chapter 2: second-order differential equations

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 :


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


The characteristic

equation:


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


The general sol.:

。 Example 2.7:

Characteristic eq. :

The repeated root:


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:


○ 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)


◎ 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


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


and

The homogeneous solutions:


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


* 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


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

2.6.3. Superposition

Let be a solution of

is a solution of (A)

47

1 2( ) ( ) ( ) ( ) ( )Ny P x y Q x y f x f x f x

pjyjy Py Qy f

1 2 ...p p pNy y y

1 2 1 2

1 2

( ... ) ( )( ... )

( )( ... )

p p pN p p pN

p p pN

y y y P x y y y

Q x y y y

1 1 1( ) ... ( )p p p pN pN pNy Py Qy y Py Qy

1 2 ... Nf f f

。 Example 2.25:


where homogeneous solutions

48

24 2 xy y x e 2

21 2

(1) 4 , (2) 4 2

/ 44

x

xp p

y y x y y e

xy y e

224

1 2

1 ( )

4 4

xx

p p p

xy y y e x e

21 2

1( ) cos2 sin 2 ( )

4xy x c x c x x e

cos2 , sin 2 :x x

chapter 2: second-order differential equations

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