Chapter 2 Second Order Linear ODEs 2.1 Homogeneous Linear Equations of 2nd Order

A 2nd order linear equation:

)()(')(" xryxqyxpy =++(1)

is called nonhomogeneous. )0)(( ≠xr(1)

0)(')(" =++ yxqyxpy(2)

is called homogeneous.

For instance a nonhomogeneous lin. ODE

xeyy x cos25 −=+′′

A homogeneous lin. ODE

0' =++′′ xyyyxA nonlinear ODE

02 =′+′′ yyy

Superposition Principle for the Homogeneous Linear ODEs:

Ex. 1 xyxy sin and cos ==are solutions of 0" =+ yy

Then xcxcy sincos 21 +=

is also a solution of the above Eq.

---superposition or linearity principle

Theorem 1. If 1y and 2y are solutions of (2), then 2211 ycycy += is a solution of (2).

Proof: 0)()( 111 =+′+′′ yxqyxpy0)()( 222 =+′+′′ yxqyxpy

We have

0)()()()'()"(

)()(

22221111

221122112211

=+′+′′++′+′′=+++++=

+′+′′

qyypycqyypycycycqycycpycyc

yxqyxpy

Ex. 2 Verify by substitution that xyxy sin1 and cos1 +=+=

are solutions of .1" =+ yy

xxyxy sincos2or )cos1(5 ++=+=

are not solutions of !1" =+ yy

Ex. 3 Verify that 1 and 2 == yxy

are solutions of the nonlinear ODE 0=′−′′ yxyy

22 1or xyxy +=−=

are not solutions of 0=′−′′ yxyy

IVP. Basis. General Solution Initial conditions: (4) 1000 )( ,)( KxyKxy =′=

(4) with (2) is called IVP for the second order homogenous linear ODE.

Ex. 4 Solve the initial value problem .5.0)0(' ,0.3)0( ,0" −===+ yyyy

Soluton. Step 1 xcxcy sincos 21 +=

is the general solution Step 2 particular solution

5.0)0(' and 0.3)0( 21 −==== cycy

xxy sin5.0cos0.3 −=

Definition 1. Two solutions 1y and 2y

of (2) are called a basis of (2) iff constant/ 21 ≠yy

Definition 2. Two solutions 1y and 2y

of (2) are a basis of (2), then

2211 ycycy +=

is called a general solution of (2).

A particular solution of (2) is obtained if we assign specific values to 1c and 2cin the general solution.

We call linearly independent if 21 and yy0 ,0 0 212211 ==⇒=+ ccycyc(7)

Definition 1’. A pair solutions 21 and yyof (2) are called a basis of (2) iff 21 and yy

are linear independent.

Ex. 5 cos x and sin x in Ex. 4 form a basis of solutions of y”+y=0 for all x. y=3.0cos x-0.5sin x is a particular solution.

Ex.6 Verify xx ee − and are solutions of .0" =−yy Then solve the IVP

2)0(' ,6)0( −== yy

Solution. ( ) ( ) 0 and 0 =−″

=−″ −− xxxx eeee

xx ececy −+= 21 is general sol.

4 ,22)0(' and 6)0(

21

2121

==⇒−=−==+=

ccccyccy

xx eey −+= 42 is particular sol.

A nontrivial solution of (2) 1y is obtained by some method, how to get the other 2y

to form a basis of (2)?

Set 12 uyy = then we have

112 '' uyyuy +=′ and

1112 "''2"" uyyuyuy ++= This gives

0)''("''2"

111

111

=+++++

quyuyyupuyyuyu

hence

.0')'2(",0)'"(')'2("

111

111

111

=++=+++

++

upyyyuqypyyu

upyyyu

Let "' ,' uUuU == then

.0)/'2(U' 11 =++ Upyy

dxpey

U

dxpyU

dxpyy

∫−=

−−=

+−=

∫

21

1

11

1

,||ln2||ln

,)/'2(U

dU

dxuuU ∫== U ,' dxUyuyy ∫== 112

(?)/ 12 constdxUuyy ≠== ∫form a basis of (2). 21 and yy

Ex. 7 Find a basis of solutions of

.0'")( 2 =+−− yxyyxxSolution. xy =1 is one of the solution

Set uxuyy == 12 then we have uxuy +=′ '2 and

'2""2 uxuy += This gives

0)'()'2")(( 2 =++−+− uxuxuxuxuxx

hence

0')2(")( 2 =−+− uxuxx Let 'uv =

0)2(')( 2 =−+− vxvxx

dxxx

dxxx

xvdv

−

−=

−−

−=2

112

2

|||1|ln||ln2|1|ln||ln

xxxxv −

=−−=

∫ +==−=−

=x

xvdxuxxx

xv 1||ln ,11122

Finally, we have

.1||ln2 +== xxuxy

,1 xy =with

They form a basis of the solutions.

2.2 2nd-Order Homogeneous Equations with Constant Coefficients

0'" =++ byayy(1) kxey −= is a solution of .0' =+kyy

To try as a solution of (1) the function xey λ=(2)

xey λλ=' and xey λλ2"= (1) becomes

0)( 2 =++ xeba λλλ

We got a characteristic equation of (1)

02 =++ baλλ(3)

2/)4( 22,1 baa −±−=λ(4)

xey 11

λ= and xey 2

2λ=

are solutions of (1).

042 >− ba042 =− ba

042 <− ba

Case I two real roots if

Case II a real double root if

Case III complex roots if

There are 3 cases for the roots.

Case I. Two Distinct Real Roots xey 1

1λ= and

xey 22

λ=

constant/ )(21

21 ≠= − xeyy λλ

The general solution is

xx ececy 2121

λλ +=

Ex. 1 in Ex. 6 of Section 2.1 .0" =−yySolution. The characteristic equation is

012 =−λIt has the roots 1±=λ

Hence the general solution xx ececy −+= 21

Ex. 2 Solve the IVP .5)0(' ,4)0( ,02'" −===−+ yyyyy

Solution. The characteristic equation is 022 =−+ λλ

It has the roots 2 ,1 21 −== λλ

Hence the general solution xx ececy 2

21−+=

xx ececy 221 2' −−=

According to the initial condition:

3 ,152)0(' and 4)0(

21

2121

==⇒−=−==+=

ccccyccy

Case II. Real Double Root

042 =− ba 2/21 a−=== λλλhence only one solution xaey )2/(

1−=

12 uyy =Set 112 '' uyyuy +=′

1112 "''2"" uyyuyuy ++=

0)''("''2"

111

111

=+++++

buyuyyuauyyuyu

.,0"

)2/' (since 0",0')'2("

,0)'"(')'2("

21

111

111

111

111

cxcuu

yayyuuayyyu

byayyuuayyyu

+==

−===++=+++

++

xu =

xaxexyy )2/(12

−==

0/ 12 ≠= xyy

The general solution is xaecxcy )2/(

21 )( −+=

Warning. (7) will not be a solution of (1) for the simple root of (4).

(7)

Ex. 3 Solve 016'8" =++ yyy

Solution. The characteristic equation is 01682 =++ λλ

It has the double root 4−=λand hence the general solution is

xecxcy 421 )( −+=

Ex. 4 Solve the initial value problem 1)0(' ,3)0( ,04'4" ===+− yyyyy

Solution. The characteristic equation is

0442 =+− λλIt has the double root 2=λand hence the general solution is

xexccy 221 )( +=

xx ecexccy 22

221 )(2' ++=

5 ,3,12)0('

,3)0(

21

21

1

−===+=

==

ccccy

cy

and the answer is

xexy 2)53( −=

Case III. Complex Roots

If (2) has a pair of complex roots 4/ ,2/ , 2

2,1 abai −=−=±= ωαωαλ

xiaey )2/(1

ω+−= and xiaey )2/(2

ω−−=

are solutions of (1).

To find the real solutions, we can obtain from (5) that

)sin(cos

),sin(cos2/)2/(

2

2/)2/(1

xixeeyxixeey

axxia

axxia

ωω

ωωω

ω

−==

+==−−−

−+−

and hence that xe ax ωcos2/− and

xe ax ωsin2/− are solutions of (1).

The corresponding general solution is

)sincos(2/ xBxAey ax ωω += −(9)

Ex. 5. Solve the initial value problem 3)0(' ,0)0( ,004.9'4.0" ===++ yyyyy

Solution. The characteristic equation is

004.94.02 =++ λλIt has the roots ,32.02,1 i±−=λand a general solution is

)3sin3cos(2.0 xBxAey x += −

The 1st initial condition gives

0)0( == Ay

It concludes that .3sin2.0 xBey x−=

)3cos33sin2.0(' 2.0 xxBey x +−= −

From this and the 2nd initial condition we get .133)0(' =⇒== BBy The answer is

xey x 3sin2.0−=

Ex. 6 A general solution of 0 where,0" 2 ≠=+ ωω yy

is

xBxAy ωω sincos +=

Ex. 7 Solve the boundary value problem

3)2/( ,3)0( ,0" −===+ πyyyy

Solution. The general solution is known as xBxAy sincos +=

So we have 3)2/( ,3)0( −==== ByAy π

The answer is xxy sin3cos3 −=

2.5 Euler-Cauchy Equation

0'"2 =++ byaxyyx(1)

is called the Euler-Cauchy equation. Let mxy = is a solution of (1), we find that

0)1( 122 =++− −− mmm bxaxmxxmmx or

0)1(2 =+−+ bmam(3)

(3) is called the characteristic equation of (1).

Case I. Two Distinct Real Roots If (3) has two distinct real roots 2,1mwe get a basis of solutions 21

21 and mm xyxy ==

constant/ 2121 ≠= −mmxyy

and a corresponding general solution of (1) is as follows

2121

mm xcxcy +=(4)

Ex. 1 Solve the Euler-Cauchy equation 05.0'5.1"2 =−+ yxyyx

Solution. The characteristic equation is 05.05.02 =−+ mm

?) 05.05.1(not 2 =−+ mmThe roots are m=0.5 and -1, this gives

xy =1 and xy /12 =and the general solution

0 ,/21 >∀+= xxcxcy

Case II. Real Double Root

If (3) has a double root, then 2/)1(2,1 am −=

hence only one solution .2/)1(1

axy −=

Set ,12 uyy = then we have

112 '' uyyuy +=′

1112 "''2"" uyyuyuy ++= This gives

0)''()"''2"(

111

1112

=+++++buyuyyuax

uyyuyux

hence

.ln||ln|'|ln ,'"

)2/)1(' (since "

,0')'2("

,0)'"(

')'2("

11

112

1112

1112

1112

xuxuuxu

yaxyxu'yyxu

uayxyxyxubyaxyyxu

xuayxyyxu

=−=−=

−=−=

=++

=+++

++

Thus ,ln12 xyy = where constxyy ≠= ln/ 12

The general solution is

2/)1(21 )ln( axxccy −+=

Ex. 2 Solve 09'5"2 =+− yxyyx

Solution. The characteristic equation 0962 =+− mm

has the double root m=3

Hence the general solution is

321 )ln( xxccy +=

Case III. Complex Conjugate Roots

If (3) has a pair of complex roots im νµ ±=2,1

then xii exxxy ln2,1

νµνµ ±± == are solutions of (1). To find the real solutions, we have

)]lnsin()ln[cos(

)],lnsin()ln[cos(

2

1

xixxyxixxy

νν

ννµ

µ

−=

+=

and hence that )lncos( xx νµ and

)lnsin( xx νµ are solutions of (1).

The corresponding general solution is

)]lnsin()lncos([ xBxAxy ννµ +=

Ex. 3 Solve 004.16'6.0"2 =++ yxyyx

Solution. The characteristic equation 004.164.02 =+− mm

has two complex conjugate roots

im 42.02,1 ±=

Hence the general solution is

)]ln4sin()ln4cos([2.0 xBxAxy +=

2.6 Existence and Uniqueness Theory Wronskian A general homogeneous linear equation

0)(')(" =++ yxqyxpy(1)

with continuous coefficients. The existence of a general solution

2211 ycycy +=and two initial conditions

1000 )(' ,)( KxyKxy ==

(2)

(3)

Theorem 1. If p(x) and q(x) are continuous on some open interval I, , then the initial value problem consisting of (1) and (3) has a unique solution on I.

Ix ∈0

Wronski determinant (or Wronskian) of two solutions of (1) as and 21 yy

122121

2121 ''

''),( yyyy

yyyy

yyW −==

We call linearly independent on I if 21 and yy0 ,0 on 0 212211 ==⇒=+ ccIycyc(4)

We call linearly dependent if 21 and yy

(4) holds for constants c1 and c2 not both 0.

Theorem 2. If p(x) and q(x) are continuous on some open interval I. Then two solutions of (1) on I are linearly dependent iff and 21 yy

. ,0))(),(( 00201 IxxyxyW ∈=

Furthermore, if . ,0))(),(( 00201 IxxyxyW ∈=

; ,0))(),((then 21 IxxyxyW ∈∀≡ hence if

, ,0))(),(( 11211 IxxyxyW ∈≠ then and 21 yy

are linearly independent on I.

Proof: (a) If and 21 yy are linearly dependent

on I, then there exists k, s.t. 21 kyy = and

0''''''

),( 222222

22

21

2121 ≡−=== kyyyky

ykyyky

yyyy

yyW

(b) Conversely, we assume that

IxxyxyW ∈= 00201 ,0))(),(( and show that

and 21 yy are linearly dependent on I.

We consider the linear equations

0)(')(',0)()(

022011

022011

=+=+

xykxykxykxyk

in the unknowns and 21 kk

(7)

(7) is homogeneous.

Its determinant is exactly 0))(),(( 0201 =xyxyW

hence the system has a nontrivial solution and 21 kk

Then we have a function )()()( 2211 xykxykxy +=

By Theorem 1 in Sec. 2.1, y(x) is a solution of (1) on I. We know that 0)(' ,0)( 00 == xyxy

It concludes by Theorem 1 that . ,0 Ixy ∈∀≡

Ixxykxyk ∈∀=+ ,0)()( 2211

this means linear dependence of and 21 yyon I.

(c) If 0))(),(( 0201 =xyxyW then and 21 yyare linear dependent, and then by (a),

.0),( 21 ≡yyW Hence 0))(),(( 1211 ≠xyxyW

implies linear independence of and 21 yy

Ex. 1 xy ωcos1 = and xy ωsin2 =

are solutions of .0" 2 =+ yy ω The Wronskian is

ωωωωωωω

=−

=xx

xxyyW

cossinsincos

),( 21

They are linearly independent iff .0≠ω0=ωIf then ,02 =y

hence they are linear dependence.

Ex. 2 A general solution of 0'2" =+− yyy

on any interval is .)( 21xexccy +=

The Wronskian is

0)1(

),( 2 ≠=+

= xxx

xxxx e

exexee

xeeW

Thus xe and xxe are linearly independent.

Theorem 3 If p(x) and q(x) of (1) are continuous on an open interval I, then (1) has a general solution on I.

Proof. By Theorem 1, (1) has solutions and 21 yy on I satisfying the initial conditions:

1)(' ,0)( 0101 == xyxy and 0)(' ,1)( 0202 == xyxy

Then and 21 yy are linearly independent on I. Thus they form a basis of (1) and then

2211 ycycy += is a general solution of (1).

Theorem 4 If p(x) and q(x) of (1) are continuous on an open interval I, then for every solution y=Y(x) of (1) is of the form

)()()( 2211 xyCxyCxY +=

where and 21 yy form a basis of (1) on I

1C 2Cand , are suitable constants.

Proof. A general solution of (1) is given as .2211 ycycy += We choose the suitable

and 21 cc such that )(')(' ),()( 0000 xYxyxYxy ==

),()()( 0022011 xYxycxyc =+ )(')(')(' 0022011 xYxycxyc =+

The Wronskian of and 21 yy is nonzero,

(10)

such that (10) has a unique solution 11 Cc =

and .22 Cc =

Thus a particular solution

)()()(* 2211 xyCxyCxy += of (1) satisfies (10). From this and the uniqueness theorem 1 we conclude that everywhere on I. )()(* xYxy =

2.7 Nonhomogeneous Equations

The nonhomogeneous linear equation

)()(')(" xryxqyxpy =++(1)

Its corresponding homogeneous equation is

0)(')(" =++ yxqyxpy(2)

Theorem 1. (a) If are solutions of (1) on I, then is a solution of (2) on I. (b) If y is a solution of (1) on I, and Y is a

solution of (2) on I, then is a solution of (1) on I.

and 21 yy21 yy −

Yy +

Proof. (a) Denote that then ,21 yyY −=

0)()()'"()'"()()'()"(

'"

222111

212121

≡−=++−++=−+−+−=

++

xrxrqypyyqypyy

yyqyypyyqYpYY

(b) Denote that then ,* yYy +=

)()(0)'"()'"()()'()"(

**)'("*)(

xrxrqypyyqYpYYyYqyYpyY

qyypy

=+=+++++=+++++=

++

Definition. A general solution of (1) on I is )()()( xyxyxy ph +=(3)

where 2211 ycycyh += is a general solution of (2), and )(xy p is any solution of (1).

A particular solution of (1) on I is a solution obtained from (3) by assigning specific values to the arbitrary constants in and 21 cc

.2211 ycycyh +=

Theorem 2. Suppose that the coefficients and r(x) in (1) are continuous on I. Then every solution of (1) on I is obtained from (3) by assigning suitable values to . and 21 cc

Proof. Let y*(x) be any solution of (1) on I. Let (3) be any general solution of (1) on I. Then is a solution of (2) by Theorem 1(a). By theorem 4 in Sec. 2.6, Y is obtained by assigning suitable values to the arbitrary constants From this and the statement follows.

pyyY −= *

. and 21 cc,* pyYy +=

Rules for the method of undetermined coefficients A.Basic Rule. Choosing in the Table. B. Modification Rule. Multiplying by

where j=1 or 2, if or is a solution of the corresponding homogeneous equation (or double roots).

C.Sum Rule. If r(x) is a sum of functions in the table, then is chosen as the sum of corresponding functions.

pyjx

xeγ xe x ωα cos

py

Ex. 1. Solve the equation

Solution. Solve the corresponding homogenous equation. We have

5.1)0(',0)0(,001.0" 2 ===+ yyxyy

,012 =+λ i±=2,1λ and then .sincos xBxAyh +=

We find that 0 is not a root of the characteristic equation. So

012

2 kxkxkyp ++= is chosen. Then

212 2",2 kykxky pp =+=′

.001.0)(2 201

222 xkxkxkk =+++

Compare with the coefficients above, we have

.02 ,0 ,001.0 2012 =+== kkkk Thus

.002.02 ,0 ,001.0 0012 −=−=== kkkk

Hence .002.0001.0 2 −= xyp And

002.0001.0sincos 2 −++= xxBxAy

Setting x=0, y(0)=A-0.002=0, hence A=0.002. y’(0)=B=1.5. Thus gives the answer

002.0001.0sin5.1cos002.0 2 −++= xxxy

Ex. 2. Solve the initial value problem 0)0(' ,1)0( ,1025.2'3" 5.1 ==−=++ − yyeyyy x

Solution. The characteristic equation is .025.232 =++ λλ It has the double root

.5.1−=λ and a general solution is

.)( 5.121

xexccy −+= At first, xp Cey 5.1−=

by the table. According to the rule, we choose

.5.12 xp eCxy −= Hence x

p exxCy 5.12 )25.1(' −+−=

.)25.262(" 5.12 xp exxCy −+−= Substitution gives

xxxx eeCxexxCexxC 5.15.125.125.12 1025.2)5.12(3)25.262( −−−− −=+−++−

,102 5.15.1 xx eCe −− −= 5−=CThus gives the general solution is

xexxccy 5.1221 )5( −−+=

,1)0( 1 −== cy)5.7105.15.1(' 2

225.1 xxxccey x +−−−= −

5.105.1)0(' 22 =⇒=−= ccyThis gives the answer xexxy 5.12 )55.11( −−+=

Ex. 3 Solve the initial value problem

08.40)0(' ,16.0)0( ,10sin19010cos405'2" 2/

==−+=++

yyxxeyyy x

Solution. The characteristic equation is .0522 =++ λλ It has the roots

,212,1 i±−=λ and a general solution is

).2sin2cos( xBxAey x += −

We choose .10sin10cos2/ xMxKCey x

p ++=

.10cos1010sin102/' 2/ xMxKCey xp +−=

.10sin10010cos1004/" 2/ xMxKCey xp −−=

Substitute this into the given equation.

16.0)514/1( 2/2/ =⇒=++ CeCe xx

.2,01909520,409520

.10sin19010cos4010cos)2095(10sin)2095(

10cos)520100(10sin)520100(

==⇒−=−−=−⇒

−=+−+−−=

+−−+++−

MKMKKM

xxxMKxKM

xKMKxMKM

This gives the general solution

10sin216.0)2sin2cos( 2/ xexBxAey xx +++= −

From the initial conditions, we have

.1008.402008.02)0('10cos2008.0)2cos22sin('

016.016.0)0(2/

=⇒=++=+++−=

=⇒=+=−

BByxexBxBey

AAyxx

We thus obtain the answer

10sin216.02sin10 2/ xexey xx ++= −

2.10 Solution by Variation of Parameters A general nonhomogeneous linear equation

)()(')(" xryxqyxpy =++(1) with continuous ).(),(),( xrxqxpMethod of variation of parameters,

dxrydxryy p ∫∫ +−=Wy

Wy 1

22

1(2) where 21, yy form a basis of the corresponding homogeneous equation of (1)

0)(')(" =++ yxqyxpy(3)

1221 '' yyyyW −= is the Wronskian of 21, yy

Ex. 1. Solve xyy sec" =+Solution. A basis of the homogeneous equation is Thus .sin and cos 21 xyxy ==

.1)sin(sincoscos'' 1221 =−−=−= xxxxyyyyW

xxxxxdxxxxdxxx

dxrydxryyp

sin|cos|lncosseccossinsecsincos

Wy

Wy 1

22

1

+=+−=

+−=

∫∫∫∫

xxcxxcxxxxxcxcy

sin)(cos|)cos|ln( sin|cos|lncossincos

21

21+++=

+++=We then obtain the answer

How to get the method? )()()( xyxyxy ph += with 2211 ycycyh +=

Let be a solution of (1). )()()()( 21 xyxvxyxuy p +=

2121 ''''' vyuyyvyuy p +++=0''Set 21 =+ yvyu then 21 ''' vyuyy p +=

2121 ""''''" vyuyyvyuy p +++=

Substituting to (1), and collecting terms

⇒=+++++++

=+++++++

ryvyuqypyyvqypyyu

rvyuyqvyuypvyuyyvyu

21

222111

21

'212121

'''')'"()'"(

)()'()""''''(

ryvyu =+ 21 '''' with 0'' 21 =+ yvyu

So we have W

ryvW

ryu 12 ' ,' =−=

dxW

ryvdxW

ryu ∫∫ =−= 12 ' ,

dxrydxryy p ∫∫ +−=Wy

Wy 1

22

1

Pages 103-104c

Continued

Pages 103-104d