct 
1 s
F  , c > 0
c c
t

f  t    g   d
F sG s
0
 t  c 
e  cs
f  n  t 
s n F  s   s n1 f  0  
 t 
F  n  s 
n
f t 
 f  n1  0 
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Section 6.3: Step Functions
Definition of the unit step function (or Heaviside function):
0, t  c,
uc  t   
c0
1, t  c,
e  cs
The Laplace transform of uc  t  is
.
s
Theorem 6.3.1: Laplace transform of a horizontally translated function.
If F  s   L  f  t   exists for s  a  0 , and if c is a positive constant, then
L uc  t  f  t  c    e  cs L  f  t    e  cs F  s  .
Conversely if f  t   L 1  F  s   , then uc  t  f  t  c   L 1 e  cs F  s   .
Theorem 6.3.2: Inverse Laplace transform of a horizontally translated transform.
If F  s   L  f  t   exists for s  a  0 , and if c is a constant, then
L ect f  t    F  s  c  ,
s > a + c.
Conversely if f  t   L 1  F  s   , then ect f  t   L 1  F  s  c   .
Section 6.4: Differential Equations with Discontinuous Forcing Functions
Pertinent relations from section 6.3:
e cs
L uc  t   
s
L uc  t  f  t  c    e  cs L  f  t    e  cs F  s 
uc  t  f  t  c   L 1 e  cs F  s  

L ect f  t    F  s  c 

ect f  t   L 1  F  s  c  
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Section 6.5: Impulse Functions
Recall from Physics and calc 2 applications that an impulse is the time integral of a force, which
results in a change in momentum:
dp
F  ma 
dt
pf
tf
pi
ti
 dp   F  t  dt
tf
p p  p   F  t  dt
pf
i
ti
In this section, we will examine “the” impulse function, which models a force of a very large
magnitude that acts over a very short time.
1
 ,   t  
Let d  t    2
. Note: lim d  t   0, t  0
 0
0,    t  
Then the impulse imparted by this force is:
I   

 d  t  dt


lim I    lim 
 0
 0

1
1
dt  lim  2  1


0
2
2
Define the Dirac delta function (or an idealized unit impulse function) as:
  t   0, t  0

   t  dt  1

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Note: The Dirac delta function can be shifted, which results in a description of an impulse
occurring at a later time:
  t  t0   0, t  t0

   t  t  dt  1
0

The Laplace Transform of the Dirac delta function can be derived by looking once again at the
limit of d  t  .
L   t  t0    e st
0
Note:

   t  t  f  t  dt  f t 
0
0

Section 6.6: The Convolution Integral
Theorem 6.6.1: The Convolution Theorem
If F  s  = L  f  t   and G  s  = L  g  t   both exist for s > a  0, then
H  s  = F  s  G  s  = L  h  t   , s  a,
where
t
t
0
0
h  t    f  t    g   d   f   g  t    d .
The function h is known as the convolution of f and g. The two integrals above are known as
convolution integrals.
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Section 7.1: Introduction to Systems of First Order Linear Equations
A second-degree or higher differential can be converted to a system of first order equations by
letting x1 = u, x2 = u, etc.
General Notes:
 In this chapter, we will examine systems of first-order differential equations:
x1  F1  t , x1 , x2 , , xn 

x2  F2  t , x1 , x2 ,
, xn 
xn  Fn  t , x1 , x2 ,
, xn 
A solution of this system on an interval I :   t   is a set of n functions that are
differentiable in I and that satisfy the above system of equations. An IVP is formed when
n initial conditions are stated.
x1  t   1  t  , x2  t   2  t  ,
, xn t   n t 
x1  t0   x10 , x2 t0   x20 ,

, xn t0   xn0
Theorem 7.1.1 Existence and Uniqueness of a Solution to a System of First-Order
Equations
Let each of the functions F1, …, Fn and the partial derivatives F1 x1 , …, F1 xn ,
…, Fn x1 , …, Fn xn be continuous in an open region R of the tx1x2…xn space
defined by   t   , 1  x1  1 ,  2  x2  2 , …,  n  xn   n , and let the point
t , x , x ,
0
0
1
0
2
, xn0  be in R. Then there is an interval t  t0  h in which there exists a
unique solution x1  1  t  , , xn  n  t  of the system of differential equations that also
satisfies the initial conditions.



In this chapter, the discussion will be restricted to systems where the functions F1, …, Fn
are linear functions of x1, …, xn. Thus, the most general system we will examine is:
x1  p11  t  x1 
 p1n  t  xn  g1  t 
x2  p21  t  x1 
 p2 n  t  xn  g 2  t 
xn  pn1  t  x1 
 pnn  t  xn  g n  t 
If all the g1(t), …, gn(t) are zero, then the system is homogeneous. Otherwise, it is
nonhomogeneous.
Theorem 7.1.2 Existence and Uniqueness of a Solution to a System of Linear FirstOrder Equations
Let each of the functions p11, p12, …, pnn, g1(t), …, gn(t) are continuous on an open
interval I :   t   , then there exists a unique solution x1  1  t  , , xn  n  t  of the
system that also satisfies the initial conditions. Moreover, the solution exists throughout
I.
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Section 7.2: Review of Matrices
A matrix is a rectangular array of numbers (called elements) arranged in m rows and n columns:
a1n 
 a11 a12
a
a22
a2 n 
21

A




amn 
 am1 am 2
aij is used to denote the element in row i and column j.
(aij) is used to denote the matrix whose generic element is aij.
The transpose of a matrix is a new matrix that is formed by interchanging the rows and columns
of a given matrix. The transpose is denoted with a superscript T:
1 2 3
1 4 7 


T
if A  4 5 6 , then A   2 5 8  ; if A = (aij), then AT = (aji).




7 8 9 
 3 6 9 
The conjugate of a matrix is a new matrix that is formed by taking the conjugate of every
element of a given matrix. The conjugate is denoted with a bar over the matrix name.
 1  i 2  3i 
 1  i 2  3i 
if A  
, then A  

 ; if A   aij  , then A   aij  .
 4  5i 6  7i 
 4  5i 6  7i 
The adjoint of a matrix is the transpose of the conjugate of a given matrix. The adjoint is
denoted with a superscript asterisk:
 1  i 2  3i 
 1  i 4  5i 
if A  
, then A*  

.
 4  5i 6  7i 
 2  3i 6  7i 
Properties of Matrices
1. Equality: Two matrices A and B are equal only if all corresponding elements are equal;
i.e., aij = bij for all i and j.
2. Zero: The matrix with zero for every element is denoted as 0.
3. Addition: The sum of two m  n matrices A and B is computed by adding corresponding
elements:
A  B   aij    bij    aij  bij 
1 2  5 6  6 8 
3 4   7 8   10 12 

 
 

a. Matrix addition is commutative and associative:
A  B  B  A ; A   B  C   A  B   C
4. Multiplication by a number: The product of a complex number and a matrix is
computed by multiplying every element by the number:
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 A    aij    aij 
1 2  2  i 4  2i 


3 4 6  3i 8  4i 
a. Multiplication by a constant obeys the distributive laws
  A  B    A   B ;     A   A   A
b. The negative of a matrix is negative one times the matrix:
A   1 A
2  i 
5. Subtraction: The difference of two m  n matrices A and B is computed by subtracting
corresponding elements:
A  B   aij    bij    aij  bij 
1 2 5 6  4 4
1 1
3 4  7 8    4 4   4  1 1 . Note: the last step is not required, but it

 
 



illustrates a simplifying technique that is sometimes used.
6. Multiplication of matrices: Matrix multiplication is only defined when the number of
columns in the first factor equals the number of rows in the second factor.
a. The product of an m  p and a p  n matrix is an m  n matrix.
b. If AB  C , then element cij is computed by taking the sum of the products of
corresponding elements from row i in matrix A and column j of matrix B:
p
cij   aik bkj
k 1
c. Matrix multiplication is associative:
 AB C  A  BC
d. Matrix multiplication is distributive:
A  B  C  AB  AC
e. Matrix multiplication is usually NOT commutative:
AB  BA (usually)
7. Multiplication of vectors:
n
a. The dot product: xT y   xi yi
i 1
b. Properties of the dot product:
xT y  yT x ; xT  y  z   xT y  xT z ;  xT  y   xT y  xT  y 
n
c. The scalar (inner) product:  x, y    xi yi  xT y
i 1
d. The magnitude, or length, of a vector x can be represented with the inner product:
x   x, x  .
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e. Note: Two vectors are orthogonal if  x, y   0 ; the (orthogonal) unit vectors are
1 
0 
0




i  0  , j  1  , and k  0  .
0 
0 
1 
f. Properties of the scalar product:
 x, y    y, x  ,  x, y  z    x, y    y, z  ,  x, y     x, y  ,  x, y     x, y 
8. Identity: The multiplicative identity for matrices, denoted by I, is the square matrix with
one in every diagonal element and zero elsewhere. The diagonal runs from the top left to
bottom right, only.
0
1 0
0 1
0 

I




1
0 0
a. Multiplication by the identity matrix is commutative for square matrices:
AI  IA  A .
9. Inverse: For some square matrices, there exists another unique matrix such that their
product is the identity. Such matrices are said to be nonsingular, or invertible. Matrices
that are not invertible are called noninvertible or singular. The “other unique matrix” is
called the (multiplicative) inverse, and is denoted with a superscript -1:
If A is invertible, then AA-1 = A-1A = I.
a. The determinant is a quantity that can be computed for any square matrix, and
whose value determines if a matrix is singular or not. A zero determinant means
the matrix is noninvertible. The determinant of a matrix A is denoted as:
det(A) = |A|.
b. The determinant of any square matrix can be computed by taking the sum of the
products of all the elements and their cofactors from any one row or column of
the matrix.
c. The cofactor Cij of a given element aij is the product of the minor Mij of the
element and an appropriate factor of -1:
Cij = (-1)i+jMij .
d. The minor Mij of an element aij is the determinant of the matrix obtained by
eliminating the ith row and jth column from the original matrix.
e. Thus, a determinant can be computed using a chain of smaller determinants. It is
useful to know the following formula for the last step in the chain:
a11 a12
a b
 ad  bc
 a11a22  a12 a21 
c d
a21 a22
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f. If B = A-1, then bij 
C ji
det  A 
Page 51 of 56
, where (Cji) is the transpose of the cofactor matrix of
A.
g. This formula is extremely inefficient for computing the inverse. The preferred
method for computing the inverse of a matrix A is to form the augmented matrix
A | I, and then perform elementary row operations on the augmented matrix until
A is transformed to the identity matrix. That will leave I transformed to A-1. This
process is called row reduction or Gaussian elimination.
h. The three elementary row operations are:
i. Interchange any two rows.
ii. Multiply any row by a nonzero number.
iii. Add any multiple of one row to another row.
10. Matrix Functions: Are matrices with functions for the elements. They can be integrated
and differentiated element-by-element…
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Section 7.3: Linear Algebraic Equations; Linear Independence, Eigenvalues, Eigenvectors
Systems of Linear Algebraic Equations
A system of n linear equations in n variables can be written in matrix form as follows:
a1n   x1   b1 
 a11 a12
 a11 x1  a12 x2   a1n xn  b1
a
a22
a2 n   x2  b2 

21


 Ax  B








a x  a x   a x  b

   
nn n
n
 n1 1 n 2 2
amn   xn  bn 
 am1 am 2
If B  0 , then the system is homogeneous; otherwise, it is nonhomogeneous.
If A is nonsingular (i.e., det(A)  0), then A-1 exists, and the unique solution to the system is:
x  A 1B
A picture for the 3  3 system is:
y  z  2
 x 

 x  2 y  3z  12
2 x  2 y  z  9

Solution: (-1, 2, -3)
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If A is singular, then either there are no solutions, or there are an infinite number of solutions.
Example pictures for 3  3 systems are:
Three parallel planes  No Solution
z  0


z  2


z  4

Note: Inconsistent systems yield false
equations (like 0 = 2 or 0 = 4) after trying to
solve them.
Planes intersect in three parallel lines No
Solution
 2
x

y
 2

x  y
 2

Note: Inconsistent systems yield a false
equation (like 0 = -6) after trying to solve
them.
All three planes intersect along same line 
Infinite number of solutions
1
x  y  z 

2
x  y  z 
x
 0.5

Note: This type of dependent system yields one
equation of 0 = 0 after row operations.
All three planes are the same Infinite
number of solutions
y 
z  1
 x 

2 x  2 y  2 z  2
 3x  3 y  3z  3

Note: This type of dependent system yields
two equations of 0 = 0 after row operations.
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If A is singular, then the homogeneous system Ax = 0 will have infinitely many solutions (in
addition to the trivial solution).
If A is singular, then the nonhomogeneous system Ax = B will have infinitely many solutions
when (B, y) = 0 for all vectors y satisfying A*y = 0 (recall A* is the adjoint of A). These
solutions will always take the form x = x(0) + , where x(0) is the particular solution, and  is the
general form for the corresponding homogeneous solution.
In practice, linear systems are solved by performing Gaussian elimination on the augmented
matrix A | B.
Linear Independence
1
k
A set of k vectors x  , , x  are said to be linearly dependent if there exist k complex numbers
c1 , , ck not all zero, such that c1x1   ck x k   0 . This term is used because if it is true that
not all the constants are zero, then one of the vectors depends on one or more of the other
vectors: ci xi   c1x1   ci 1xi 1  ci 1xi 1   ck x k  .
On the other hand, if the only values of c1 ,
c1  c2 
1
 ck  0 , then the vectors x  ,
, ck that make c1x1 
 ck x k   0 be true are
, x k  are said to be linearly independent.
The test for linear dependence or independence can be represented with matrix arithmetic:
Consider n vectors with n components. Let xij be the ith component of vector x(j), and let
X = (xij). Then:
 x11 c1   x1 n cn   x11c1   x1n cn 

 



  Xc  0

 1
 n  
 xn c1   xn cn   xn1c1   xnn cn 
If det(X) = 0, then c = 0, and thus the system is linearly independent.
If det(X)  0, then there are nonzero values of ci, and thus the system is linearly dependent.
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Note: Frequently, the columns of a matrix A are thought of as vectors.
The columns of vectors are linearly independent iff det(A)  0.
If C = AB, it happens to be true that det(C) = det(A)det(B). Thus, if the columns of A and B are
linearly independent, then so are the columns of C.
Eigenvalues and Eigenvectors
The equation Ax = y is a linear transformation that maps a given vector x onto a new vector y.
Special vectors that map onto multiples of themselves are very important in many applications,
because those vectors tend to correspond to “preferred modes” of behavior represented by the
vectors. Such vectors are called eigenvectors (German for “proper” vectors), and the multiple
for a given eigenvector is called its eignevalue.
To find eigenvalues and eigenvectors, we start with the definition:
Ax  x , which can be written as
 A  I  x  0 , which has solutions iff
det  A  I   0
The values of  that satisfy the above determinant equation are the eigenvalues, and those
eigenvalues can then be plugged back into the defining equation Ax  x to find the
eigenvectors.
You will see that eigenvectors are only determined up to an arbitrary factor; choosing the factor
is called normalizing the vector. The most common factor to choose is the one that results in the
eigenvector having a length of 1.
Notes:
 In these examples, you can see that finding the eigenvalues of an n  n matrix involved
solving a polynomial equation of degree n, which means that there are always n
eigenvalues for an n  n matrix.
 Also in these two examples, all the eigenvalues were distinct. However, that is not
always the case. If a given eigenvalue appears m times as a root of the polynomial
equation, then that eigenvalue is said to have algebraic multiplicity m.
 Every eigenvalue will have q linearly independent eigenvectors, where 1  q  m. The
number q is called the geometric multiplicity of the eigenvalue.
 Thus, if each eigenvalue of A is simple (has algebraic multiplicity m = 1), then each
eigenvalue also has geometric multiplicity q = 1.
 If 1 and 2 are distinct eigenvalues of a matrix A, then their corresponding eigenvectors
are linearly independent.
 So, if all the eigenvalues of an n  n matrix are simple, then all its eigenvectors are
linearly independent. However, if there are repeated eigenvalues, then there may be less
than n linearly independent eigenvectors, which will pose complications later on when
solving systems of differential equations (which we won’t have time to get to…).
 Symmetric matrices are a subset of Hermitian, or self-adjoint matrices: A*= A; a ji  aij

Hermitian matrices have the following properties:
 All eigenvalues are real.
 There are always n linearly independent eigenvectors, regardless of multiplicities.
 All eigenvectors of distinct eigenvalues are orthogonal.
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
If an eigenvalue has algebraic multiplicty m, then it is always possible to choose
m mutually orthogonal eigenvectors.
Section 7.4: Basic Theory of Systems of First Order Linear Equations
Section 7.5: Homogeneous Linear Systems with Constant Coefficients
Assume a solution of the form x  ξe rt , plug it into the system, then find the eigenvalues for r
and the eigenvectors for .
Section 7.6: Complex Eigenvalues
If the eigenvalues come in complex conjugate pairs, then the solution will break down into an
exponential times vectors with sines and cosines for elements.