중앙대학교 건설환경플랜트공학과 교수

김 진 홍

- 6주차 강의 내용 -

2.6 Existence and Uniqueness of Solutions, Wronskian

- Theorem 2 ; Existence and Uniqueness Theorem for IVP

If are continuous functions on some interval I andand is in I,

then IVP consisting of (1) and (2) has a unique solution

)(xp )(xq 0x

(1) 0)()( yxqyxpy

(2) ,)( 00 Kxy 10)( Kxy

- Theorem 3 ; Linear Dependence and Independence of Solutions

Let ODE (1) has continuous coefficients on an open interval I. )(xp )(xqand

Then two solutions of (1) on I are linearly dependent on I if and1y

only if their "Wronskian",

2yand

- Theorem 1 ; Existence of a General Solution

If )(xp )(xqand are continuous functions on an open interval I, then ODE (1)

has a general solution.

''),( 122121 yyyyyyW

is 0 at some in I. 0x

Ex. 1) wxy cos1 wxy sin2 and are solutions of .02 ywy

Wronskian is,

wwxwwxwwxwwxw

wxwxwxwxW

22 sincos

cossin

sincos)sin,(cos

→ linearly independent if and only if 0w

For.tan/ 12 wxyy 0,0 2 yw and 0/ 12 yy

→ linearly dependent

Ex. 2) General solution of 02 yyy is .)( 21

xexccy

andxe and xxe are linearly independent.

0)1()1(

),( 222

xxx

xx

xx

xx exeexexe

xeexeeW

Its Wronskian is not 0,

Namely,

where, ''''

),( 1221

21

21

21 yyyyyy

yyyyW

⇐ Wronski determinant or Wronskian

Hence, if there is an in I where, then,0W are linearly 1y 2yand1x

independent on I.

?)( 1 y

2.7 Nonhomogeneous ODEs

- (1)

(2)

- General solution of the nonhomogeneous ODE (1) on an open interval I is,

)()()( xyxyxy ph (3)

where, ; general solution of (2) on an open interval I 2211 ycycyh

py ; any solution of (1) containing no arbitrary constant

- Theorem 1 ; Relations of solutions of (1) to those of (2)

(a) The sum of solution of (1) on some open interval I and a solution

y~ of (2) on I is a solution of (1) on I.

In particular, (3) is a solution of (1) on I.

(b) The difference of two solutions of (1) on I is a solution of (2) on I.

0)(),()()( xrxryxqyxpy

0)()( yxqyxpy

y

- Method of Undetermined Coefficients ; method to find

This method is suitable for linear ODEs with constant coefficients a .b

(4)

is an exponential function, a power of)(xr ,x cosine or sine,

or sum or products of such functions.

itself.)(xr

Method to find py

Term in r(x) Choices for yp(x)

xkexCe

),1,0( nkxn01

1

1 KxKxKxK n

n

n

n

wxkwxk sin,cos wxMwxK sincos

,sin,cos wxkewxke axax )sincos( wxMwxKeax

py

These functions have derivatives similar to

)(xrbyyay

and

where

- Choice Rules for Method of Undetermined Coefficients

(a) Basic Rule ; py is determined using the above table and coefficients are

determined by substituting and its derivatives into (4).py

(b) Modification Rule ; If a term happens to be a solution of py

homogeneous ODE corresponding to (4), is multiplied topy (or byx 2x

if this solution corresponds to a double root of the characteristic

equation of the homogeneous ODE)

(c) Sum Rule ; If is a sum of functions in the first column of the above )(xr

table, is the sum of the functions in the corresponding lines of the py

second column.

Some Examples

22 xyyy xx

h ececyyyy 2

21

2 ,02,02

,01

2

2 KxKxKyp ,2 12 KxKyp 22Kyp

2

01

2

2122 )(2)2(22 xKxKxKKxKKyyy ppp

4/3,022,2/1,2/1 0012122 KKKKKKK

75.05.05.0 2 xxyp

1)

75.05.05.0 22

21 xxececyyy xx

ph

xeyy 34 2)

,04 yy ,042 xBxAyh 2sin2cos

xeyy 34 13/1,149 CCCx

p ey 3

13

1

x

ph exBxAyyy 3

13

12sin2cos

,3x

p Cey ,3 3x

p Cey x

p Cey 39

Some Examples

xyyy sin2 xx

h ececyyyy 2

21

2 ,02,02

,sincos xMxKyp ,cossin xMxKyp xMxKyp sincos

xxMxKxMxKxMxKyyy ppp sinsin2cos2cossinsincos2

13,03 KMKM

xxyp sin1.0cos1.0

3)

xxececyyy xx

ph sin1.0cos1.02

21

xeyyy x 3cos134 24)

,0134 yyy ,01342 )3sin3cos(2 xBxAey x

h

),3sin3cos(2 xMxKxey x

p

1.0,1.0 KM

,32 i

ph yyy

Ex. 5) Solve IVP, 5.1)0(,0)0(,001.0 2 yyxyy

xBxAyyy h sincos,0

,01

2

2 KxKxKyp ,2 12 KxKyp 22Kyp

,001.02 2

01

2

22 xKxKxKKyy pp

002.02,02,0,001.0 200212 KKKKKK

002.0001.0 2 xyp002.0001.0sincos 2 xxBxAyyy ph

002.0,0002.0)0( AAy

5.1)0(,002.0cossin ByxxBxAy

002.0001.0sin5.1cos002.0 2 xxxy

Ex. 6) Solve IVP, ,1025.23 5.1 xeyyy 0)0(,1)0( yy

,025.23 yy ,0)5.1(25.23 22 x

h exccy 5.1

21 )(

,5.12 x

p eCxy ,)5.12( 5.12 x

p exxCy x

p exxxCy 5.12)25.2332(

xeyyy 5.11025.23

,1025.2)5.12(3)25.262( 222 CxxxCxxC 5C

,5 5.12 x

p exy xx

ph exexccyyy 5.125.1

21 5)(

,1)0( 1 cy xxx exxeexcccy 5.125.15.1

212 5.710)5.15.1(

,05.1)0( 12 ccy 5.15.1 12 cc

xxx exxexexy 5.125.125.1 )55.11(5)5.11(

Ex. 7) Solve IVP,

,09.0sin25.0cos275.02 xxxyyy ,78.2)0( y 43.0)0( y

,075.02 yyy ,075.022 2/3,2/1 2/3

2

2/

1

xx

h ececy

,21 ppp yyy ,cossin,sincos 11 xMxKyxMxKy pp

xMxKyp sincos1

1,0 MK

32.012.02 xyp

32.012.0sin2/3

2

2/

121 xxececyyyy xx

pph

78.232.0)0( 21 ccy

12.0cos5.15.0 2/3

2

2/

1 xececy xx

0,1.3,43.012.015.15.0)0( 2121 ccccy

32.012.0sin1.3 2/ xxey x

xyp sin1

,275.02 KMK 25.075.02 MKM

0,, 212012 ppp yKyKxKy

0275.0,09.075.0 101 KKK 32.0,12.0 01 KK

2.10 Solutions by Variation of Parameters

- Method of undetermined coefficients is efficient for obtaining in (1), if py

)(xr is not complicated.

(1)

This method is restricted to functions )(xr whose derivatives are similar

form to )(xr itself.

- The method of variation of parameters is generally used for obtaining py

where are assumed to be continuous on open interval I.

(2) dxW

ryydx

W

ryyxyp

12

21)( ; variation of parameters

where form a basis of solutions of the corresponding homogeneous 21, yy

ODE

)()()( xryxqyxpy

(3)

and is the Wronskian of 21, yyW

(4) 1221 yyyyW

0)()( yxqyxpy

rqp ,,

- Formula (2) is possible only if formula (1) is a standard form. And it may

often cause difficulties when the integration in (2) is not simple or (1) has

variable coefficients.

Ex. 1) Solve, xyy sec

,sin,cos 21 xyxy 1)sin(sincoscos),( 21 xxxxyyW

xxcxxcyyy ph sin)(cos)cosln( 21

xdxxxxdxxxxyp seccossinsecsincos)( xxxx sincoslncos

Ex. 2) Solve, xyy tan

,sin,cos 21 xyxy 1)sin(sincoscos),( 21 xxxxyyW

xxxxcxcyyy ph tanseclncossincos 21

xdxxxxdxxxxyp tancossintansincos)(

xdxxdxx

xx sinsin

cos

sincos

2

xxdxx

xx cossin

cos

1coscos

2

xxdxxxx cossin)sec(coscos

xxxxxx cossin)tansecln(sincos xxx tanseclncos

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