Chapter 1 First-Order ODEs
C
Chhaapptteerr 22 SSeeccoonndd--O
Orrddeerr L
Liinneeaarr O
OD
DE
Ess ((二
二階
階線
線性
性常
常微
微分
分方
方程
程式
式))
Chapter 3 Higher-Order Linear ODEs
Chapter 5 Series Solutions of ODEs
Chapter 6 Laplace Transforms
‧ Ordinary differential equations may be divided into two large classes, linear (線性) and
nonlinear (非線性) ODEs. Where nonlinear ODEs are difficult to solve, linear ODEs are
much simpler because there are standard methods for solving many of these equations.
22..11 H
Hoom
mooggeenneeoouuss L
Liinneeaarr O
OD
DE
Ess ooff SSeeccoonndd O
Orrddeerr ((二
二階
階線
線性
性齊
齊性
性微
微分
分方
方程
程式
式))
‧ A second-order ODE is called linear (線性的) if it can be written as
y   p( x ) y   q( x ) y  r ( x ) .
(1)
 線性：方程式的每一項都不得出現 y ( x ) 和其導數( y  , y  ,…)的乘積或自乘
In case r ( x )  0 , the equation is called homogeneous (齊性的).
In case r ( x )  0 , the equation is called nonhomogeneous (非齊性的).
The functions p ( x ) and q( x ) are called the coefficients of the ODEs.
‧ Theorem 1
Superposition Principle for the Homogeneous Linear ODE (適用於線性
齊性常微分方程式的疊加原理)
If both y1 ( x ) and y2 ( x ) are solutions of the homogeneous linear ODE
y   p( x ) y   q( x ) y  0 ,
(2)
then a linear combination (線性組合) of y1 and y 2 , say c1 y1 ( x )  c2 y2 ( x ) , is also a
solution of the differential equation.
Proof –
- 19 -
Let y1 and y 2 be solutions of equation (2). It means that
y1  py1  qy1  0 and y2  py2  qy2  0 .
Then by substituting y  c1 y1  c2 y2 into (2), we get
y  py  qy  (c1 y1  c2 y2 )  p(c1 y1  c2 y2 )  q(c1 y1  c2 y2 )
 c1 y1  c2 y2  p(c1 y1  c2 y2 )  q(c1 y1  c2 y2 )
 c1 ( y1  py1  qy1 )  c2 ( y2  py2  qy2 )
0
This shows that c1 y1  c2 y2 is a solution of (2).
Example 1
A Homogeneous Linear ODE (線性齊性微分方程式)
Example 2
A Nonhomogeneous Linear ODE (線性非齊性微分方程式)
Example 3
A homogeneous Nonlinear ODE (非線性齊性微分方程式)
- 20 -
‧
For a second-order homogeneous linear ODE, an initial value problem (初始值問題)
consists of the equation
y   p( x ) y   q( x ) y  0
and two initial conditions (兩個初始條件)
y ( x0 )  K0 and y( x0 )  K1 .
The two conditions are used to determine the two arbitrary constants c1 and c2 in a
general solution
y  c1 y1  c2 y2
of the ODE.
Example 4
Initial Value Problem
Solve the initial value problem
y   y  0 with y (0)  3.0 , y (0)  0.5 .
- 21 -
‧ A general solution (通解) of the ODE y   p( x ) y   q( x ) y  0 on an interval I can be
expressed as y  c1 y1  c2 y2 where y1 and y 2 are solutions of the ODE on I and are
linearly independent (線性獨立) or not proportional (非等比例), and c1 and c2 are arbitrary constants.
A particular solution (特解) of the ODE is obtained if we assign specific values to c1
and c2 .
These y1 , y 2 are called a basis of solutions or a fundamental set of solutions (一組基
礎解). A basis of solutions of y   p( x ) y   q( x ) y  0 is a pair of linearly independent
solutions of the ODE. (一組基礎解就是一對彼此線性獨立的解)
Example 5
A Basis of Solutions, General solution and Particular Solution
Example 6
A Basis of Solutions, General solution and Particular Solution
Verify by substitution that y1  e x and y2  e  x are solutions of the ODE
y   y  0 .
Then solve the initial value problem
y   y  0 with y (0)  6 , y (0)  2 .
- 22 -
Find a Basis of Solutions if One Solution is Known. (已知一解時，可使用降階法求得另
一線性獨立的解)
‧ It happens quite often that one solution can be found by inspection or in some other way.
Then a second linearly independent solution can be obtained by solving a first-order ODE.
This is called the method of reduction of order (降階法).
Method of Reduction of Order (降階法)
‧ Assume a first solution y1 to be known. To get a second linearly independent solution
y1 . We let
y2 ( x)  u( x)  y1 ( x) .
Then compute y 2  u   y1  u  y1 , y 2  u   y1  2u   y1  u  y1 and substitute them
into y   p( x ) y   q( x ) y  0 to give
(u   y1  2u   y1  u  y1)  p( x)(u   y1  u  y1 )  q( x)u  y1  0
 u   y1  u (2 y1  py1 )  u( y1  py1  qy1 )  0 .
 u  y1  u(2 y1  py1 )  0
(
y1  py1  qy1  0 )
In case y1  0 , we have
u  
Let v ( x )  u ( x ) and g ( x ) 
2 y1  py1
u  0 .
y1
2 y1 ( x )  p( x ) y1 ( x )
, then we have
y1 ( x )
v   g ( x )  v  0 . (a first-order linear ODE)
The general solution of this first-order linear differential equation is
 g ( x ) dx
.
v( x )  C  e 
We may take C  1 since we need only one second solution y 2 . Thus a second linearly
independent solution
 g ( x ) dx
dx ]  y1 ( x )
y 2 ( x )  u( x )  y1 ( x )  [  v( x )dx ]  y1 ( x )  [  e 
is found. Since y1 and y 2 form a basis of solutions of the ODE. Thus the general solution is c1 y1  c2 y2 .
‧ We do not recommend memorizing formulas for g ( x ) , v (x ) , u(x ) and y 2 ( x ) . Instead,
- 23 -
the following procedures are recommended:
1) Given a first solution y1 .
2) Substitute y 2  u  y1 into y   p( x ) y   q( x ) y  0 .
3) After some cancellations, solve the resulting first-order linear differential equation.
Example 7
Find a Basis of solutions if a First Solution is Known
Find a basis of solutions of the ODE
( x 2  x ) y  xy  y  0 .
‧ Homework for sec.2.1  #1, 3, 5, 7, 9, 11, 15, 16
- 24 -
22..22 H
Hoom
mooggeenneeoouuss L
Liinneeaarr O
OD
DE
Ess w
wiitthh C
Coonnssttaanntt C
Cooeeffffiicciieennttss ((常
常係
係數
數線
線性
性齊
齊
性
性常
常微
微分
分方
方程
程式
式))
‧ Consider second-order homogeneous linear ODEs whose coefficients a and b are constant,
y   ay   by  0 .
(1)
These equations have important applications, especially in connection with mechanical
and electrical vibrations.
將常係數微分方程式轉換成特徵方程式(Characteristic Equations)
‧ How to solve equation (1)? We remember from Sec. 1.5 that the solution of the first-order
linear ODE with a constant coefficient k
y   ky  0 .
is an exponential function y  ce kx . This gives us the idea to try
y  e x .
as a solution
Substituting y  e x and its derivatives y   e x and y   2e x into equation (1),
we obtain
( 2  a  b)e x  0 .
  2  a  b  0 , which is called a characteristic equation (特徵方程式)
 1,2  12 (a  a 2  4b )
It means that
y1  e1x and y2  e 2 x
are two solutions of equation (1).
- 25 -
Case I: a 2  4b  0 - Two Distinct Real Roots, 1 and 2
‧ In this case, a basis of solutions on any interval is
y1  e1x and y2  e 2 x .
The corresponding general solution is
y ( x )  c1e1x  c2e 2 x .
Example 1
Find the general solution of y   y   0 .
The characteristic equation is  2    0   (  1)  0    0 or   1.
Thus a basis of solutions is y1  1 and y2  e x . And the general solution is
y  c1 y1  c2 y2  c1  c2e x .
Example 2
Solve the initial value problem y   y   2 y  0 , y (0)  4 , y (0)  5 .
- 26 -
Case II: a 2  4b  0 - A Real Double Root,    a2
‧ If a 2  4b  0 , we get only one root, hence a first solution is
y1  e
 a2 x
.
To obtain a second linearly independent solution y 2 (to form a basis), we use the method
of reduction of order. Set y2  uy1  ue
 12 aue
 a2 x
, y2  ue
 a2 x
 aue
 a2 x
 14 a 2ue
 a2 x
 a2 x
and compute the derivatives y2  ue
 a2 x
. Then substitute them into
y   ay   by  0
 ue
 a2 x
 aue
 a2 x
 14 a 2ue
 a2 x
 a (ue
 e
 a2 x
[u  (  14 a 2  b)u]  0
 e
 a2 x
u  0 since  14 a 2  b  0
 u  0 since e
 u( x )  c3 x  c4 ,
 12 ax
 a2 x
 12 aue
 a2 x
)  bue
 a2 x
0
0
c3 , c4 are constants
For simplicity, we choose u ( x )  x , and obtain a second solution y2  xy1  xe
 a2 x
.
Since y1 and y 2 are linearly independent, the general solution is
y ( x )  c1e
Example 3
 a2 x
 c2 xe
 a2 x
.
Find the general solution of y   6 y   9 y  0 .
The characteristic equation is  2  6  9  0  (  3)2  0    3 (a double root)
Thus a basis of solutions is y1  e 3 x and y2  xe 3 x . And the general solution is
y  c1e 3 x  c2 xe 3 x .
Example 4
Solve the initial value problem y   y   0.25 y  0 , y (0)  3.0 , y (0)  3.5 .
- 27 -
Case III: a 2  4b  0 - Two Complex Roots, 1   12 a  i and 2   12 a  i
‧ In this case, we have two complex distinct roots, say
  p  iq with p   12 a and q 
1
2
4b  a 2 .
The general solution can be written as
y( x)  c1e piq x  c2e piq x (in complex form)
or y ( x )  e px [c3 cos( qx )  c4 sin( qx )] (in real form) (ˇ)
where c1 , c2 are complex constants while c3 , c4 are real constants.
‧ Relations between c1 , c2 and c3 , c4
By Euler’s formula e  ix  cos x  i sin x , we have
y ( x )  c1e( p iq ) x  c2e ( p iq ) x
 c1e px e  iqx  c2e px e  iqx )
 e px (c1e  iqx  c2e  iqx )
 e px [c1 (cos( qx )  i sin( qx ))  c2 (cos(  qx )  i sin(  qx ))]
 e px [c1 (cos( qx)  i sin( qx))  c2 (cos( qx)  i sin( qx ))]
 e px [( c1  c2 ) cos( qx )  i(c1  c2 ) sin( qx )]
Comparing it with y h  e px [c3 cos( qx )  c4 sin( qx )] , we will have the relations between
c1 , c2 and c3 , c4 as
 c  1 (c  ic4 )
 c3  c1  c2
or  1 21 3
.

c4  i(c1  c2 )
c2  2 (c3  ic4 )
- 28 -
Example 5
Solve the initial value problem y   0.4 y   9.04 y  0 , y (0)  0 , y (0)  3 .
Example 6
Complex Roots
Summary of Cases I-III
‧ Homework for sec.2.2  #1, 5, 9, 21, 25, 29, 33
- 29 -
22..44 M
Mooddeelliinngg:: FFrreeee O
Osscciillllaattiioonnss ((自
自由
由震
震盪
盪)) //
M
Maassss--SSpprriinngg SSyysstteem
m ((重
重物
物--彈
彈簧
簧系
系統
統))
Undamped Motion (無阻尼運動) or Harmonic Motion (簡諧運動)
‧ Consider Fig. 32. The spring is first unstretched. We now attach the body. This stretches the
spring by an amount s0 shown in the figure. The extension s0 is such that au upward
F0  ks0 (by Hooke’s law/虎克定律) in the spring balances the weight W  mg of the
body. k is the called the spring constant (彈簧常數) or spring modulus (彈簧模數) and
m is the mass of body.
By Newton’s second law (牛頓第二運動定律)
Mass  Acceleration = Force ,
it gives the model my    ky or
y  
k
y 0.
m
( m  0, k  0 )
(1)
The corresponding characteristic equation is  2  mk  0 . It gives two complex roots

k
m
and we obtain as a general solution.
y(t )  A cos(0t )  B sin(0t ) , 0 
k
m
.
(2)
The corresponding motion is called a harmonic oscillation with a natural frequency (自
然頻率) 0 2 .
- 30 -
The sum in equation (2) can be combined into a phase-shifted cosine with amplitude (振
幅) C 
A2  B 2 and phase angle (相位角)   tan 1 ( BA ) ,
y (t ) 
A2  B 2 cos(0t   ) .
(3)
Equation (2) is simpler in connection with initial value problems, whereas equation (3) is
physically more informative because it exhibits the amplitude and phase of the oscillation.
Example 1
Complex Roots
- 31 -
Damped Motion (有阻尼運動)
‧ We now add a damping force (阻滯力) cy  to our model, so that we have my  
 ky  cy  or
y  
c
k
y  y  0 .
m
m
( m  0, k  0 , c  0 )
(4)
c is called the damping constant (阻滯常數) and is always positive.
Equation (4) is homogeneous linear and has constant coefficients. Hence we can solve it
by the method in Sec. 2.2. The characteristic equation is
2 
c
k
   0.
m
m
By the usual formula for the roots of a quadratic equation we obtain

c
1

c 2  4mk .
2m 2m
Thus depending on the amount of damping, there will be three types of motion corresponding to the three cases I-III in Sec. 2.2:
Case I: Overdamping (過阻尼)
In this case the corresponding general solution of (4) is
y(t )  c1e
 21m ( c  c2 4 mk ) t
 c2e
 21m ( c  c2 4 mk ) t
.
We see that damping takes out energy so quickly that the body does not oscillate. For t  0
- 32 -
both exponents in the solution are negative. Hence both terms in the solution approach zero
as t   .
Figure 36 shows solutions for some typical initial conditions.
Case II: Critical Damping (臨界阻尼)
Critical damping is the border case between nonoscillatory motions (Case I) and oscillations (Case III). It occurs if c 2  4mk . In this case the corresponding general solution is
y (t )  c1e
 2cm t
 c2te
 2cm t
or (c1  c2t )e
 2cm t
.
The solution can pass through the equilibrium position y  0 at most once because e
 2cm
is never zero and c1  c2t can have at most one positive zero (say t   cc12 ). If both c1
and c2 are positive (or both negative), it has no positive zero, so that y does not pass
through 0 at all.
Fig. 37. Critical damping
- 33 -
Figure 37 shows solutions for some typical initial conditions. They look almost like those
in the figure 36.
Case III: Underdamping (阻尼不足)
It occurs if the damping constant c is so small that c 2  4mk . Then the corresponding
general solution is
y (t )  e
 2cm t
(c1 cos(0t )  c2 sin(0t )) , 0 
1
2m
4mk  c 2 .
This presents damped oscillations (阻滯性振盪). Their curve lies between the dashed
curves y  Ce
 2cm t
, C  c12  c2 2 in Fig. 38.
If c  0 , then 0 
k
m
. It gives the harmonic oscilltion of equation (2) with a natural
frequency f 0  0 2 .
Fig. 38
Example 2
The Three Cases of Damped Motion
- 34 -
Damped oscillation in Case III
‧ Homework for sec.2.4  #8(a), 8(b), 8(c)三題任選一題
- 35 -
22..55 E
Euulleerr--C
Caauucchhyy E
Eqquuaattiioonnss
‧ Euler-Cauchy equations are ODEs of the form
x 2 y  axy  by  0
(1)
with given constant a and b.
‧ Euler-Cauchy equation can be transformed to a constant-coefficient one through a change
of variables:
Let x  e t and y( x )  y(et )  Y (t ) .
Compute y ( x ) 
y ( x ) 
d
dx
d
dx
Y (t ) 
dY ( t )
dt
dt
 dx
 Y (t )  1x ,
[Y (t )  1x ]  Y (t )  (  x12 )  [ dxd Y (t )]  1x
 Y (t )  (  x12 )  [Y (t )  1x ]  1x 
1
x2
[Y (t )  Y (t )]
Substitute into equation (1) to get
[Y  (t ) Y  (t )] a Y (t ) b Y (t )
0
 Y (t )  (a  1)Y (t )  bY (t )  0
(2)
This is a constant-coefficient homogeneous linear second-order equation for
Y (t ) . Solve this equation for Y (t ) , then let t  ln x in the solution Y (t )
to obtain y ( x ) .
Example 1
Two Real Roots
The Euler-Cauchy equation
x 2 y  1.5xy  0.5 y  0
has the auxiliary equation Y   (1.5  1)Y   0.5Y  0 , which has the characteristic equation
 2  0.5  0.5  0
 (  0.5)(  1)  0
   0.5 or   1
 Y1 (t )  e0.5t  (et )0.5 , Y2 (t )  e 1t  ( et ) 1
et  x  y1 ( x )  x 0.5 , y2 ( x )  x 1 .
Thus the corresponding general solution is
y ( x )  c1 x  c2
1
.
x
- 36 -
Example 2
A Double Root
The Euler-Cauchy equation
x 2 y  5xy  9 y  0
has the auxiliary equation Y   ( 5  1)Y   9Y  0 , which has the characteristic equation
 2  6  9  0
 (  3)2  0
   3 (a double root)
 Y1 (t )  e3t  ( et )3 , Y2 (t )  te3t  t ( et )3
et  x  y1 ( x )  x 3 , y2 ( x )  (ln x ) x 3 .
Thus the corresponding general solution is
y ( x )  c1 x 3  c2 (ln x ) x 3 .
Example 3
Two Roots
The Euler-Cauchy equation
x 2 y  0.6 xy  16.04 y  0
has the auxiliary equation Y   (0.6  1)Y   16.04Y  0 , which has the characteristic equation
 2  0.4  16.04  0
   0.2  4i (complex roots), i  1
 Y (t )  e0.2 t (c1 cos(4t )  c2 sin(4t ))
et  x and t  ln x
 y ( x )  x 0.2 (c1 cos(4 ln x )  c2 sin(4 ln x)) .
is the corresponding general solution.
‧ Homework for sec.2.5  #1, 5, 9, 11, 15
- 37 -
22..77 N
Noonnhhoom
mooggeenneeoouuss O
OD
DE
Ess ((非
非齊
齊性
性常
常微
微分
分方
方程
程式
式))
‧ The linear ODE
y   p( x ) y   q( x ) y  r ( x )
(1)
is called nonhomogeneous (非齊性) if r ( x )  0 .
Homogeneous Solution (齊性解 y h ) and Particular Solution (特解 y p )
‧ A general solution of the nonhomogeneous ODE (1) on an open interval I is a solution of
the form
y ( x )  yh ( x )  y p ( x )
 general
(2)
solution  =  homogeneous solution  +  particular solution 
where yh  c1 y1  c2 y2 is a general solution of the homogeneous ODE y   p ( x ) y 
 q( x ) y  0 , and y p is any particular solution of (1).
CHECK: Substituting y ( x )  yh ( x )  y p ( x ) into equation (1), we have
( yh  y p )  p( x )( yh  y p )  q( x )( yh  y p )  r ( x )
 ( yh  p( x) yh  q( x) yh )  ( y p  p( x) y p  q( x) y p )  r( x)
 0  r( x)  r( x)
Hence yh ( x )  y p ( x ) is a general solution of (1).
Theorem 2: A General Solution of a Nonhomogeneous ODE includes All Solutions
‧ If the coefficients p ( x ) , q( x ) and r ( x ) in (1) are continuous on some open interval I,
then every solution of (1) on I can be obtained by assigning suitable values to the arbitrary
constants c1 and c2 in y h .
Steps to Find a General Solution
‧ Step 1：Find yh :
it is to solve y '' p  x  y ' q  x  y  0 .
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Step 2：Find y p :
two methods will be introduced
Sec. 2.7
Method of Undetermined Coefficients (未定係數法)
Sec. 2.10
Method of Variation of Parameters (參數變換法)
Method of Undetermined Coefficients (未定係數法)
‧ This method is much simpler than another. It is frequently used in engineering.
The method of undetermined coefficients is suitable for linear ODEs with constant coefficients a and b. (未定係數法僅適用於常係數、線性、非齊性 ODE)
y   ay   by  r ( x )
(3)
To find y p , we choose a form for y p similar to r ( x ) , but with unknown coefficients to
be determined. Table 2.1 shows the choice of y p for practically important forms of
r( x) .
Table 2.1 Method of Undetermined Coefficients
Choice Rules for the method of Undetermined Coefficients
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Example 1
Applications of the Basic Rule (a)
Solve the initial value problem
y  y  0.001x 2 , y (0)  0 , y (0)  1.5 .
Example 2
Applications of the Modification Rule (b)
Solve the initial value problem
y  3 y  2.25 y  10e1.5 x , y (0)  1 , y (0)  0 .
- 40 -
Example 3
Applications of the Sum Rule (c)
Solve the initial value problem
y  2 y  5 y  e0.5 x  40cos(10 x )  190sin(10 x) , y (0)  0.16 , y (0)  40.08 .
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Stability(穩定性)
‧ If all the roots of the characteristic equation of the constant-coefficient homogeneous ODE
y   ay   by  0 are negative, or have a negative real part, then the corresponding ho-
mogeneous yh  0 as x   . Thus y h is a transient solution (暫態解), whereas y P
is a steady-state solution (穩態解). In this case the physical or other system modeled by
the ODE is called stable (穩定); otherwise it is called unstable (非穩定).
‧ Homework for sec.2.7  #1, 5, 7, 13, 15, 19
- 42 -
22..88 FFoorrcceedd O
Osscciillllaattiioonnss aanndd R
Reessoonnaannccee((共
共振
振))
mass/spring system
y  
c
k
f (t )
y  y 
m
m
m
RCL circuit
is analog to
q(t ) 
R
1
1
q(t ) 
q( t )  E ( t )
L
LC
L
.
displacement y
velocity y 
driving force f (t )
to
to
to
charge q
current i
electromotive force E (t )
mass m
damping constant c
spring modulus k
to
inductance L
resistance R
reciprocal of capacitance 1 C
Step 1: Solve y  
to
to
c
k
y   y  0 for y h .
m
m
The corresponding characteristic equation is  2 
It is obtained   
c
k
   0.
m
m
c
1

c 2  4km .
2m 2m
Case 1: c 2  4km  0 (overdamping)
 We have two negative real roots 1 and 2 .
 yh  c1e1t  c2e2t , (see Fig.2.5)
which is a transient solution because lim yh  0 .
t 
Case 2: c 2  4km  0 (critical damping)
 We have a negative double root 1 .
 yh  c1e1t  c2te1t , (see Fig.2.6)
- 43 -
which is a transient solution because lim yh  0 .
t 
Case 3: c 2  4km  0 (underdamping)
 We have two complex roots, say 1,2  p  iq where p  
c
4km  c 2
, q
.
2m
2m
 yh  e pt [c1 cos( qt )  c2 sin( qt )] , (see Fig.2.7)
which is a transient solution too because lim yh  0 .
t 
Step 2: Solve y  
c
k
A
y   y  cos(t ) for y p .
m
m
m
Note that the driving force f (t )  A cos(t ) is assigned for convenience, and  is
called the input frequency(輸入頻率).
To find y p , we can use the method of undetermined coefficient, mentioned in Sec.2.6.2,
and let y p  a cos(t )  b sin(t ) . a and b can be solved to get
yp 
A(k  m 2 )
Ac
cos(t ) 
sin(t )
2 2
2 2
(k  m )   c
(k  m 2 )2   2c 2
k
, which is called the nature frequency(自然頻率).
m
mA( 2   2 )
Ac
 2 2 02 2
cos(t )  2 2
sin(t ) ,
2 2
m (0   )   c
m (0   2 )2   2c 2
let 0 
 A steady-state solution is obtained for y p .
‧ Resonance (共振)
In the absence of damping ( c  0 ), an interesting phenomenon called resonance can occur.
A
cos(t )
c  0  yp 
2
m(0   2 )
The closer are the natural and input frequencies, the larger is the amplitude of the
cos(t ) term in the solution, that is, lim y p   (see Figure 2.11).
0
- 44 -
‧ Homework for sec.2.8  無
- 45 -
22..1100 M
Meetthhoodd ooff V
Vaarriiaattiioonn ooff PPaarraam
meetteerrss((參
參數
數變
變換
換法
法))
‧ Suppose we can find a fundamental set of solutions y1 and y 2 for the homogeneous
equation.
Try y p  uy1  vy2（試誤法(trial-and-error method)：找到無限多個特解中的任何一個）
 y p  uy1 + uy1 + v y2 + vy2
let uy1 + v y2 =0 …… (1)
 y p  uy1 + vy2
 y p "  u ' y1 ' uy1 " v ' y2 ' vy2 "
Substitute into the non-homogeneous equation
 u ' y1 ' uy1 " v ' y2 ' vy2 " p uy1 ' vy2 '  q uy1  vy2   f  x 
 u  y1 " py1 ' qy1   v  y2 " py2 ' qy2   u ' y1 ' v ' y2 '  f  x 
 u ' y1 ' v ' y2 '  f  x  ……（2）
1 y2 '  2 y2  u '  y1 y2 ' y2 y1 '   y2 f  x 
 u  
y2 f ( x)
 u   u dx
y1 y 2  y1 y 2
1 y1 '  2 y1  v '  y2 y1 ' y1 y2 '   y1 f  x 
 v 
(消去 v  )
(消去 u )
y1 f ( x )
 v   v dx
y1 y 2  y1 y 2
For any non-vanishing pair of u and v, we can use them to find y p .
‧ 定義 ─ The Wronskian of the two solutions y1 and y 2 is defined as
W ( x) 
y1 ( x )
y1 ( x )
y2 ( x)
 y1 ( x )  y 2 ( x )  y1 ( x )  y2 ( x )
y2 ( x )
Wronskian Test ─
- 46 -
The two solutions y1 and y 2 are linearly dependent on I
if and only if
W ( x )  0 for all x in I.
Example 2.15
- 47 -
Example 2.16
‧ Homework for sec.2.10  無
- 48 -