Chapter 1 First-Order ODEs
C C h h a a p p t t e e r r 22 S S e e c c o o n n d d --O O r r d d e e r r L L i i n n e e a a r r O O
D D
E E s s ((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 H o o m m o o g g e e n n e e o o u u s s L L i i n n e e a a r r O O D D E E s s o o f f S S e e c c o o n n d d O O r r d d e e r r ((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 ()0r x =, the equation is called homogeneous (齊性的).
In case ()0r x ≠, 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 1()y x and 2()y x are solutions of the homogeneous linear ODE
()()0y p x y q x y '''++=,
(2)
then a linear combination (線性組合) of 1y and 2y , say 1122()()c y x c y x +, is also a solution of the differential equation. Proof –
Let 1y and 2y be solutions of equation (2). It means that
1
110y py qy '''++= and 2220y py qy '''++=. Then by substituting 1122y c y c y =+ into (2), we get
112211221122()()()y py qy c y c y p c y c y q c y c y ''''''++=+++++
11
2211221122()()c y c y p c y c y q c y c y ''''''=+++++ 11
112222()()c y py qy c y py qy ''''''=+++++ 0=
This shows that 1122c y c y + is a solution of (2).
A Homogeneous Linear ODE (線性齊性微分方程式)
A Nonhomogeneous Linear ODE (線性非齊性微分方程式)
A homogeneous Nonlinear ODE (非線性齊性微分方程式)
? For a second-order homogeneous linear ODE, an initial value problem (初始值問題) consists of the equation
()()0y p x y q x y '''++=
and two initial conditions (兩個初始條件)
00()y x K = and 01()y x K '=.
The two conditions are used to determine the two arbitrary constants 1c and 2c in a general solution
1122y c y c y =+
of the ODE.
Initial Value Problem Solve the initial value problem
0y y ''+= with (0) 3.0y =, (0)0.5y '=-.
? A general solution (通解) of the ODE ()()0y p x y q x y '''++= on an interval I can be expressed as 1122y c y c y =+ where 1y and 2y are solutions of the ODE on I and are linearly independent (線性獨立) or not proportional (非等比例), and 1c and 2c are ar-bitrary constants.
A particular solution (特解) of the ODE is obtained if we assign specific values to 1c and 2c .
These 1y , 2y are called a basis of solutions or a fundamental set of solutions (一組基礎解). A basis of solutions of ()()0y p x y q x y '''++= 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)
A Basis of Solutions, General solution and Particular Solution
A Basis of Solutions, General solution and Particular Solution Verify by substitution that 1x y e = and 2x y e -= are solutions of the ODE
0y y ''-=.
Then solve the initial value problem
0y y ''-= with (0)6y =, (0)2y '=-.
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 1y to be known. To get a second linearly independent solution
1y . We let
)()()(12x y x u x y ?=.
Then compute 112
y u y u y '?+?'=', 11122y u y u y u y ''?+'?'+?''='' and substitute them into 0)()(=+'+''y x q y x p y to give
0)())(()2(11111
1=?+'?+?'+''?+'?'+?''y u x q y u y u x p y u y u y u ? 0)()2(11111
1=+'+''++''+?''qy y p y u py y u y u . ? 11
1(2)0u y u y py ''''?++= ( 1110y py qy '''++=) In case 01≠y , we have
021
11
='+'+
''u y py y u . Let )()(x u x v '= and )
()()()(2)(111
x y x y x p x y x g +'=
, then we have
0)(=?+'v x g v . (a first-order linear ODE)
The general solution of this first-order linear differential equation is
?
?=-dx
x g e C x v )()(. We may take 1=C since we need only one second solution 2y . Thus a second linearly independent solution
=?=?=?)(])([)()()(112x y dx x v x y x u x y )(][1)(x y dx e dx
x g ??
?- is found. Since 1y and 2y form a basis of solutions of the ODE. Thus the general solu-tion is 1122c y c y +.
? We do not recommend memorizing formulas for )(x g , )(x v , )(x u and )(2x y . Instead,
the following procedures are recommended: 1) Given a first solution 1y .
2) Substitute 12y u y ?= into 0)()(=+'+''y x q y x p y .
3) After some cancellations, solve the resulting first-order linear differential equation.
Find a Basis of solutions if a First Solution is Known Find a basis of solutions of the ODE
2()0x x y xy y '''--+=.
? Homework for sec.2.1 ? #1, 3, 5, 7, 9, 11, 15, 16
22..22 H H o o m m o o g g e e n n e e o o u u s s L L i i n n e e a a r r O O D D E E s s w w i i t t h h C C o o n n s s t t a a n n t t C C o o e e f f f f i i c c i i e e n n t t s s ((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,
0y ay by '''++=.
(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
0y ky '+=.
is an exponential function kx y ce -=. This gives us the idea to try
x y e λ=.
as a solution
Substituting x y e λ= and its derivatives x y e λλ'= and 2x y e λλ''= into equation (1), we obtain
2()0x a b e λλλ++=.
? 20a b λλ++=, which is called a characteristic equation (特徵方程式)
? 11,22(a λ=-±
It means that
11x y e λ= and 22x y e λ=
are two solutions of equation (1).
Case I : 240a b -> - Two Distinct Real Roots , 1λ and 2λ ? In this case, a basis of solutions on any interval is
11x y e λ= and 22x y e λ=.
The corresponding general solution is
1212()x x y x c e c e λλ=+.
Find the general solution of 0y y '''-=.
The characteristic equation is 20λλ-= ? (1)0λλ-= ? 0λ= or 1λ=. Thus a basis of solutions is 11y = and 2x y e =. And the general solution is
112212x y c y c y c c e =+=+.
Solve the initial value problem 20y y y '''+-=, (0)4y =, (0)5y '=-.
Case II : 240a b -= - A Real Double Root , 2a λ=- ? If 240a b -=, we get only one root, hence a first solution is
2
1a x y e
-=.
To obtain a second linearly independent solution 2y (to form a basis), we use the method of reduction of order. Set 2
21a x y uy ue
-== and compute the derivatives 2
2a x y u e
-''=
2
12a x aue
--, 2
2
2
2
124a a a x x x y u e
au e
a ue
---'''''=-+. Then substitute them into
0y ay by '''++=
? 22222221142
()0a
a
a
a
a
a
x x x x x x u e au e a ue a u e aue bue ------''''-++-+= ? 22
14[()]0a x e u a b u -''+-+= ? 2
0a x e
u -''= since 2140a b -+=
? 0u ''= since 12
0ax e
-≠
? 34()u x c x c =+, 3c , 4c are constants
For simplicity, we choose x x u =)(, and obtain a second solution 2
21a x y xy xe -==.
Since 1y and 2y are linearly independent, the general solution is
2
2
12()a a x x y x c e
c xe
--=+.
Find the general solution of 690y y y '''++=.
The characteristic equation is 2690λλ++= ? 2(3)0λ+= ? 3λ=- (a double root) Thus a basis of solutions is 31x y e -= and 32x y xe -=. And the general solution is
3312x x y c e c xe --=+.
Solve the initial value problem 0.250y y y '''++=, (0) 3.0y =, (0) 3.5y '=-.
Case III : 240a b -< - Two Complex Roots , 112a i λω=-+ and 1
22a i λω=--
? In this case, we have two complex distinct roots, say
iq p ±=λ with 12p a =- and q =The general solution can be written as
(
)(
)12()p iq x
p iq x
y x c e c e +-=+ (in complex form)
or 34()[cos()sin()]px y x e c qx c qx =+ (in real form) (ˇ)
where 1c , 2c are complex constants while 3c , 4c are real constants.
? Relations between 1c , 2c and 3c , 4c
By Euler ’s formula cos sin ix e x i x +=+, we have
()()12()p iq x p iq x y x c e c e +-=+
12)px iqx px iqx c e e c e e +-=+ 12()px iqx iqx e c e c e +-=+
12[(cos()sin())(cos()sin())]px e c qx i qx c qx i qx =++-+-
))]s i n ()(c o s ())sin()(cos([21qx i qx c qx i qx c e px -++=
)]sin()()cos()[(2121qx c c i qx c c e px -++=
Comparing it with )]sin()cos([43qx c qx c e y px h +=, we will have the relations between
1c , 2c and 3c , 4c as
??
?-=+=)(2
142
13c c i c c c c or ???+=-=)()(43
21
243211ic c c ic c c .
Solve the initial value problem 0.49.040y y y '''++=, (0)0y =, (0)3y '=.
Complex Roots
Summary of Cases I-III
? Homework for sec.2.2 ? #1, 5, 9, 21, 25, 29, 33
22..44 M M o o d d e e l l i i n n g g :: F F r r e e e e O O s s c c i i l l l l a a t t i i o o n n s s ((自自由由震震盪盪)) //
M M a a s s s s --S S p p r r i i n n g g S S y y s s t t e e m 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 0s shown in the figure. The extension 0s is such that au upward
00F ks =- (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
0k
y y m
''+
=. (0m >, 0k >) (1)
The corresponding characteristic equation is 20k m
λ+
=. It gives two complex roots
λ=and we obtain as a general solution.
00()cos()sin()y t A t B t ωω=+, 0ω=
(2)
The corresponding motion is called a harmonic oscillation with a natural frequency (自然頻率) 0ωπ.
The sum in equation (2) can be combined into a phase-shifted cosine with amplitude (振
幅) C =
and phase angle (相位角) 1tan ()B A δ-=,
0())y t t ωδ=-.
(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.
Complex Roots
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
0c k
y y y m m
'''+
+=. (0m >, 0k >, 0c >) (4)
c is calle
d th
e 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
20c k
m m
λλ+
+=. By the usual formula for the roots of a quadratic equation we obtain
2c m λ=-
Thus depending on the amount of damping, there will be three types of motion corre-sponding 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
1122((12()m
m
c t c t y t c e
c e
--=+.
We see that damping takes out energy so quickly that the body does not oscillate. For 0t >
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 oscilla-tions (Case III). It occurs if 24c mk =. In this case the corresponding general solution is
2212()c c m
m
t t y t c e
c te
--=+ or 212()c m
t c c t e
-+.
The solution can pass through the equilibrium position 0y = at most once because 2c m
e -
is never zero and 12c c t + can have at most one positive zero (say 1
2
c c t =-). If both 1c an
d 2c ar
e positive (or both negative), it has no positive zero, so that y does not pass through 0 at all.
Fig. 37. Critical damping
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 24c mk <. Then the corresponding general solution is
21020()(cos()sin())c m
t y t e
c t c t ωω-=+, 0ω=This presents dampe
d oscillations (阻滯性振盪). Their curv
e lies between the dashed
curves 2c m
t y Ce -=±, C =in Fig. 38.
If 0c →, then 0ωfrequency 002f ωπ=.
Fig. 38 Damped oscillation in Case III
The Three Cases of Damped Motion
?Homework for sec.2.4 #8(a), 8(b), 8(c)三題任選一題
22..55 E E u u l l e e r r --C C a a u u c c h h y y E E q q u u a a t t i i o o n n s s
? Euler-Cauchy equations are ODEs of the form
20x y axy by '''++=
(1)
with given constant a and b .
? Euler-Cauchy equation can be transformed to a constant-coefficient one through a change of variables: Let t e x = and )()()(t Y e y x y t ==.
Compute =?=
=
'dx dt dt t dY dx d t Y x y )()()(x t Y 1)(?',
x dx d x x dx
d t Y t Y t Y x y 111)]([)()(])([)(2?'+-?'=?'=
''
=
??''+-?'=x x x t Y t Y 1
11])([)()(2)]()([2
1
t Y t Y x '-''
Substitute into equation (1) to get
[()()]()()
Y t Y t a Y t b Y t ''''-++= ? ()(1)()()0Y t a Y t bY t '''+-+= (2)
This is a constant-coefficient homogeneous linear second-order equation for
)(t Y . Solve this equation for )(t Y , then let x t ln = in the solution )(t Y
to obtain )(x y .
Two Real Roots The Euler-Cauchy equation
2 1.50.50x y xy y '''+-=
has the auxiliary equation (1.51)0.50Y Y Y '''+--=, which has the characteristic equation
20.50.50λλ+-= ? (0.5)(1)0λλ-+= ? 0.5λ= or 1λ=-
? 0.50.51()()t t Y t e e ==, 112()()t t Y t e e --==
t e x = ? 0.51()y x x =, 12()y x x -=.
Thus the corresponding general solution is
2
1
()y x c c x
=.
A Double Root The Euler-Cauchy equation
2590x y xy y '''-+=
has the auxiliary equation (51)90Y Y Y '''+--+=, which has the characteristic equation
2690λλ-+= ? 2(3)0λ-= ? 3λ= (a double root)
? 331()()t t Y t e e ==, 332()()t t Y t te t e ==
t e x = ? 31()y x x =, 32()(ln )y x x x =.
Thus the corresponding general solution is
3312()(ln )y x c x c x x =+.
Two Roots The Euler-Cauchy equation
20.616.040x y xy y '''++=
has the auxiliary equation (0.61)16.040Y Y Y '''+-+=, which has the characteristic equation
20.416.040λλ-+=
? 0.24i λ=± (complex roots), i =? 0.212()(cos(4)sin(4))t Y t e c t c t =+
t e x = and ln t x = ? 0.212()(cos(4ln )sin(4ln ))y x x c x c x =+.
is the corresponding general solution.
? Homework for sec.2.5 ? #1, 5, 9, 11, 15
22..77 N N o o n n h h o o m m o o g g e e n n e e o o u u s s O O D D E E s s ((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 ()0r x ≠.
Homogeneous Solution (齊性解h y ) and Particular Solution (特解p y )
? A general solution of the nonhomogeneous ODE (1) on an open interval I is a solution of the form
()()()h p y x y x y x =+
(2)
()general solution =()homogeneous solution +()particular solution
where 1122h y c y c y =+ is a general solution of the homogeneous ODE ()y p x y '''+
()0q x y +=, and p y is any particular solution of (1).
CHECK: Substituting ()()()h p y x y x y x =+ into equation (1), we have
()()()()()()h p h p h p y y p x y y q x y y r x '''+++++=
? (()())(()())()h h h p p p y p x y q x y y p x y q x y r x ''''''+++++= ? 0()()r x r x +=
Hence ()()h p y x y 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 1c and 2c in h y .
Steps to Find a General Solution
? Find h y : it is to solve ()()'''0y p x y q x y ++=.