Chapter 2: Second-Order Differential Equations 2.1. Preliminary Concepts ○ Second-order differential equation e.g., Solution: A function satisfies , (I : an interval) 1 (,,, ) 0 Fxyy y 3 10 7 4 0 y y y x 12 0 y x (, (), (), ( )) 0 Fx x x x x I () x
Chapter 2: Second-Order Differential Equations. 2.1. Preliminary Concepts ○ Second-order differential equation e.g., Solution: A function satisfies , ( I : an interval). ○ Linear second-order differential equation - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Chapter 2: Second-Order Differential
Equations2.1. Preliminary Concepts
○ Second-order differential equation
e.g.,
Solution: A function satisfies
,
(I : an interval)1
( , , , ) 0F x y y y
3 10 7 4 0y y y x
12 0y x
( , ( ), ( ), ( )) 0F x x x x x I
( )x
○ Linear second-order differential equation
Nonlinear: e.g.,
2.2. Theory of Solution
○ Consider
y contains two parameters c and d
2
( ) ( ) ( )y P x y Q x y R x
( , ) ( , ) ( , )y P x y y Q x y R x y
12 0 12y x y x 2( ) 12 6y y x dx xdx x c
2 3( ) (6 ) 2y y x dx x c dx x cx d
The graph of
Given the initial condition
The graph of
3
32y x cx d
(0) 3y 3 (0) 2 0 0 3y c d
32 3 , 3y x cx d
Given another initial condition
The graph of
◎ The initial value problem:
○ Theorem 2.1: : continuous on I,
has a unique solution
4
(0) 1y 2 (0) 6 0 1, 1y c c
32 3y x x
0 0( ) , ( )( ) ( ) ( ) ; y x a y x by P x y Q x y R x
, , P Q R
0 0 0( ) , ( ) , ,y x a y x b x I
( ) ( ) ( );y P x y Q x y R x
x I
2.2.1.Homogeous Equation
○ Theorem 2.2: : solutions of Eq. (2.2)
solution of Eq. (2.2)
: real numbers
Proof:
5
( ) ( ) 0 (2.2)y P x y Q x y
1 2, y y
1 1 2 2( ) ( ) :c y x c y x 1 2,c c
( ) ( )y P x y Q x y
1 1 2 2 1 1 2 2 1 1 2 2( ) ( )( ) ( )( )c y c y P x c y c y Q x c y c y
1 1 2 2 1 1 2 2( ) ( )c y c y c P x y c P x y
1 1 2 2( ) ( )c Q x y c Q x y
1 1 1 1 2 2 2 2[ ( ) ( ) ] [ ( ) ( ) ]c y P x y Q x y c y P x y Q x y
0 0 0
6
Two solutions are linearly independent.
Their linear combination provides an
infinity of new solutions
○ Definition 2.1: f , g : linearly dependent
If s.t. or ;
otherwise f , g : linearly independent
In other words, f and g are linearly
dependent only if for
,c ( ) ( )f x cg x ( ) ( )g x cf x
※
1 2 0c f c g 1 2 0c c
○ Wronskian test -- Test whether two solutions of a
homogeneous differential equation are linearly
independent
Define: Wronskian of solutions to be the 2
by 2 determinant
7
1 2, y y
1 2 1 2( ) ( ) ( ) ( ) ( )W x y x y x y x y x
1 2
1 2
( ) ( )
( ) ( )
y x y x
y x y x
○ Let
If : linear dep., then or
Assume
8
1 1 2 2 0 ( )c y c y A
1 1 2 2 0 ( )c y c y B
1 1 2 1 2 1 2(A) ( ) ( ) 0y B y c y y y y
2 2 1 1 2 1 2(A) ( ) ( ) 0,y B y c y y y y
1 2, y y1 0c 2 0c
1 1 2 1 20 0c y y y y
○ Theorem 2.3:
1) Either or
2) : linearly independent iff
Proof (2):
(i) (if : linear indep. (P), then
(Q) if ( Q) ,
then : linear dep. ( P) )
: linear dep. 9
( ) 0, W x x I ( ) 0, W x x I
1 2, y y ( ) 0W x
1 2, y y
( ) 0W x ( ) 0W x
1 2, y y
1 2 1 2 1 21 2
1 1 0 ,W y y y y y y
y y
1 2 1 2 21 2
1 1, ln ln ln lny y y y c cy
y y
1 2 1 2, , y cy y y
(ii) (if (P), then : linear indep. (Q)
if : linear dep. ( Q), then
( P))
: linear dep.,
※ Test at just one point of I to determine
linear dependency of the solutions
10
( ) 0W x 1 2, y y
1 2, y y ( ) 0W x
1 2, y y 1 2, c y cy
1 2 1 2 2 2 2 2 0W y y y y cy y cy y
( )W x
11
1 2( ) cos , ( ) siny x x y x x
0y y
1 2
1 2
2 2
( ) ( ) cos sin( )
( ) ( ) sin cos
cos sin 1 0
y x y x x xW x
y x y x x x
x x
1 2 , y y
。 Example 2.2:
are solutions of
: linearly independent
。 Example 2.3:
Solve by a power series method
The Wronskian of at nonzero x
would be difficult to evaluate, but at x = 0
are linearly independent
12
0y xy
3 6 91
4 7 102
1 1 1( ) 1
6 180 129601 1 1
( )12 504 45360
y x x x x
y x x x x x
1 2, y y
1 2
1 2
(0) (0) 1 0(0) 1 0
(0) (0) 0 1
y yW
y y
1 2, y y
◎ Find all solutions
○ Definition 2.2:
1. : linearly independent
: fundamental set of solutions
2. : general solution
: constant
○ Theorem 2.4:
: linearly independent solutions on I
Any solution is a linear combination of
13
1 2, y y
1 2{ , }y y
1 1 2 2c y c y
1 2, c c
1 2, y y
1 2,y y
Proof: Let be a solution.
Show s.t.
Let and
Then, is the unique solution on I of the initial
value problem
14
1 2, c c 1 1 2 2( ) ( ) ( )x y x y xc c
0x I 0 0( ) , ( )x a x b
0 0( ) ( ) 0, ( ) , ( )y P x y Q x y y x a y x b
0 0 ( ) : solution, ( ) , ( )x x a x b
2 0 2 0 1 0 1 01 2
0 0
( ) ( ) ( ) ( ),
( ) ( )
ay x by x by x ay x
W x W xc c
0 1 1 0 2 2 0
0 1 1 0 2 2 0
( ) ( ) ( )From ,
( ) ( ) ( )
x a c y x c y x a
x b c y x c y x b
2.2.2. Nonhomogeneous Equation
○ Theorem 2.5:
: linearly independent homogeneous
solutions of
: a nonhomogeneous solution of
any solution has the form
15
( ) ( ) ( )y P x y Q x y R x
1 2, y y
( ) ( ) 0y P x y Q x y
py
( ) ( ) ( )y P x y Q x y R x
1 1 2 2 pc y c y y
Proof: Given , solutions
: a homogenous solution of
: linearly independent homogenous solutions
(Theorem 2.4)
16
:py
( ) ( ) ( )
( )
0
p p p
p p p
y P y Q y
P Q y Py Qy
R R
py ( ) ( ) 0y P x y Q x y
1 2, y y
1 1 2 2py c y c y
1 1 2 2 pc y c y y
○ Steps:
1. Find the general homogeneous solutions
of
2. Find any nonhomogeneous solution of
3. The general solution of
is
2.3. Reduction of Order
-- A method for finding the second independent
homogeneous solution when given the first one
17
1 2, y y 0y Py Qy
py
y Py Qy R 0y Py Qy
1 1 2 2 pc y c y y
2y 1y
○ Let
Substituting into
( : a homogeneous
solution )
Let (separable)
18
2 1( ) ( ) ( )y x u x y x
2 1 1 2 1 1 1, 2y u y uy y u y u y uy
( ) ( ) 0y P x y Q x y
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1
2 [ ] 0
[2 ] [ ] 0
[2 ] 0
u y uy P u y uy Quy
u y Py u y Py Qy
u y Py
u y
u y
u y
1y
1 1
1
20 ( ) 0
y Pyu u u G x u
y
( ) 0v u v G x v
( )( )
G x dxv x ce
For symlicity, let c = 1,
: independent solutions
19
( )0
G x dxu v e
( )
( )G x dx
u x dxe
( )
2 1 1( ) ( ) ( ) ( )G x dx
y x u x y x y x dxe
1 2 1 2 1 1 1 1 1
2 21 11 1 1 1 1 1
( )
0
( ) y y y y uy u y y uy
u y vy
W x y
y uy y u y y uy
1 2 1, y y uy
。 Example 2.4:
: a solution
Let
4 4 0, ( )y y y A 2( ) xy x e
22 1( ) ( ) ( ) ( ) xy x u x y x u x e
Substituting into (A),
For simplicity, take c = 1, d = 0
: independent
The general solution: 20
2 22 2 ,x xy u e e u 2 2 2
2 4 4x x xy u e e u u e
2 2 2 2 2 24( ) 4 04 4 2x x x x x xuu e e u u e u e e u e 2 20, 0, 0, ( )x x u x cx du e e u
( )u x x 2 2
2( ) ( ) x xy x u x e xe 2 2
4
2 2 20
2 ( )
2
x xx
x x xx
e xeW x e
e e e
2 21 2( ) x xc xy x c e e
2 21 2, x xy y xe e
2.4. Constant Coefficient Homogeneous
A, B : numbers ----- (2.4)
The derivative of is a constant (i.e., ) multiple
of
Constant multiples of derivatives of y , which has form
, must sum to 0 for (2,4)
○ Let
Substituting into (2,4),
(characteristic equation)
21
0y Ay By
( ) xy x e2 0x x xA Be e e
xe xe
xe
2 0A B 2 4 ) / 2( A A B
i)
Solutions :
: linearly independent
The general solution:
22
2 4 0BA 2 2
1 24 ) / 2, 4 ) / 2( (A A B A A B 1 2
1 2, x xy ye e
1 2 2 1
1 2
1 2 1 2 2 1
( ) 22 1
( )
( ) 4 0
x x x x
x Ax
W x y y y
A B
y e e e e
e e
1 2, y y
1 21 2( ) x xy x c ce e
。 Example 2.6:
Let , Then
Substituting into (A),
The characteristic
equation:
The general solution:
23
6 0 ( )y y y A
( ) xy x e 2,x xy e y e
2 6 0x x xe e e
2 6 0
1 2( 2)( 3) 0, 2, 3
2 31 2( ) x xy x c ce e
ii)
By the reduction of order method,
Let
Substituting into (2.4)
24
2 214 0 , ( )
2
AxAB y x eA
22 1( ) ( )
Ax
y u x y u x e
22 2 2
2 2 2
4
( ) 02
Ax Ax Ax
Ax Ax Ax
Aue Au e u e
AA ue u e Bue
2
( ) 0 0, ( )4
Au B u u u x cx d
Choose
: linearly independent
The general sol.:
。 Example 2.7:
Characteristic eq. :
The repeated root:
The general solution:
25
2 22( ) ( )
Ax Ax
u x x y u x e xe
1 2, y y
2 21 2( )
Ax Ax
y x c c xe e
6 9 0y y y
2 26 9 ( 3) 0
3
31 2( ) ( ) xy x c c x e
iii)
Let
The general sol.:
26
2 4 0BA 2 2
1 2
4 4 ,
2 2
A B A i A B A i
21, 4
2 2
Ap q B A
( ) ( )1 2 ( ) , ( )p qi x p qi xy x e y x e
( ) ( )
( ) ( )( )
( ) ( )
p qi x p qi x
p qi x p qi x
e eW x
p qi e p qi e
2 2 2( ) ( ) 2 0px px pxp qi e p iq e iqe
( ) ( )1 2( ) (2.5)p qi x p gi xy x c e c e
。 Example 2.8:
Characteristic equation:
Roots:
The general solution:
○ Find the real-valued general solution
。 Euler’s formula:
27
2 6 0y y y
2 2 6 0
1 21 5 , 1 5i i
( 1 5 ) ( 1 5 )1 2( ) i x i xy x c ce e
cos sin , cos sinix ixe x i x e x i x
Maclaurin expansions:
28
2 3
0
2 2 4 6
0
2 1 3 5 7
0
1 1 11
! 2! 3!
( 1) 1 1 1cos
(2 )! 2! 4! 6!
( 1) 1 1 1sin 1
(2 1)! 3! 5! 7!
nx
n
nn
n
nn
n
e x x x xn
x x x x x xn
x x x x xn
2 3 4
2 3 4 5
2 4 3 5
1 1 1 1 ( ) ( ) ( ) ( )
2! 3! 4!1 1 1 1
12! 3! 4! 5!
1 1 1 1(1 ) ( )
2! 4! 3! 5!cos sin
ixe ix ix ix ix
ix x ix x ix
x x i x x x
x i x
。 Eq. (2.5),
29
( ) ( )1 2( ) p qi x p gi xy x c e c e
( ) (cos sin )p qi x px qxi pxe e e e qx i qx
cos sinpx pxe qx ie qx
( ) cos sinp qi x px pxe e qx ie qx
1( ) ( cos sin )px pxy x c e qx ie qx
2 ( cos sin )px pxc e qx ie qx
1 2 1 2( ) cos ( ) sinpx pxc c e qx c c ie qx
Find any two independent solutions
Take
Take
30
1 2
1 ( ) cos
2pxc c y x e qx
1 2
1 1, ( ) sin
2 2pxc c y x e qx
i i
( )
cos sin
cos sin sin cos
px px
px px px px
W x
e qx e qx
pe qx qe qx pe qx qe qx
2 0pxe
1 2( ) ( cos sin )pxy x e c qx c qx The general sol.:
2.5. Euler’s Equation
, A , B : constants -----(2.7)
Transform (2.7) to a constant coefficient equation
by letting
31
2
1 10y Ay By
x x
tx e1
( ) ( ) ( ), , t t dty x y e Y t dx e dt xdt
dx x
1( ) ( )
dY dty x Y t
dt dx x
2 2
1( ) ( ) ( ( ))
1 1 1 1 ( ) ( ) ( )
d dy x y x Y t
dx dx xd dY dt
Y t Y t Y tx x dx x x dt dx
2 2
1 1 1 1( ) ( ) ( ( ) ( ))Y t Y t Y t Y t
x x x x
Substituting into Eq. (2.7), i.e.,
--------(2.8)
Steps: (1) Solve
(2) Substitute
(3) Obtain
32
, ,y y y
2
1 10y Ay By
x x
2 2
1 1 ( ( ) ( )) ( ) ( ) 0
A BY t Y t Y t Y t
x x x x
( ) ( ) ( ) ( ) 0Y t Y t AY t BY x
( 1) 0Y A Y BY
( )Y t
t ln x
( )y x
。 Example 2.11: ------(A)
-------(B)
(i) Let
Substituting into (A)
Characteristic equation:
Roots:
General solution: 33
2 2 6 0x y xy y
2
1 12 6 0, 0y y y x
x x
tx e
2
1 1 ( ) ( ), , ( )y x Y t y Y y Y Y
x x
, ,y y y 2
2
1 1( ) 2 6 0x Y Y x Y Y
x x
2 6 0 , 6 0Y Y Y Y Y Y Y 2 6 0
3, 2 3 2
1 2( ) t tY t c e c e
34
, ,tx e t ln x 3 2 3 21 2 1 2( ) ,ln x ln xy x c e c e c x c x
0x
○ Solutions of constant coefficient linear equation have the
forms:
Solutions of Euler’s equation have the forms:
, , sin , cosx x x xe xe e x e x
, , cos( ), sin( )r r p px x ln x x qln x x qln x
2.6. Nonhomogeneous Linear Equation
------(2.9)
The general solution:
◎ Two methods for finding
(1) Variation of parameters
-- Replace with in the general homogeneous solution
Let
Assume ------(2.10)
Compute 35
( ) ( ) ( )y P x y Q x y R x
h py y y py
1 2, c c ( ), ( )u x v x
1 2( ) ( ) ( ) ( )py u x y x v x y x
1 2 1 2 py uy vy u y v y
1 2 0u y v y
1 2 py uy vy
1 2 1 2py u y v y uy vy
Substituting into (2.9),
-----------(2.11)
Solve (2.10) and (2.11) for
.
Likewise,
36
1 2 1 2 1 2( )u y v y uy vy P uy vy
1 2( )Q uy vy R
1 1 1 2 2 2 1 1 2[ ] [ ]u y Py Qy v y Py Qy u y v y R
1 2u y v y R , u v
1 1(10) (11)y y
1 1 2 1 1 1 2 1 1( )u y y v y y u y y v y y Ry
2 1 2 1 1 2 1 2 1 1, ( )v y y v y y Ry v y y y y Ry
1 1
2 1 2 1
Ry Ryv
y y y y W
2Ry
uW
1 2, Ry Ry
v uW W
。 Example 2.15: ------(A)
i) General homogeneous solution :
Let . Substitute into (A)
The characteristic equation:
Complex solutions:
Real solutions:
:independent37
4 secy y x
hy
xy e2 4 0, 2i
( ) 21
( ) 22
( )
( )
p qi x ix
p qi x ix
y x e e
y x e e
1
2
( ) cos cos2
( ) sin sin 2
px
px
y x e qx x
y x e qx x
cos2 sin 2
( ) 22sin 2 2cos2
x xW x
x x
1 2 , y y
ii) Nonhomogeneous solution
Let
38
:p
y
1 2py uy vy 1
22
1 1cos2 sec 2cos sin sec
2 2 sin
1 1sin 2 sec (2cos 1)sec
2 21
cos sec2
y Rx x x x x
Wx
y Rx x x x
W
x x
1( ) sin cosRy
v x xdx xW
2 1 1( ) cos sec sin sec tan
2 2
Ryu x xdx xdx x ln x x
W
iii) The general solution:
39
1 2
1( ) (sin sec tan )cos2
2py x uy vy x ln x x x
1 2( ) ( ) ( ) cos2 sin 2
1 cos sin 2 (sin | sec
2 tan |)cos2
h py x y x y x c x c x
x x x ln x
x x
cos sin 2x x
(2) Undetermined coefficients
Apply to
A, B: constants
Guess the form of from that of R
e.g. : a polynomial
Try a polynomial for
: an exponential for
Try an exponential for
40
( )y Ay By R x
py
( )R x
( )R x
py
py
。 Example 2.19: ---(A)
It’s derivatives can be multiples of
or
Try
Compute
Substituting into (A),
41
5 6 3sin 2y y y x ( ) 3sin 2R x x
sin 2x
cos2xcos2 sin 2py c x d x
2 sin 2 2 cos2py c x d x 4 cos2 4 sin 2py c x d x
4 cos2 4 sin 2 5( 2 sin 2 2 cos2 )c x d x c x d x
6( cos2 sin 2 ) 3sin 2c x d x x
(2 10 3)sin 2 ( 2 10 )cos2d c x c d x
: linearly independent
and
The homogeneous solutions:
The general solution:
42
sin 2 , cos2x x
2 10 3 0d c 2 10 0c d 15 3
, 52 52
c d
3 15cos2 sin 2
52 52py x x
3 2, x xe e
3 21 2
15 3cos2 sin 2
52 52x xy c e c e x x
。 Example 2.20: ------(A)
, try
Substituting into (A),
* This is because the guessed
contains a homogeneous solution
Strategy: If a homogeneous solution appears in any term of
, multiply this term by x. If the modified term
still occurs in a homogeneous solution, multiply
by x again
43
2 3 8 xy y y e
( ) 8 xR x e xpy ce
2 3 0 8x x x xce ce ce e x
py ce
xe
py
Try
Substituting into (A),
44
xp
y cxe
, 2x x x xp p
y ce cxe y ce cxe
2 2( ) 3
4 8
x x x x x
x x
ce cxe ce cxe cxe
ce e
2 and 2 xpc y xe
○ Steps of undetermined coefficients:
(1) Find homogeneous solutions
(2) From R(x), guess the form of
If a homogeneous solution appears in
any term of , multiply this term by
x. If the modified term still occurs in a homogeneous
solution, multiply by x again
(3) Substitute the resultant into
and solve for its coefficients
45
py
( )y Ay By R x
py
py
○ Guess from
Let : a given polynomial
, : polynomials with unknown coefficients
46
py ( )R x
( )P x
( )Q x ( )S x
Guessed ( )R x py
( )P x ( )Q x
axce axde
cos sinbx or bx cos sinc bx d bx
( ) axP x e ( ) axQ x e
( )cos ( )sinP x bx or P x bx ( )cos ( )sinQ x bx S x bx
( ) cos ( ) sinax axP x e bx or P x e bx ( ) cos ( ) sinax axQ x e bx S x e bx