1. FIRST-ORDER DIFFERENTIAL EQUATIONS
1.1 Preliminary Concepts
1. General and particular solutions: For F ( x, y, y ' )  0 , any equation involving a first
derivative; y   (x) such that F  0 .
Example: y   y  2 
xy    y 
y( x)  2  ce  x
y ( x)  c / x
y   cos x  0 
y ( x)  sin x  c
2. Implicitly defined solutions
Example: y   
2 xy3  2
3x 2 y 2  8e 4 y

x 2 y 3  2 x  2e 4 y  c
3. Integral curves: a graph of a solution
4. The initial value problem: F ( x, y, y ' )  0 , initial condition: y( x0 )  y 0
Example: y   3 y,
y(0)  5.7 
y( x)  5.7e 3 x
5. Direction fields: F ( x, y, y ' )  0  y '  F ( x, y )
1.2 Separable Equations
1. Separable differential equation: y '  A( x) B( y )
Example: y   y 2 e  x

y
1
e c
x
1
RC circuits: Charging: E  IR 
Discharging: IR 
Q
C
Q
C
 Q  CE (1  e t / RC )
 Q  Q0 e t / RC
1.3 Linear Differential Equations: y ' ( x)  p( x) y ( x)  q( x) , integrating factor: e 
Example: y   y  sin( x) 
1
y  [sin( x)  cos( x)]  Ce  x .
2
p ( x ) dx
y   3x 2 
y
, y (1)  5 .
x
1.4 Exact Differential Equations
1. Potential function: For M ( x, y )  N ( x, y ) y '  0 , we can find a  ( x, y ) such that
and

M
x

 N ;  is the potential function; M ( x, y )  N ( x, y ) y '  0 is
y
exact.
2. Exact differential equation: a potential function exists; general solution:  ( x, y )  c .
2 xy3  2
Example: y    2 2
.
3x y  8e 4 y
3. Theorem: Test for exactness:
M N

y
x
Example: , x 2  3xy  (4 xy  2 x) y   0 .
e x s i ny  2 x  (e x c o sy  1) y   0 .
1.5 Integrating Factors
1. Integrating factor:  ( x, y )  0 such that M ( x, y )  N ( x, y ) y '  0 is exact.
Example: y 2  6 xy  (3xy  6 x 2 ) y   0 .
2
2. How to find integrating factor:
( M ) ( N )

y
x
Example: x  xy  y   0 .
3. Separable equations and integrating factors:  
4. Linear equations and integrating factors:   e 
1
B
p ( x ) dx
1.6 Homogeneous and Bernoulli Equations
y
1. Homogeneous differential equation: y '  f ( ) ; let y  ux  separable.
x
Example: xy  
y2
 y.
x
2. Bernoulli equation: y' P( x) y  R( x) y  ;   0  linear;   1  separable; otherwise,
let v  y1  linear
Example: y  
y
 3x 2 y 3 .
x
3
2. SECOND-ORDER DIFFERENTIAL EQUTIONS
2.1 Preliminary Concepts
1. F ( x, y, y , y )  0 , an equation that contains a second derivative, but no higher derivative.
2. Linear second-order differential equations: R( x) y" P( x) y 'Q( x) y  S ( x) .
2.2 Theory of Solutions
1. The initial value problem: y" p( x) y ' q( x) y  f ( x) ; y( x0 )  A , y' ( x0 )  B .
Example: y   12 x  0, y(0)  3, y (0)  1

y  2x 3  x  3 .
2. The homogeneous linear ODEs of 2nd order: y" p( x) y ' q( x) y  0 .
3. Theorem: Let y1 and y 2 be solutions of y" p( x) y ' q( x) y  0 on an interval I. Then
any linear combination of these solutions, i.e., y  c1 y1  c2 y2 , is also a solution.
4. Linear dependence: Two functions f and g are linearly dependent on an open interval I if, for
some constant c, either f ( x)  cg ( x) for all x in I, or g ( x)  cf ( x) for all x in I.
Linear
independence: If f and g are not linearly dependent on I.
Example: y  y  0

y1  cos x, y2  sin x .
5. Wronskian Test: Let y1 and y 2 be solutions of y" p( x) y ' q( x) y  0 on an open interval
I. Then, (1) Either W ( x)  0 for all x in I, or W ( x)  0 for all x in I. (2) y1 and y 2
are linearly independent on I if and only if W ( x)  0 on I, where W ( x) 
4
y1
y1
y2
.
y 2
y   xy  0

Example:
1 3
1 6
1
x 
x 
x 9  ,
6
180
12960
.
1 4
1 7
1
10
y2  x  x 
x 
x 
12
504
45360
y1  1 
6. Theorem: Let y1 and y 2 be linearly independent solutions of y" p( x) y ' q( x) y  0 on
an open interval I. Then, every solution of this differential equation on I is a linear
combination of y1 and y 2 .
7. Definition: Let y1 and y 2 be solutions of y" p( x) y ' q( x) y  0 on an open interval I.
(1) y1 and y 2 form a fundamental set (or a basis) of solutions on I if y1 and y 2 are
linearly independent on I. (2) When y1 and y 2 form a fundamental set of solutions, we
call c1 y1  c2 y 2 , with c1 and c2 arbitrary constants, the general solution of the
differential equation on I.
8. The nonhomogeneous equations: y" p( x) y ' q( x) y  f ( x) .
9. Theorem: Let y1 and y 2 be a fundamental set of solutions of y" p( x) y ' q( x) y  0 on an
open interval I.
Let y p be any solution of y" p( x) y ' q( x) y  f ( x) on I. Then, for
any solution  of y" p( x) y ' q( x) y  f ( x) , there exist numbers c1 and c2 such that
  c1 y1  c 2 y 2  y p .
2.3 Reduction of Order: Given y" p( x) y ' q( x) y  0 , if we know a first solution y1 , then a
second solution can be the form y 2  u( x) y1 .
Example: y   4 y   4 y  0, y1  e 2 x

y 2  xe2 x .
2.4 The Constant Coefficient Homogeneous Linear Equation: y" Ay' By  0 , A and B are
numbers.
5
1. Characteristic equation: 2  A  B  0 obtained by substituting y  e x into
y" Ay' By  0 .
 A  A2  4B
2. Case 1. A  4 B  0 : The general solution is y( x)  c1e  c2 e ; a 
,
2
2
ax
bx
 A  A2  4B
.
b
2
Example: y   y   6 y  0

y  c1e 2 x  c2 e 3 x .
3. Case 2. A 2  4 B  0 : The general solution is y( x)  c1e ax  c2 xeax ; a  
Example: y   6 y   9 y  0

A
.
2
y  c1e 3 x  c2 xe3 x .
4. Case 3. A 2  4 B  0 : The general solution is y( x)  c1e ( p iq) x  c2 e ( p iq) x ; p  
q
A
,
2
4B  A2
.
2
Example: y  2 y  6 y  0

y  c1e ( 1
5i ) x
 c2 e ( 1
5i ) x
.
5. An alternative general solution in the complex root case: y( x)  e px (c1 cos(qx)  c2 sin( qx)) .



1
(1) n 2 n
(1) n 2 n1
Maclaurin expansions: e x   x n , cos x  
x , sin x  
x .
n  0 n!
n 0 (2n)!
n 0 (2n  1)!
Euler’ formula: e ix  cos x  i sin x .
Example: y   2 y   6 y  0

y  e  x (c1 cos( 5 x)  c 2 sin( 5 x)) .
2.5 Euler’s (Euler-Cauchy) Equation: x 2 y" Axy' By  0 , let (i) y  x   Characteristic
equation: 2  ( A  1)  B  0 , or (ii) let x  e t , t  ln x , Y (t )  y (e t ) 
Y "( A  1)Y ' BY  0 .
6
1. Case 1. ( A  1) 2  4B  0 : The general solution is y( x)  c1 x a  c2 x b ;
a
(1  A)  ( A  1) 2  4 B
(1  A)  ( A  1) 2  4 B
, b
.
2
2
Example: x 2 y   2 xy   6 y  0
y  c1 x 3  c2 x 2 .

2. Case 2. ( A  1) 2  4B  0 : The general solution is y( x)  c1 x a  c2 x a ln x ; a 
Example: x 2 y   5 xy   9 y  0

1 A
.
2
y  c1 x 3  c2 x 3 ln x .
3. Case 3. ( A  1) 2  4B  0 : The general solution is y( x)  x p (c1 cos( q ln x)  c2 sin( q ln x)) ;
1 A
p
, q
2
Example: x 2 y   0.6 xy   16.04 y  0

4 B  ( A  1) 2
.
2
y  x 0.2 (c1 cos(4 ln x)  c2 sin( 4 ln x)) .
2.6 The Nonhomogeneous Equation: y" p( x) y ' q( x) y  f ( x) , general solution y  y h  y p .
1. The method of variation of parameters: let y p  uy1  vy2 , then simultaneously solve
 u y1  v y 2  0
.

u y1  v y 2  f
y   4 y  sec x,   / 4  x   / 4
Example:

.
1
y  c1 cos 2 x  c 2 sin 2 x  cos x cos 2 x  (sin x  ln sec x  tan x ) sin 2 x
2
2. The method of undetermined coefficients: only applied while p(x) and q(x) are constant, i.e.,
y" Ay' By  f ( x) .
7
Example: y   4 y  8 x 2  2 x

y  c1e 2 x  c 2 e  2 x  2 x 2 
1
x 1 .
2
-- Modification Rule: If a term in your choice for y p happens to be a solution of the
homogeneous ODE, multiply your choice of y p by x (or by x 2 if this solution
corresponds to a double root of the characteristic equation of the homogeneous ODE).
Example: y   2 y   3 y  8e x
y   6 y   9 y  5e 3 x

y  c1e 3 x  c2 e x  2 xex .

y  c1e 3 x  c 2 xe3 x 
5 2 3x
x e
2
3. The principle of superposition: y" p( x) y'q( x) y  f1 ( x)  f 2 ( x)    f n ( x) , y pj is a
solution of y" p ( x) y ' q( x) y  f j ( x) , then y p1  y p 2    y pn is a solution.
Example: y   4 y  x  2e  2 x

y  c1 cos 2 x  c 2 sin 2 x 
8
1
( x  e 2 x ) .
4
3. HIGHER ORDER LINEAR ODES
3.1 Homogeneous Linear ODEs
1. F ( x, y, y , , y ( n ) )  0 , a nth order ODE if the nth derivative y ( n ) 
dny
of the unknown
dy n
function y (x) is the highest occurring derivative.
2. Linear ODE: y ( n )  p n 1 ( x) y ( n 1)    p1 ( x) y   p0 ( x) y  g ( x) .
3. Homogeneous linear ODE: y ( n )  p n 1 ( x) y ( n 1)    p1 ( x) y   p0 ( x) y  0 .
4. Theorem: Fundamental Theorem for the Homogeneous Linear ODE: For a homogeneous
linear ODE, sums and constant multiples of solutions on some open interval I are again
solutions on I. (This does not hold for a nonhomogeneous or nonlinear ODE!).
5. General solution: y  c1 y1    cn y n , where y1 , , y n is a basis (or fundamental system)
of solutions on I; that is, these solutions are linearly independent on I.
6. Linear independence and dependence: n functions y1 , , y n are called linearly independent
on some interval I where they are defined if the equation k1 y1    k n y n  0 on I implies
that all k1 , , k n are zero. These functions are called linearly dependent on I if this
equation also holds on I for some k1 , , k n not all zero.
Example:
d4y
d2y

5
 4y  0 .
dx 4
dx 2
Sol.: y  c1e 2 x  c 2 e  x  c3 e x  c 4 e 2 x .
9
7. Theorem: Let the homogeneous linear ODE have continuous coefficients p0 ( x),  , p n1 ( x)
on an open interval I. Then n solutions y1 , , y n on I are linearly dependent on I if and
only if their Wronskian is zero for some x  x0 in I. Furthermore, if W is zero for x  x0 ,
then W is identically zero on I. Hence if there is an x1 in I at which W is not zero, then
y1 , , y n are linearly independent on I, so that they form a basis of solutions of
the homogeneous linear ODE on I.
y1
y1
Wronskian: W ( y1 , , y n ) 

( n 1)
1
y
y
y2
y 2





( n 1)
 yn
( n 1)
2
yn
y n
8. Initial value problem: An ODE with n initial conditions y ( x0 )  K 0 , y ( x0 )  K1 ,  ,
y ( n 1) ( x0 )  K n 1 .
3.2 Homogeneous Linear ODEs with Constant Coefficients
1. y ( n )  a n 1 y ( n 1)    a1 y   a0 y  0 : Substituting y  e x , we obtain the characteristic
equation n  a n 1n 1    a1  a 0  0 .
(i) Distinct real roots: The general solution is y  c1e 1x    c n e n x
Example: y   2 y   y   2 y  0 . Sol.: y  c1e  x  c 2 e x  c3 e 2 x .
(ii) Simple complex roots:   p  qi , y1  e px cos( qx) , y 2  e px sin( qx) .
Example: y   y   100 y   100 y  0 .
Sol.: y  c1e x  c 2 cos 10 x  c3 sin 10 x .
10
(iii) Multiple real roots: If  is a real root of order m, then m corresponding linearly
independent solutions are: e x , xex , x 2 e x ,  , x m 1e x .
Example: y (5)  3 y ( 4)  3 y   y   0 .
Sol.: y  c1  c 2 x  (c3  c 4 x  c5 x 2 )e x .
(iv) Multiple complex roots: If   p  qi are complex double roots, the corresponding
linearly independent solutions are:
e px cos(qx) ,
e px sin( qx) ,
xe px cos(qx) ,
xe px sin( qx) .
2. Convert the higher-order differential equation to a system of first-order equations.
Example:
d6y
d4y
dy

4
 2  15 y  0 .
6
4
dx
dx
dx
3.3 Nonhomogeneous Linear ODEs
1. y ( n )  p n 1 ( x) y ( n 1)    p1 ( x) y   p0 ( x) y  g ( x) , the general solution is of the form:
y  y h  y p , where y h is the homogeneous solution and y p is a particular solution.
2. Method of undermined coefficients
Example: y   3 y   3 y   y  30e  x .
Sol.: y  (c1  c 2 x  c3 x 2 )e  x  5 x 3 e  x .
3. Method of variation of parameters: y p  u1 y1    u n y n , where u k 
Example: x 3 y   3x 2 y   6 xy   6 y  x 4 ln x .
11
Wk
, k  1,  , n .
W
Sol.: y  c1  c 2 x  c3 x 2 
1 4
11
x (ln x  ) .
6
6
4. LAPLACE TRANSFORM
4.1 Definition and Basic Properties: initial value problem  algebra problem  solution of the
algebra problem  solution of the initial value problem

1. Definition (Laplace Transform): The Laplace transform L[ f ]( s)   e st f (t )dt  F (s) , for
0
all s such that this integral converges.
Examples: f (t )  e at  L[ f ]( s ) 
1
, s  a.
sa
g (t )  sin t  L[ f ]( s ) 
1
.
s 1
2
2. Table of Laplace transform of functions
3. Theorem (Linearity of the Laplace transform): Suppose L[ f ]( s ) and L[ g ]( s ) are defined
for s  a , and  and  are real numbers. Then
L[f  g ]( s )  F ( s )  G ( s ) for s  a .
12
4. Definition (Inverse Laplace transform): Given a function G, a function g such that L[ g ]  G
is called an inverse Laplace transform of G.
In this event, we write
g  L1 [G ] .
5. Theorem (Lerch): Let f and g be continuous on [0, ) and suppose that L[ f ]  L[ g ] .
Then f  g .
6. Theorem: If L1[ F ]  f
and L1 [G ]  g and  and 
are real numbers, then
L1[F  G]( s)  f  g .
4.2 Solution of Initial Value Problems Using the Laplace Transform
1. Theorem (Laplace transform of a derivative): Let f be continuous on [0, ) and suppose f '
is piecewise continuous on [0, k ] for every positive k.
Suppose also that
lim e  sk f (k )  0 if s  0 . Then L[ f ' ]( s)  sF ( s )  f (0) .
k 
2. Theorem (Laplace transform of a higher derivative): Suppose f , f ' , , f n1 are continuous
on [0, ) and f (n ) is piecewise continuous on [0, k ] for every positive k.
Suppose also that lim e  sk f ( j ) (k )  0 for s  0 and for j  1, 2, , n  1 . Then
k 
L[ f ( n ) ]( s)  s n F (s)  s n1 f (0)  s n2 f ' (0)    sf
Examples: y '4 y  1; y (0)  1  y 
( n 2)
(0)  f ( n1) (0) .
5 4t 1
e  .
4
4
1
3
7
y"4 y'3 y  e t ; y(0)  0, y' (0)  2  y  e t  e t  e 3t .
8
4
8
4.3 Shifting Theorems and the Heaviside Function
1. Theorem (First shifting theorem, or shifting in the s variable): Let L[ f ]( s )  F ( s ) for
13
s b  0.
Let a be any number.
Then L[e at f ]( s)  F ( s  a) for s  a  b .
 L1[ F ( s  a)]  e at L1[ F ( s)]  e at f (t ) .
Examples: L[e at cos(bt )] 
sa
.
(s  a) 2  b 2
4


Find L1  2
 e 2t sin 4t .

 s  4 s  20 
2. Definition (Heaviside function): The Heaviside function (or unit step function) H is defined
0 if t  0
0 if t  a
by H (t )  
.  H (t  a)  
1 if t  0
1 if t  a
 0 if t  a
 0 if t  a
-- On-off effect: H (t  a) g (t )  
, H (t  a) g (t  a)  
 g (t ) if t  a
 g (t  a) if t  a
3. Definition (Pulse): A pulse is a function of the form H (t  a)  H (t  b) , in which a  b .
14
4. Theorem (Second shifting theorem, or shifting in the t variable): Let L[ f ]( s )  F ( s ) for
s  b . Then L[ H (t  a) f (t  a)]( s)  e  as F (s) for s  b . 
L1[e  as F ( s)]  H (t  a) f (t  a)
Examples: L[ H (t  a)] 
e  as
.
s
L-1 [
se 3 s
]  H (t  3) xos(2(t  3)) .
s2  4
Compute L[g ] , where g (t )  0 for 0  t  2 and g (t )  t 2  1 for t  2 .
4 5
2
 e 2 s  3  2   .
s
s
s
Solve
y   4 y  f (t )
,
y (0)  y (0)  0 ,
0 if t  3
f (t )  
 t if t  3
.

3
1
3 1

y  H (t  3)   (t  3)  cos( 2(t  3))  sin( 2(t  3))  .
4
8
4 4

4.4 Convolution
1. Definition (Convolution): If f and g are defined on [0, ) , then the convolution f  g of f
t
with g is the function defined by ( f  g )(t )   f (t   ) g ( )d for t  0 .
0
2. Theorem (Convolution theorem): If f  g is defined, then
L[ f  g ]  L[ f ]L[ g ]  F ( s)G ( s) .
3. Theorem: Let L1[ F ]  f and L1 [G ]  g . Then L1[ FG]  f  g .


1 4t 1 4t 1
1
te  e  .

Example: L1 
2 
4
16
16
 s( s  4) 
t
2
Determine f such that f (t )  2t 2   f (t   )e  d  2t 2  t 3 .
0
3
4. Theorem: If f  g is defined, so is g  f , and f  g  g  f .
15
Example: Solve y"2 y '8 y  f (t );
y (0)  1, y ' (0)  0 
1
1
1
2
f  e 4t  f  e  2t  e 4t  e  2t .
6
6
3
3
4.5 Unit Impulses and the Dirac’s Delta Function
1. Dirac’s delta function:  (t )  lim   (t ) , where   (t ) 
 0 
1

[ H (t )  H (t   )] ;
L[ (t  a)]  e  as ; L[ (t )]  1 .
2. Theorem (Filtering property): Let a  0 and let f be integrable on [0, ) and continuous at
a. Then


0
-- Let f (t )  e  st 
f (t ) (t  a)dt  f (a)


0

f (t ) (t  a)dt   e st (t  a)dt  e sa  f (a)  the definition
0
of the Laplace transformation of the delta function.
Example: Solve y"2 y '2 y   (t  3);
y (0)  y ' (0)  0  y  H (t  3)e  (t 3) sin( t  3) .
4.6 Laplace Transform Solution of Systems
x"2 x'3 y '2 y  4,
7
1
10
x    3t  e  2t  et
2
6
3

Example: Solve the system: 2 y ' x'3 y  0,
1  2t 2 t
y  1  e  e
x(0)  x' (0)  y (0)  0.
3
3
16
4.7 Differential Equations with Polynomial Coefficients
1. Theorem: Let L[ f ]( s )  F ( s ) for s  b and suppose that F is differentiable.
Then
L[tf (t )]( s)   F ' ( s) for s  b .
2. Corollary: Let L[ f ]( s )  F ( s ) for s  b and let n be a positive integer. Suppose F is n
times differentiable.
Then L[t n f (t )]( s)  (1) n
Example: ty"(4t  2) y '4 y  0;
dn
F ( s) for s  b .
ds n
y (0)  1 
1
1
1
 1

y  e  4t  2te  4t  c 
 t  e  4t  te  4t  .
32
16
 32 16

3. Theorem: Let f be piecewise continuous on [0, k ] for every positive number k and suppose
there are numbers M and b such that f (t )  Me bt for t  0 . Let L[ f ]( s )  F ( s ) .
Then lim F ( s )  0 .
s 
Example: y"2ty'4 y  1;
y (0)  y ' (0)  0  y 
17
1 2
t .
2
5. SERIES SOLUTIONS
5.1 Power Series Solutions of Initial Value Problems
1. Definition (Analytic function): A function f is analytical at x 0 if f (x) has a power series

representation in some open interval about x 0 : f ( x)   a n ( x  x0 ) n in some
n 0
interval ( x0  h, x0  h) .

Example: Taylor series: f ( x)  
n 0
f ( n ) ( x0 )
f ( n ) ( x0 )
n
.
( x  x0 ) , a n 
n!
n!

Maclaurin series: f ( x)  
n 0
f ( n ) (0) n
x , i.e., x0  0 in Taylor series.
n!
(1) n 2 n1
at x  0 .
sin x  
x
n 0 (2n  1)!

2. Theorem: Let p and q be analytic at x 0 . Then the initial value problem y ' p( x) y  q( x) ;
y( x0 )  y 0 has a solution that is analytical at x 0 .
Example: y'e x y  x 2 ; y (0)  4  y ( x)  4  4 x  x 3 
3. Theorem: Let p, q and f be analytic at x 0 .
x4

12
Then the initial value problem
y" p( x) y ' q( x) y  f ( x) ; y ( x0 )  A , y' ( x0 )  B has a unique solution that is
also analytical at x 0 .
Examples: y" xy'e x y  4 ; y (0)  1, y ' (0)  4  y( x)  1  4 x 
18
3 2 x3
x 
 .
2
6
y" cos( x) y '4 y  2 x  1 
y ( x)  a  bx 
 1  4a  b 2 4a  3b  3 3
x 
x   , a  y (0) , b  y (0) .
2
6
5.2 Power Series Solutions Using Recurrence Relations
1. Coefficients developed to be a recurrence relation
Example: y" x 2 y  0 at x  0  a2  0 , a3  0 , a n  
y  a 0 (1 
1
a n 4 , n  4, 5,  ,
n(n  1)
1 4
1 8
1 5
1 9
x 
x  )  a1 ( x 
x 
x  ) .
12
672
20
1440
2. Two-term recurrence relation
Example: y" x 2 y'4 y  1  x 2 at x  0  a 2 
a4  
1
2
 2a 0 , a 3   a 1 ,
2
3
4a  (n  3)a n 3
1 2
1
 a 0  a1 , an   n2
, n  5, 6,  .
4 3
12
n(n  1)
5.3 Singular Points and the Method of Frobenius
1. Definition (Ordinary and singular points): x 0 is an ordinary point of equation
P( x) y"Q( x) y ' R( x) y  F ( x) if P( x0 )  0 and Q( x) / P( x) , R ( x) / P ( x) , and
F ( x) / P( x) are analytic at
x0 .
x0
is a singular point of equation
P( x) y"Q( x) y ' R( x) y  F ( x) if x 0 is not an ordinary point.
Example: x 3 ( x  2) 2 y"5( x  2)( x  2) y'3x 2 y  0  x  0 , x  2 are singular points.
19
2. Definition (Regular and irregular singular points): x 0 is a regular singular point of
P( x) y"Q( x) y ' R( x) y  0 if
( x  x0 )
x0
is a singular point, and the functions
R( x)
Q( x)
and ( x  x0 ) 2
are analytic at x 0 . A singular point that is
P( x)
P( x)
not regular is said to be an irregular singular point.

Example: x 3 ( x  2) 2 y"5( x  2)( x  2) y'3x 2 y  0
x  0 is an irregular singular
point, x  2 is a regular singular points.

3. Frobenius series: y ( x)   c n ( x  x0 ) n  r .
n 0
4. Theorem (Method of Frobenius): Suppose
P( x) y"Q( x) y ' R( x) y  0 .

y ( x)   c n ( x  x0 ) n  r
n 0
( x  x0 )
x 0 is a regular singular point of
Then there exists at least one Frobenius solution
with c0  0 .
Further, if the Taylor expansions of
R( x)
Q( x)
and ( x  x0 ) 2
about x 0 converge in an open interval
P( x)
P( x)
( x0  h, x0  h) , then this Frobenius series also converges in this interval, except
perhaps at x 0 itself.
There will be an indicial equation used to determine the values of r.
1
1
Example: x 2 y" x(2 x  ) y ' ( x  ) y  0 
2
2
r  1 : cn  
1
2n  2 
r   : c n  
c , n  1, 2, 
3 n 1
2
n( n  )
2
20
2n  1
c , n  1, 2,  ;
3 n 1
n( n  )
2
5.4 Second Solutions and Logarithm Factors
1. Theorem (A second solution in the method of Frobenius): Suppose 0 is a regular singular
point of P( x) y"Q( x) y ' R( x) y  0 .
Let r1 and r2 be roots of the indicial
If these are real, suppose r1  r2 . Then (a) If r1  r2 is not an integer,
equation.

there are two linearly independent Frobenius solutions: y1 ( x)   c n x n  r1 and
n 0

y 2 ( x)   c n* x n  r2 , with c0  0 and c 0*  0 .
These solutions are valid in some
n 0
(b) If r1  r2  0 , there is a Frobenius solution
interval (0, h) or (h, 0) .

y1 ( x)   c n x n  r1
with
n 0
c0  0
as
well
as
a
second
solution:

y 2 ( x)  y1 ( x) ln x   c n* x n  r1 .
n 1
Further, y1 and y 2 form a fundamental set of
solutions on some interval (0, h) .
(c) If r1  r2 is a positive integer, then there

is a Frobenius solutions: y1 ( x)   c n x n  r1 .
In this case there is a second
n 0

solution of the form y 2 ( x)  ky1 ( x) ln x   c n* x n  r2 . If k  0 this is a second
n 0
Frobenius series solution; if not, the solution contains a logarithm term.
In either
event, y1 and y 2 form a fundamental set on some interval (0, h) .
Examples: x 2 y"5xy'( x  4) y  0 . 

r  2 : y1 ( x)  c0  (1) n
n 0
c n  
1 
2(1)
c 
, n  2, 3,  .
2 n 1
n
n(n!) 2
n
x 2 y" x 2 y'2 y  0  r  2 or r  1 .
21
1
x n2 , c1  2 ,
2
(n!)
6. FOURIER SERIES
6.1 The Fourier Series of a Function
1. f ( x) 

1
nx
nx
a 0   a n cos(
)  bn sin(
),  L  x  L,
2
L
L
n 1
L
2. Lemma 13.1: If n and m are nonnegative integers, i.  cos(
L
L
ii.  cos(
L
L
L
L
f ( x)dx exists.
nx
mx
) sin(
)dx  0 ;
L
L
L
nx
mx
nx
mx
) cos(
)dx   sin(
) sin(
)dx  0 , if n  m ;

L
L
L
L
L
iii.  cos 2 (
L

L
nx
nx
)dx   sin 2 (
)dx  L , if n  0 .

L
L
L
3. Definition 13.1: Let f be a Riemann integrable function on [  L, L ] , then Fourier series of f
on [  L, L ] :

1
nx
nx
a 0   a n cos(
)  bn sin(
) ; Fourier coefficients of f
2
L
L
n 1
on [  L, L ] : a n 
1 L
nx
1 L
nx
f ( x) cos(
)dx , bn   f ( x) sin(
)dx for

L L
L
L L
L
n  0, 1, 2,  .
4. Definition 13.2: Even and odd functions:
f is an even function on [  L, L ] if f ( x)  f ( x) for  L  x  L ; f is an odd function
on [  L, L ] if f ( x)   f ( x) for  L  x  L ; even  even  even ; odd  odd  even ;
even  odd  odd .

L
L
f ( x)dx  0 if f is odd on [  L, L ] ;

L
L
6.2 Convergence of Fourier Series
22
L
f ( x)dx  2 f ( x)dx if f is even on [  L, L ] .
0
1. Definition 13.3: Piecewise continuous function
f is piecewise continuous on [a, b] if
1. f is continuous on [a, b] except perhaps at finitely many points.
2. Both lim f ( x) and lim f ( x ) exist and are finite.
xa 
x b 
3. If x 0 is in (a, b) and f is not continuous at x 0 , then lim f ( x) and lim f ( x) exist
x  x0 
x  x0 
and are finite.
2. Definition 13.4: Piecewise smooth function
f is piecewise smooth on [a, b] if f and f ' are piecewise continuous on [a, b].
3. Theorem 13.1: Convergence of Fourier series
Let f is piecewise smooth on [  L, L ] . Then for  L  x  L , the Fourier series of f on
1
[  L, L ] converge to ( f ( x )  f ( x )) .
2
4. Convergence at the endpoints
5. Definition 13.5: Right derivative f R ' (c)  lim
h 0 
6. Definition 13.6: Left derivative f L ' (c)  lim
h 0 
f (c  h )  f ( c  )
h
f (c  h )  f (c )
h
7. Theorem 13.2: Let f is piecewise smooth on [  L, L ] . Then,
i. If  L  x  L and f has a left and right derivative at x, then the Fourier series of f on
1
[  L, L ] converge at x to ( f ( x )  f ( x )) .
2
23
ii. If f R ' (L) and f L ' ( L) exist, then at both L and  L , the Fourier series of f on
1
[  L, L ] converge to ( f ( L )  f ( L )) .
2
8. Partial sums of Fourier series
6.3 Fourier Cosine and Sine Series
1. The Fourier cosine series of a function
 f ( x), for 0  x  L
Let f be integrable on the half-interval [0, L]: f e ( x)  
, f e is an
 f ( x), for  L  x  0
even function and called even extension of f on [  L, L ] .
Fourier cosine series of f on [0, L]:
on [0, L]: a n 

1
nx
a 0   a n cos(
) ; Fourier cosine coefficients of f
2
L
n 1
2 L
nx
2 L
nx
f
(
x
)
cos(
)
dx

f ( x) cos(
)dx .
e


0
0
L
L
L
L
2. Theorem 13.3: Convergence of Fourier cosine series
Let f is piecewise continuous on [0, L] . Then,
i. If 0  x  L and f has a left and right derivative at x, then the Fourier cosine series of f
1
on [0, L] converges at x to ( f ( x )  f ( x )) .
2
ii. If f has a right derivative at 0, then the Fourier cosine series of f on [0, L] converges at
0 to f (0) .
iii. If f has a left derivative at L, then the Fourier cosine series of f on [0, L] converges at
24
L to f ( L ) .
3. The Fourier sine series of a function
 f ( x), for 0  x  L
Let f be integrable on the half-interval [0, L]: f o ( x)  
, f o is
 f ( x), for  L  x  0
an odd function and called odd extension of f on [  L, L ] .

Fourier sine series of f on [0, L]:
b
n 1
bn 
n
sin(
nx
) ; Fourier sine coefficients of f on [0, L]:
L
2 L
nx
2 L
nx
f
(
x
)
sin(
)
dx

f ( x) sin(
)dx .
o


0
0
L
L
L
L
4. Theorem 13.4: Convergence of Fourier sine series
Let f is piecewise continuous on [0, L] . Then,
i. If 0  x  L and f has a left and right derivative at x, then the Fourier sine series of f on
1
[0, L] converges at x to ( f ( x )  f ( x )) .
2
ii. At 0 and L, the Fourier sine series of f on [0, L] converges to 0.
6.4 Integration and Differentiation of Fourier Series
1. Theorem 13.5: Integration of Fourier series
Let f be piecewise continuous on [  L, L ] , with Fourier series

1
nx
nx
a 0   a n cos(
)  bn sin(
).
2
L
L
n 1

x
L
f (t )dt 
Then, for any x on [  L, L ] ,
1
L  1
nx
nx


a0 ( x  L)   a n sin(
)  bn  cos(
)  (1) n  .
2
 n 1 n 
L
L


25
2. Theorem 13.6: Differentiation of Fourier series
Let f be continuous on [  L, L ] and suppose f ( L)  f ( L) .
continuous on [  L, L ] . Then, f ( x) 
Let f ' be piecewise

1
nx
nx
a 0   a n cos(
)  bn sin(
) , and at each
2
L
L
n 1
point in ( L, L) where f " ( x ) exists,
n
n 1 L

f ' ( x)  
nx
nx 

 a n sin( L )  bn cos( L ) .
3. Theorem 13.7: Bessel’s inequalities: i. The coefficients in the Fourier sine expansion of f on

[0, L] satisfy
b
n 1
2
n

of f on [0, L] satisfy
2 L 2
f ( x)dx ; ii. The coefficients in the Fourier cosine expansion
L 0
1 2  2 2 L 2
a 0   a n   f ( x)dx ; iii. If f is integrable on [  L, L ] , then
2
L 0
n 1
the Fourier coefficients of f on [  L, L ] satisfy
1 2  2
1 L
a 0   (a n  bn2 )   f 2 ( x)dx
2
L L
n 1
4. Theorem 13.8: Uniform and absolute convergence of Fourier series: Let f be continuous on
[  L, L ] and let f ' be piecewise continuous. Suppose f ( L)  f ( L) . Then, the Fourier
series of f on [  L, L ] converges absolutely and uniformly to f (x) on [  L, L ] .
5. Theorem 13.9: Parseval’s theorem: Let f be continuous on [  L, L ] and let f ' be piecewise
continuous. Suppose f ( L)  f ( L) . Then, the Fourier coefficients of f on [  L, L ]
satisfy
1 2  2
1 L
a 0   (a n  bn2 )   f 2 ( x)dx
2
L L
n 1
6.5 The Phase Angle Form of a Fourier Series
1. Periodic, fundamental period
26
2. Definition 13.7: Phase angle form: Let f have fundamental period p. Then the phase angle
form of the Fourier series of f is

1
2
a 0   c n cos( n 0 x   n ) , in which  0 
,
2
p
n 1
 b
cn  an2  bn2 , and  n  tan 1   n
 an

 for n  1, 2,  .

Harmonic form, nth harmonic, harmonic amplitude, phase angle.
6.6 Complex Fourier Series and the Frequency Spectrum
1. Conjugate, magnitude, argument, polar form.
2. Definition 13.8: Complex Fourier series: Let f have fundamental period p. Let  0 

Then the complex Fourier series of f is
d
n  
n
e in0 x , where d n 
2
.
p
1 p/2
f (t )e in0t dt for


p
/
2
p
n  0,  1,  2,  . The numbers d n are the complex Fourier coefficients of f.
3. Theorem 13.10: Let f be periodic with fundamental period p. Let f be piecewise smooth on
[ p / 2, p / 2] . Then at each x the complex Fourier series converges to
4. Amplitude spectrum; frequency spectrum.
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1
( f ( x )  f ( x )) .
2