Chapter 2: Second-Order Differential
Equations
2.1. Preliminary Concepts
○ Second-order differential equation
F ( x, y , y , y )  0
e.g., y  3 y  10 y  7 x  4  0
y  12 x  0
Solution: A function  ( x) satisfies
F ( x, ( x), ( x), ( x))  0 , x  I
(I : an interval)
1
○ Linear second-order differential equation
y  P( x) y  Q( x) y  R( x)
Nonlinear: e.g., y  P( x, y ) y  Q( x, y )  R( x, y )
2.2. Theory of Solution
○ Consider y  12 x  0  y  12 x
y   y( x)dx   12 xdx  6 x 2  c
y   y( x)dx   (6 x 2  c)dx  2 x 3  cx  d
y contains two parameters c and d
2
The graph of y  2 x 3  cx  d
Given the initial condition y (0)  3
 y (0)  2  03  c  0  d  3
The graph of y  2 x3  cx  3 , d  3
3
Given another initial condition y(0)  1
 y(0)  6  02  c  1, c  1
The graph of y  2 x3  x  3
◎ The initial value problem:
y  P( x) y  Q( x) y  R( x) ; y( x0 )  a, y( x0 )  b
○ Theorem 2.1: P, Q, R : continuous on I,
 y  P( x) y  Q( x) y  R( x);
y( x )  a, y( x )  b, x  I ,
0
0
0
has a unique solution x  I
4
2.2.1.Homogeous Equation
y  P( x) y  Q( x) y  0
     (2.2)
○ Theorem 2.2: y1 , y2 : solutions of Eq. (2.2)
 c1 y1 ( x)  c2 y2 ( x) : solution of Eq. (2.2)
c1 , c2 : real numbers
Proof: y  P( x) y  Q( x) y
 (c1 y1  c2 y2 )  P( x)(c1 y1  c2 y2 )  Q( x)(c1 y1  c2 y2 )
 c1 y1  c2 y2  c1P( x) y1  c2 P( x) y2
c1Q( x) y1  c2Q ( x) y2
 c1[ y1  P( x) y1  Q( x) y1 ]  c2[ y2  P( x) y2  Q( x) y2 ]
 00  0
5
※ Two solutions are linearly independent.
Their linear combination provides an
infinity of new solutions
○ Definition 2.1: f , g : linearly dependent
If c, s.t. f ( x)  cg ( x) or g ( x)  cf ( x) ;
otherwise f , g : linearly independent
In other words, f and g are linearly
dependent only if c1  c2  0 for c1 f  c2 g  0
6
○ Wronskian test -- Test whether two solutions of a
homogeneous differential equation are linearly
independent
Define: Wronskian of solutions y1 , y2 to be the 2
by 2 determinant
W ( x)  y1 ( x) y2 ( x)  y1 ( x) y2 ( x)
y1 ( x)

y1( x)
y2 ( x)
y2 ( x)
7
○ Let
c1 y1  c2 y2  0
 c1 y1  c2 y2  0
     ( A)
   ( B )
(A)  y1  ( B)  y1  c2 ( y1 y2  y1 y2 )  0
(A)  y2  ( B)  y2  c1 ( y1 y2  y1 y2 )  0,
If y1 , y2 : linear dep., then c1  0 or c2  0
Assume c1  0  y1 y2  y1 y2  0
8
○ Theorem 2.3:
1) Either W ( x)  0, x  I or W ( x)  0, x  I
2) y1 , y2 : linearly independent iff W ( x)  0
Proof (2):
(i)  (if y1 , y2 : linear indep. (P), then
W ( x)  0 (Q)  if W ( x)  0 ( Q) ,
then y1 , y2 : linear dep. (P) )
1  1 


W  y1 y2  y1 y2  0 , y1 
y2
y1
y2
1 
1 
 y1 y1   y2 y2 , ln y1  ln y2  ln c  ln cy2
y1  cy2 , y1 , y2 : linear dep.
9
(ii)  (if W ( x)  0(P), then y1 , y2 : linear indep. (Q)
 if y1 , y2: linear dep. (Q), then W ( x)  0
(  P))
y1 , y2 : linear dep., c, y1  cy2
W  y1 y2  y1 y2  cy2 y2  cy2 y2  0
※ Test W ( x) at just one point of I to determine
linear dependency of the solutions
10
。 Example 2.2: y1 ( x)  cos x, y2 ( x)  sin x
are solutions of y  y  0
y1 ( x)
W ( x) 
y1( x)
y2 ( x )
cos x sin x

y2 ( x)  sin x cos x
 cos 2 x  sin 2 x  1  0
 y1 , y2 : linearly independent
11
。 Example 2.3: y  xy  0
Solve by a power series method
1 3
1 6
1
y1 ( x)  1  x 
x 
x9    
6
180
12960
1 4
1 7
1
y2 ( x )  x 
x 
x 
x10    
12
504
45360
The Wronskian of y1 , y2 at nonzero x
would be difficult to evaluate, but at x = 0
y1 (0)
W (0) 
y1(0)
y2 (0)
1

y2 (0)
0
0
1
1 0
y1 , y2 are linearly independent
12
◎ Find all solutions
○ Definition 2.2:
1. y1 , y2 : linearly independent
{ y1 , y2 } : fundamental set of solutions
2. c1 y1  c2 y2 : general solution
c1 , c2 : constant
○ Theorem 2.4:
y1 , y2 : linearly independent solutions on I
Any solution is a linear combination of y1 , y2
13
Proof: Let  be a solution.
Show  c1 , c2 s.t.  ( x)  c1 y1 ( x)  c2 y2 ( x)
Let x  I and  ( x0 )  a, ( x0 )  b
0
Then,  is the unique solution on I of the initial
value problem
y  P( x) y  Q( x) y  0, y( x0 )  a, y( x0 )  b
 ( x) : solution,  ( x0 )  a,  ( x0 )  b
  ( x0 )  a c1 y1 ( x0 )  c2 y2 ( x0 )  a
From 
, 
 ( x0 )  b  c1 y1 ( x0 )  c2 y2 ( x0 )  b
ay2 ( x0 )  by2 ( x0 )
by1 ( x0 )  ay1( x0 )
c1 
, c2 
W ( x0 )
W ( x0 )
14
2.2.2. Nonhomogeneous Equation
y  P( x) y  Q( x) y  R ( x)
○ Theorem 2.5:
y1 , y2 : linearly independent homogeneous
solutions of y  P ( x) y  Q ( x ) y  0
yp
: a nonhomogeneous solution of
y  P( x) y  Q( x) y  R ( x)
 any solution  has the form
  c1 y1  c2 y2  y p
15
Proof: Given
,
y p : solutions
(  y p )  P(  y p )  Q(  y p )
    P   Q  ( y p  Pyp  Qy p )
RR0
   y p : a homogenous solution of
y  P ( x ) y  Q ( x ) y  0
y1 , y2 : linearly independent homogenous
solutions
  y p  c1 y1  c2 y2 (Theorem 2.4)
  c1 y1  c2 y2  y p
16
○ Steps:
1. Find the general homogeneous solutions
y1 , y2 of y  Py  Qy  0
2. Find any nonhomogeneous solution y p of
y  Py  Qy  R
3. The general solution of y  Py  Qy  0
is c1 y1  c2 y2  y p
2.3. Reduction of Order
-- A method for finding the second independent
homogeneous solution y2 when given the first one y1
17
○ Let y2 ( x)  u( x) y1 ( x)
 y2  uy1  uy1, y2  uy1  2uy1  uy1
Substituting into y  P( x) y  Q( x) y  0
uy1  2uy1  uy1  P[u y1  uy1 ]  Quy1  0
uy1  u[2 y1  Py1 ]  u[ y1  Py1  Qy1 ]  0
uy1  u[2 y1  Py1 ]  0
( y1 : a homogeneous
solution )

2 y1  Py1

u 
u  0  u  G( x)u  0
y1
Let v  u  v  G ( x)v  0 (separable)
v( x)  ce 
 G ( x ) dx
18
For symlicity, let c = 1, u  v  e 
  G ( x ) dx
 u ( x)  e
dx
 G ( x ) dx
0

y2 ( x)  u ( x) y1 ( x)  y1 ( x)  e 
 G ( x ) dx
dx
W ( x)  y1 y2  y1 y2  y1 (uy1  u y1 )  y1uy1
 y1uy1  y1uy1  y1uy1  u y12  vy12  0
y1, y2  uy1 : independent solutions
。 Example 2.4: y  4 y  4 y  0,      ( A)
y( x)  e2 x : a solution
Let y2 ( x)  u ( x) y1 ( x)  u ( x)e 2 x
19
 y2  ue 2 x  2e 2 xu , y2  ue 2 x  4e 2 xu  4ue 2 x
Substituting into (A),
ue2 x  4e2 xu  4ue 2 x  4(ue 2 x  2e 2 xu )  4ue 2 x  0
ue2 x  0,
e2 x  0, u  0, u( x)  cx  d
For simplicity, take c = 1, d = 0  u ( x)  x
y2 ( x)  u( x)e2 x  xe2 x
e2 x
W ( x) 
2e2 x
xe2 x
e2 x  2 xe2 x
 e4 x  0
y1  e2 x , y2  xe2 x : independent
The general solution: y( x)  c1e2 x  c2 xe2 x
20
2.4. Constant Coefficient Homogeneous
y  Ay  By  0 A, B : numbers ----- (2.4)
The derivative of
e
 x is
a constant (i.e., ) multiple
of e  x
Constant multiples of derivatives of y , which has form
e x , must sum to 0 for (2,4)
○ Let y ( x)  e x
Substituting into (2,4),  2e x  Ae x  Be x  0
 2  A  B  0
(characteristic equation)
  ( A  A2  4B ) / 2
21
i) A2  4B  0
 1  ( A  A2  4 B ) / 2, 2  ( A  A2  4 B ) / 2
Solutions : y1  e1x , y2  e2 x
W ( x)  y1 y2  y1 y2  e1x 2 e2 x  e2 x 1e1x
 (2  1 )e( 1 2 ) x   A2  4 B e  Ax  0
y , y : linearly independent
1
2
The general solution: y ( x )  c1e 1 x  c2 e 2 x
22
。 Example 2.6: y  y  6 y  0
     ( A)
Let y ( x)  e x , Then y   e  x , y   2 e  x
Substituting into (A),
 2e x  e x  6e x  0
The characteristic equation:  2    6  0
(  2)(  3)  0,
1  2,
2  3
The general solution: y ( x)  c1e 2 x  c2 e3 x
23
Ax

A
ii) A2  4 B  0     , y1 ( x)  e 2
2
By the reduction of order method,
Let y2  u ( x) y1  u ( x)e

Ax
2
Substituting into (2.4)
Ax
Ax


A2  Ax2
ue
 Au e 2  u e 2 
4
Ax
Ax


A  Ax2
A( ue
 u e 2 )  Bue 2  0
2
A2
u  ( B  )u  0  u  0, u ( x)  cx  d
4
24
Choose u ( x)  x  y2  u ( x)e

Ax
2
 xe

Ax
2
y , y : linearly independent
1
2
The general sol.: y ( x)  c1e

Ax
2
 c2 xe

Ax
2
。 Example 2.7: y  6 y  9 y  0
Characteristic eq. :
 2  6  9  (  3) 2  0
The repeated root:
 3
The general solution: y( x)  (c1  c2 x)e3 x
25
iii) A2  4 B  0
 1 
A 
4 B  A2 i
A 
, 2 
2
A
1
, q
Let p 
2
2
4 B  A2 i
2
4 B  A2
 y1 ( x)  e( p  qi ) x , y2 ( x)  e( p qi ) x
W ( x) 
e( p  qi ) x
( p  qi )e
( p  qi ) x
e( p  qi ) x
( p  qi )e
( p  qi ) x
 ( p  qi )e 2 px  ( p  iq )e 2 px  2iqe 2 px  0
The general sol.:
y( x)  c1e( p qi ) x  c2e( p  gi ) x
   (2.5)
26
。 Example 2.8: y  2 y  6 y  0
Characteristic equation:  2  2  6  0
Roots:
1  1  5i, 2  1  5i
The general solution: y ( x )  c1e( 1
5i ) x
 c2e( 1
5i ) x
○ Find the real-valued general solution
。 Euler’s formula:
eix  cos x  i sin x, e  ix  cos x  i sin x
27
Maclaurin expansions:

1 n
1 2
1 3
e 
x  1 x 
x 
x  
2!
3!
n 0 n!
x
( 1) n 2 n
1 2
1 4
1 6
cos x  
x  x
x 
x 
x  
2!
4!
6!
n  0 (2 n )!

( 1) n
1 3
1 5
1 7
sin x  
x 2 n 1  1 
x 
x 
x  
3!
5!
7!
n  0 (2 n  1)!

1
1
1
2
3
e  1  (ix)  (ix)  (ix)  (ix) 4    
2!
3!
4!
1 2 1 3 1 4 1 5
 1  ix 
x  ix 
x  ix    
2!
3!
4!
5!
1 2 1 4
1 3 1 5
 (1 
x 
x  )  i ( x  x  x  )
2!
4!
3!
5!
 cos x  i sin x
ix
28
。 Eq. (2.5),
e
( p  qi ) x
y ( x)  c1e
e e
px qxi
( p  qi ) x
 c2e
( p  gi ) x
 e (cos qx  i sin qx)
px
 e cos qx  ie sin qx
px
px
e( p qi ) x  e px cos qx  ie px sin qx
y ( x)  c1 (e cos qx  ie sin qx) 
px
px
c2 (e cos qx  ie sin qx)
px
px
 (c1  c2 )e px cos qx  (c1  c2 )ie px sin qx
29
Find any two independent solutions
1
 y ( x)  e px cos qx
Take c1  c2 
2
1
1
Take c1 
, c2  
 y ( x)  e px sin qx
2i
2i
W ( x) 
e px cos qx
e px sin qx
pe px cos qx  qe px sin qx
pe px sin qx  qe px cos qx
 e2 px  0
px
y
(
x
)

e
(c1 cos qx  c2 sin qx)
The general sol.:
30
2.5. Euler’s Equation
1
1
y  Ay  2 By  0 , A , B : constants -----(2.7)
x
x
Transform (2.7) to a constant coefficient equation
by letting x  et
dt 1
 y ( x)  y (et )  Y (t ), dx  et dt  xdt ,

dx x
dY dt
1

y ( x) 
 Y (t )
dt dx x
d
d 1
y( x) 
y( x)  ( Y (t ))
dx
dx x
1
1 d
1
1 dY  dt



  2 Y (t ) 
Y (t )   2 Y (t ) 
x
x dx
x
x dt dx
1
1
1 1


  2 Y (t )  Y (t )  2 (Y (t )  Y (t ))
x
x
x x
31
Substituting y, y, y into Eq. (2.7), i.e.,
y 
1
1
Ay  2 By  0
x
x

1
A1
B



(
Y
(
t
)

Y
(
t
))

Y
(
t
)

Y (t )  0
2
2
x
x x
x
Y (t )  Y (t )  AY (t )  BY ( x)  0
Y   ( A  1)Y   BY  0
Steps: (1) Solve
--------(2.8)
Y (t )
(2) Substitute
t  ln x
(3) Obtain y ( x)
32
2
。 Example 2.11: x y  2 xy  6 y  0 ------(A)
y  2
(i) Let x  et
1
1

y  6 2 y  0, x  0 -------(B)
x
x
1
1
 y ( x)  Y (t ), y  Y  , y  2 (Y   Y )
x
x
Substituting y, y, y into (A)
1
1
x 2 (Y   Y )  2 x Y   6Y  0
x
x
Y   Y   2Y   6Y  0 , Y   Y   6Y  0
2
Characteristic equation:  2    6  0
Roots:   3,   2
General solution: Y (t )  c1e 3t  c2 e 2t
33
x  et , t  ln x, y ( x)  c1e 3ln x  c2 e 2ln x  c1 x 3  c2 x 2 ,
x0
○ Solutions of constant coefficient linear equation have the
forms:
e x , xe x , e x sin  x, e x cos  x
Solutions of Euler’s equation have the forms:
x r , x r ln x, x p cos(qln x), x p sin(qln x)
34
2.6. Nonhomogeneous Linear Equation
y  P( x) y  Q( x) y  R ( x) ------(2.9)
The general solution: y  yh  y p
◎ Two methods for finding y p
(1) Variation of parameters
-- Replace c1 , c2 with u ( x), v( x) in the general
homogeneous solution
Let y p  u ( x) y1 ( x)  v( x) y2 ( x )
 y p  uy1  vy2  uy1  vy2
Assume uy1  vy2  0
------(2.10)
 y p  uy1  vy2
Compute yp  uy1  vy2  uy1  vy2
35
Substituting into (2.9), uy1  vy2  uy1  vy2  P(uy1  vy2 )
Q(uy1  vy2 )  R
u[ y1  Py1  Qy1 ]  v[ y2  Py2  Qy2 ]  u1 y1  vy2  R
uy1  vy2  R -----------(2.11)
Solve (2.10) and (2.11) for u, v
(10)  y1  (11)  y1
uy1 y1  vy2 y1  (uy1 y1  vy2 y1 )   Ry1
vy2 y1  vy2 y1  Ry1 , v( y2 y. 1  y2 y1 )  Ry1
v 
Ry1
Ry1 Likewise,
Ry2


u 
y2 y1  y2 y1
W
W
Ry1
Ry2
v
, u  
W
W
36
。 Example 2.15: y  4 y  sec x ------(A)
i) General homogeneous solution yh :
x
Let y  e . Substitute into (A)
The characteristic equation:  2  4  0,   2i
 y1 ( x )  e ( p  qi ) x  e 2ix
Complex solutions: 
( p  qi ) x
2 ix
y
(
x
)

e

e
 2
 y1 ( x)  e px cos qx  cos 2 x
Real solutions: 
px
y
(
x
)

e
sin qx  sin 2 x
 2
W ( x) 
cos 2 x
sin 2 x
2sin 2 x
2cos 2 x
2
 y1 , y2 :independent
37
ii) Nonhomogeneous solution y :
p
Let y p  uy1  vy2
y1 R 1
1
 cos 2 x sec x  2 cos x sin x sec x
W
2
2
 sin x
y2 R 1
1
 sin 2 x sec x  (2 cos 2 x  1)sec x
W
2
2
1
 cos x  sec x
2
Ry2
1
1
u ( x)  
  cos xdx   sec xdx  sin x  ln sec x  tan x
W
2
2
Ry1
v( x)   
   sin xdx  cos x
W
38
1
y p ( x)  uy1  vy2  (sin x  ln sec x  tan x )cos 2 x
2
 cos x sin 2 x
iii) The general solution:
y ( x )  yh ( x )  y p ( x )  c1 cos 2 x  c2 sin 2 x
1
  cos x sin 2 x  (sin x  ln | sec x
2
 tan x |) cos 2 x
39
(2) Undetermined coefficients
Apply to y  Ay  By  R( x)
A, B: constants
Guess the form of y p from that of R
e.g. R( x) : a polynomial
 Try a polynomial for
R( x) : an exponential for
yp
Try an exponential for y p
40
。 Example 2.19: y  5 y  6 y  3sin 2 x ---(A)
R( x)  3sin 2 x
It’s derivatives can be multiples of sin 2x
or cos 2x
Try y p  c cos 2 x  d sin 2 x
Compute y p  2c sin 2 x  2dcos 2 x
y p  4c cos 2 x  4d sin 2 x
Substituting into (A),
4c cos 2 x  4d sin 2 x  5(2c sin 2 x  2d cos 2 x)
6(c cos2 x  d sin 2 x)  3sin 2 x
(2d  10c  3)sin 2 x  (2c  10d )cos2 x
41
sin 2 x, cos 2 x : linearly independent
2d  10c  3  0 and 2c  10d  0
c
15
3
, d 
52
52
3
15
y p   cos 2 x 
sin 2 x
52
52
The homogeneous solutions:
3x
e , e
2x
The general solution:
15
3
y  c1e  c2e  cos 2 x  sin 2 x
52
52
3x
2x
42
。 Example 2.20: y  2 y  3 y  8e x ------(A)
R( x)  8ex , try yp  cex
Substituting into (A),
cex  2cex  3cex  0  8ex
* This is because the guessed
yp  cex
contains a homogeneous solution
Strategy: If a homogeneous solution appears in any term of
y , multiply this term by x. If the modified term
p
still occurs in a homogeneous solution, multiply e x
by x again
43
Try y  cxex
p
y  cex  cxex , y  2cex  cxex
p
p
Substituting into (A),
2cex  cxex  2(cex  cxex )  3cxex
 4cex  8ex
c  2 and y p  2 xex
44
○ Steps of undetermined coefficients:
(1) Find homogeneous solutions
(2) From R(x), guess the form of yp
If a homogeneous solution appears in
any term of yp , multiply this term by
x. If the modified term still occurs in a homogeneous
solution, multiply by x again
(3) Substitute the resultant yp into y  Ay  By  R ( x )
and solve for its coefficients
45
○ Guess y p from R( x)
Let P( x) : a given polynomial
Q( x) , S ( x) : polynomials with unknown coefficients
Guessedy p
R( x)
P( x)
Q( x)
ce ax
de ax
 cos bx or
 sin bx
P ( x)e ax
P( x)cos bx or
P ( x)sin bx
P( x)eax cos bx or P( x)eax sin bx
c cos bx  d sin bx
Q ( x)e ax
Q( x)cos bx  S ( x)sin bx
Q( x)e ax cos bx  S ( x)e ax sin bx
46
2.6.3. Superposition
y  P( x) y  Q( x) y  f1 ( x)  f 2 ( x)    f N ( x)
Let y pj be a solution of y  Py  Qy  f j
 y p1  y p 2  ...  y pN is a solution of (A)
( y p1  y p 2  ...  y pN )  P( x)( y p1  y p 2  ...  y pN )
 Q( x)( y p1  y p 2  ...  y pN )
 ( yp1  Pyp1  Qy p1 )  ...  ( ypN  PypN  Qy pN )
 f1  f 2  ...  f N
47
。 Example 2.25: y  4 y  x  2e 2 x
(1) y  4 y  x,
(2) y  4 y  2e 2 x

y p1

x

4
 y p  y p1  y p 2
y p 2  e 2 x / 4
2 x
x
1
4
  e  ( x  e 2 x )
4
4
The general solution:
1
y ( x)  c1 cos 2 x  c2 sin 2 x  ( x  e 2 x )
4
cos 2 x, sin 2 x : where homogeneous solutions
48