em702p second-order linear ordinary differential equations

8/14/2019 EM702p Second-order Linear Ordinary Differential Equations

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Advanced Engineering Mathematics 2. Second-order Linear ODEs 1

2.1 Homogeneous linear ODEs

2.2 Homogeneous linear ODEs with constant coefficients

2.3 Differential operators

2.5 Euler-Cauchy equation

2.6 Existence and uniqueness theory

2.7 Nonhomogeneous ODEs

2.10 Solution by variation of parameters

2. Second-order Linear Ordinary

Differential Equations


2.1 Homogeneous linear ODEs

A linear second-order DEis formed of

y + p(x)y + q(x)y = r(x)

If r(x) 0 (i.e., r(x) = 0 for allxconsidered),

then the DEis called homogeneous.

If r(x) 0 , the DEis called nonhomogeneous.

The functionsp, q and rare called the coefficients of the equation.

A solution of a second-order DEon some open interval a < x < b is a

function y(x) that satisfies the DE, for allxin that interval.

/


2/20


At first we discuss the properties of solutions of second-order linear DE,then we consider the solving strategies.

Theorem 1 (Fundamental Theorem for homogeneous linear 2nd-order DE)

For a homogeneous linear 2nd-order DE, any linear

combination of two solutions on an open interval Iis again a

solution on I. In particular, for such an equation, sums and

constant multiples of solutions are again solutions.

Proof.

Let y1 and y2 be solutions of y + py + qy= 0 on I.

Substituting y = c1y1 + c2y2 into the DE, we get

y + py + qy= (c1y1 + c2y2) +p(c1y1 + c2y2) + q(c1y1 + c2y2)

= c1y1 + c2y2 + c1py1 + c2py2 + c1qy1 + c2qy2

= c1 (y1 +py1 + qy1) + c2 (y2 +py2 + qy2)

= 0.


Ex.1.y1 = e

x and y2 = e-x are two solutions of y y= 0

y= -3y1 + 8y2 = -3ex+ 8e -xis a solution of y y= 0

since (-3ex+ 8e -x) ( -3ex+ 8e -x) = 0.

Caution! The linear combination solutions does not hold for

nonhomogeneous or nonlinear DE.

Ex.2.y1 = 1 + cosxand y2 = 1 + sinx are solutions of nonhomogeneous DE

y + y= 1.

but 2(1 + cosx) and (1 + cosx) + (1 + sinx) are not solutions of

y + y= 1.

Ex.3.y1 =x

2 and y2 = 1 are solutions of nonlinear DE yyxy = 0,

but -x2 and x2 + 1 are not solution of yyxy = 0.


3/20


A general solution of a 1st-order DEinvolved one arbitrary constant, a

general solution of a 2nd-order DEwill involve two arbitrary constant,and is of the form

y = c1y1 + c2y2 ,

where y1 and y2 are linear independent and called a basis of the DE.

A particular solution is obtained by specifying c1 and c2.

Two function y1(x) and y2(x) are linear independent, if

c1y1(x) + c2y2(x) = 0 implies c1 = c2 = 0 .


Wronskian test for linear independent

Let y1(x) and y2(x) be solutions of y +py + qy= 0 on an interval I

Let W[y1, y2] = = y1y2 y1y2 forxin I.

Then 1. Either W[ y1, y2] = 0 for allxin I

or W[y1, y2] 0 for allxin I.

2. y1(x) and y2(x) are linear independent on I

W[y1, y2] 0 for somexin I.

The proof will be given later.

'' 21

21

yy

yy


4/20


An initial value problem now consists of y +py + qy= 0 and

two initial conditions y(x0) = k0 , y(x0) = k1 .

Ex.4.

Solve the initial value problem

y y= 0, y(0) = 4 , y(0) = -2

y= c1ex+ c2e

-x

y = c1exc2e

-x

y(0) = c1 + c2 = 4 c1 = 1

y(0) = c1c2 = -2 c2 = 3

solution y = ex+ 3ex.

Problems of Section 2.1.


2.2 Homogeneous linear ODEswith constant coefficients

y + ay + by= 0,

where coefficients a and b are constant.

Solution.

Assume y = ex and substituting it into the original DEto get

(2 + a + b) e x= 0The equation 2 + a + b = 0 is called the characteristic equation of the

DE. Its roots are and .

Then the solutions are and .

Consider three cases:

case 1. two distinct real roots if a2 4b > 0.

case 2. a real double root if a2 4b = 0.

case 3. complex conjugate roots if a2 4b < 0.

+= baa 4

2

1 21

= baa 4

2

1 22

xe 1

xe 2


5/20


Case 1 (Two distinct real roots 1 and 2)

y1 = and y2 = are linear independent.

The solution is just y= .

xe 1

xe 2

xxecec 21 21

+

Ex.2. Solving y + y 2y= 0 with

y(0) = 4 and y(0) = -5.

Solution.

The characteristic equation 2 + - 2 = 0, its roots are

and

Then y = c1ex+ c2e

-2x

Since y(0) = 4 = c1 + c2y(0) = -5 = c1 - 2c2

c1 = 1 and c2 = 3

y = ex+ 3e -2x.

( ) 1912

11 =+= ( ) 2912

12 ==


Case 2 (Real double root )

Assume

Use the method of reduction of order to find another solution.

step 1. Set y2 = uy1, and try to determine the function usuch

that y2 is an independent solution.

y2 = uy1 + uy1 and y2 = uy1 + 2uy1 + uy1

step 2. Substituting y2 in the original DE

(uy1 + 2uy1 + uy1) + a (uy1 + uy1) + buy1 = 0

uy1 + u(2y1 + ay1) + u(y1 + ay1 + by1) = 0

Since y1 is a solution, y1 + ay1 + by1 = 0

Since

2y1 + ay1 = 0

uy1 = 0

2

a=

xa

ey 21

=

1

22

1 )2(2'2 yaeae

a

y

xa

xa

===

x3-6x2+11x-6 = 0,x1 = 1.


6/20


u = 0

u= c (This is the order reduced equation.)

u= c1x+ c2Simply taking c1 = 1 and c2 = 0

We get y2 =xy1 , y1 and y2 are linear independent.Thus the general solution

y= (c1 + c2x) .

Warning.

If is a simple root of y +py + qy= 0 ,

then (c1 + c2x)ex is not a solution of the DE. (Explanation on page 17)

Ex. (omitted)

xa

e 2


Case 3Complex exponential function

Euler formulae ix= cosx isinx

Then and are complex solutions of the original DE.

The solution can be derived further.

z1= = = (cos x+ isin x) and

z2 = = = (cos x- isin x).

xe 1

xe 2

xixa

ee 2

xa

e 2

xix

xxxi

xx

ixixixixixe

ix

sincos

...)!5!3

(...)!4!2

1(

...!5

)(

!4

)(

!3

)(

!2

)(1

5342

5432

+=

+++++=

++++++=

xe 1

xe 2

0where

04where,4

2

2

41

21

221

21

>==


7/20


.0)(sin)(cos2

cossinsincos

sincos

andofceindependenlinearforTesting

22

21

21

21

21

21

21

21

21

21

=+=

+

axaxax

axaxaxax

axax

exexe

xexeaxexea

xexe

yy

Then the general solution becomes

(A cos x+ B sin x).x

a

ey 2

=

A basis of real solution of the DEis

.sin2

andcos2

2212

2211 xe

i

zzyxe

zzy

xa

xa

=

==

+=


Summary of cases 1, 2, and 3

case 1. real 1 2 y=

case 2. double real

case 3. complex conjugate

.)( 22121 21

ax

exccya

+===

).sincos(

,

,

21

22

1

41

21

21

xBxAey

abia

ia

ax+=

==

+=

.21 21xx ecec

+


8/20


Initial value problem = a DE+ initial conditions.

For example, y + 2y + 5y = 0, y(0) = 1, y(0) = 5.

Ex.4. (Boundary value problem)

A boundary value problem = a DE+ boundary conditions

y + y= 0 and two boundary conditions

y(0) = 3

y() = -3.

Characteristic function 2+1 = 0, = i

y= c1cosx+ c2 sinx.

Since y(0) = 3 = c1 c1 = 3

y() = -3 = -c1 c1 = 3Thus y= 3cosx+c2sinx.



2.3 Differential operator

Introduce a method for solving DE, say operational method.

Let D denote differentiation with respect tox, write Dy = y .

D is an operator, it transforms yinto its derivative y.

For example,

D (x2) = 2x , D (sinx) = cosxD (Dy) = Dy = y,

D3y= y, .

Define another operator L, L = P(D) = D2+ aD +b , call a 2nd-orderdifferential operator. Here a and b are constant, Pmeans polynomial,and L means linear.

L [y] = P(D) [y] = (D2

+ aD +b) y= y + ay+ by.


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L is linear operator; i.e., L [ y+ z] = L[y] + L[z] ,

P(D) [e 1x] = (12 + a1+ b) e

1x= P(1) e1x= 0

e 1x is a solution of y+ay+by= 0c

P(1) = 0 ( 1 is a root of 2 + a+b = 0)

If P() has two different roots, we obtain a basis.

If P() has a double root, we need a second independent solution.

Differentiate both sides of P(D) [e x] = P() e x w.r.t. to get

P(D) [xe x] = P() e x+ P()xe x

P(D) [xe 1x] = P(1) e1x+ P(1)xe

1x= 0

xe

1x

is a solution of y+ay+by= 0cP(1) = 0 and P(1) = 0

(That is, 1 is a double root of 2 + a+b = 0)

P() / (- 1) = P() (reduction of order)


How to use differential operator to solve DE?

Ex.1.

.

32

0302

03'02'

0)3(

0)2(

0)2)(3(

0)6(

06'"Solving

23

32

2

2

2

21

1

1

xx

x

x

ececy

eyey

yyyy

yD

yD

yDD

yDD

yyy

+=

=

=

=

=

=+

=

=+

=

=+

=

=+

=+

=+



10/20


2.5 Euler-Cauchy equation

x2y + axy + by= 0 (1)

Solution.

Substituting y=xm

into Eq.(1), to get

Consider three cases

case 1. Distinct real roots

Then the general solution is

0)1( 122 =++ mmm bxaxmxxmmx

equation)(auxiliary0)1(

0if0)1(

2 =++

=++

bmam

xbammm

( ) ( )

= baam 411 2

2

1

.221

1

mmxcxcy +=


case 2. Double root

using the method of reduction of order to find another solution.Substituting y2= uy1 into Eq.(1). We obtain

x2(uy1

+ 2uy1

+ uy1

) + ax(uy1

+ uy1

) + buy1

= 0

ux2y1 + ux(2xy1 + ay1) + u(x2y1 + axy1 + by1) = 0

Since y1 is a solution,x2y1 + axy1 + by1 = 0

Since

( ) 21

121 1

a

xyam

==

( )

( )

12

1

21

21

21

221

112

1

1

1

2'2,2

1

yx

xaxa

xaxxayxyxy

a

aa

aaa

a

==

+=

+=+=

ux2y1 + uxy1 = 0

ux+ u = 0.


11/20


.,lnln

'

0for,ln'ln

1'"

2-1

1

2am mxxy

xu

xu

xxu

xuu

===

=

>=

=

Thus the general solution ( ) .ln 21

21

a

xxccy

+=

wrong.isIt

2

'

0if,ln'ln

defined.notis0,lnsince

wrong,isderivationfollowingThe

22

2

+=

=

=


12/20


2.6 Existence and uniqueness theory

Theorem 1 (Existence and uniqueness theory for initial value problems)

For an initial values problem

y +py +qy= 0

y(x0) = k0, y(x0)= k1,

ifp(x) and q(x) are continues functions on some open interval

Iandx0 is in I, then the initial value problem has a unique

solution y(x) on the interval I.

We will discuss the existence of a general solution for a DE.

For a 2nd-order DE, existing a general solution c1y1 + c2y2 means y1 andy2 are linear independent; that is, k1y1 + k2y2 = 0 k1= k2= 0.

We know that the Wronskian are linearindependent.

2121

21and0

''yy

yy

yy


Now we want to prove the statement

Theorem 2 (Wronskian test for linear independent)

Let y1(x) and y2(x) be solution of y + py + qy = 0 on an

interval I

Then (a) Either W[y1,y2] = 0 for allxin I,

or W[y1,y2] 0 for allxin I,

(b) y1 and y2 are linear independent on I

W[y1,y2] 0 for somexin I .Proof

(a) Calculate W = y1 y2 + y1y2 - y1 y2 - y1 y2

= y1y2 - y1 y2y1 + py1 + qy1 = 0

y2 + py2 + qy2 = 0

[ ] Ixyyyyyy

yyyyW inallfor, 2121

21

2121 =

=

Since


13/20


(b) (independent W 0)

Prove by showing dependent W = 0.

If y1

and y2

are linear dependent, then y2

= ky1

for

some constant k. Thus

W[y1,y2] = y1y2 - y1y2 = ky1y1 - k y 1y1 = 0 .

-y2y1 - p y 2y1 - q y 2y1 = 0

y1y2 + py1y2 + qy1y2 = 0

-y2y1 + y1y2 + p (-y2y1 + y1y2) = 0

W + p W = 0 (A linear 1st-order DE)

W = ce -pdx

if c = 0 W= 0 for allxin I,

if c0 W 0 for allxin I


(independent W 0)

prove by showing W = 0 dependent.

Assume y1 0 , y2 0. Consider the linear system of equations

k1y1(x0) + k2y2(x0) = 0

k1y1(x0) + k2y2(x0) = 0,

where k1 and k2 are unknown.

By Cramers theorem,

then the system has a non-trivial solution; that is, k1 and k2 arenot both zero. Using these k1 and k2 to construct a function

y(x) = k1y1(x) + k2y2(x), then y(x) = k1y1 + k2y2 is a solution ofy + py + qy = 0, and ysatisfies the initial conditions

y(x0) = 0 and y(x0) = 0

0=

2

1

0201

0201

)()(

)()(

k

k

xyxy

xyxy

0),.,(0if21

21 ==

Wi.eyy

yy


14/20


By the existence and uniqueness theorem, we known that the

solution is unique such that k1y1 + k2y2 = 0 on I. Since k1 and k2are not both zero, y1 and y2 are linear dependent.

A general solution of y +py + qy= 0 includes all solutions;

that is, exist a solution of form c1y1 + c2y2.

Theorem 3 (Existence of a gereral solution)

If the coefficientsp(x) and q(x) of y +py + qy= 0 are

continuous on some open interval I, then the DEhas a

general solution on I.

Theorem 4 (General solution)If the coefficientsp(x) and q(x) of y +py + qy= 0 are

continuous on some open interval, then every solution

y= y(x) on Iis of the form y(x) = c1y1(x) + c2y2(x).


How to obtain a solution basis when only one solution is known ?

Let y1 be a solution of y +py + qy= 0 on some interval I. Substitute

y2 = uy1 into the DEto get y2 = uy1 + uy1 and y2 = uy1 + 2uy1 + uy1.

Then uy1 + 2uy1 + uy1 +p(uy1 + uy1) + quy1 = 0

uy1 + u(2y1 +py1) + u(y1 +py1 + qy1) = 0

uy1 + u(2y1 +py1) = 0

Let v= u

==

=

=

=

=++

vdxyuyy

ey

v

pdxyv

py

y

v

v

py

yvv

pdx

112

21

1

1

1

1

1

0

1

ln2ln

'2'

0)'2

('

Sincev dx= u cant be a constant, y1 and y2 form a basis of solution.


15/20


Ex. Find a basis of solution for the DE

x2y xy + y= 0

Solution.

One solution is y1 =x

.0,ln1

11)1

(1

2

22

ln

>==

==

=

xxxdxx

xy

xe

x

dxxe

xv x



2.7 Nonhomogeneous ODEs

Given a nonhomogeneous equation

y +p(x)y + q(x)y= r(x) , r(x)0 ... (1)

Consider the corresponding homogenous equation

y +p(x)y + q(x)y= 0 ... (2)

Theorem 1 (Relation between solutions of Eqs.(1) and (2))

(a) The difference of two solutions of Eq.(1) on some openinterval Iis a solution of Eq.(2) on I.

(b) The sum of a solution of Eq.(1) on Iand a solution of

Eq.(2) on Iis a solution of Eq.(1) on I.


16/20


Proof.

(a) Let y1(x) and y2(x) be two solutions of y +py + qy= r.

y1 +py1 + qy1 = r and y2 +py2 + qy2 = r

(y1 +py1 + qy1) (y2 +py2 + qy2) = rr= 0

(y1 - y2) +p(y1 - y2) + q(y1 - y2) = 0.

(b) Let

.)()()(

0'"thenEq.(2),ofsolutionabe

and,'"thenEq.(1),ofsolutionabe

212121

2222

1111

ryyqyypyy

qypyyy

rqypyyy

=+++++

=++

=++

A general solution of the nonhomogeneous equation (1) on some open

interval Iis a solution of the formy(x) = yh(x) + yp(x) .. (3)

where yh(x) = c1y1(x) + c2y2(x) is a general solution of the homogeneous

equation (2) on Iand yp(x) is any solution of Eq.(1) on Icontaining no

arbitrary constants.


Theorem2 (General solution)

Suppose thatp(x), q(x) and r(x) in Eq.(1) are continuous on

some open interval I. Then every solution of Eq.(1) on Iis

obtained by assigning suitable values to the arbitrary constant

in a general solution (3) of Eq.(1) on I.

Conclusion

a general solution of nonhomogeneous equation

a general solution of corresponding homogeneous equation

+

a particular solution of the nonhomogeneous equation.


17/20


Solution by undetermined coefficients

A simple method for finding the particular solution of

y +py + qy= r .. (1)

Principle:

Choose for yp a form similar to that of r(x) and involving unknown

coefficient to be determined by substituting that choice for yp into Eq.(1).

How to choose the solution form:

(a) Basic rule: Table 2.1 on page 105

r(x) yp(x)

k evx c evx

k xn knxn + kn-1x

n-1 + + k1x + k0

kcosx, ksinx m cosx + n sinx

k e xcosx, k e xsinx e x(m cosx + n sinx)


(b) Modification rule:

If cr(x) = a solution of the corresponding homogeneous equation,

then choose yp(x) = kxr(x) or kx2r(x),

(c) Sum rule:

If r(x) is a linear combination of functions listed in the left column of

Table 2.1, then yp(x) is taken as linear combination of the

corresponding functions listed in the right column of Table 2.1.

The properties of the method

(a) The method corrects itself in the same that a false choice of yp.

(b) One choice with too few terms will lead to a contradiction.

(c) A choice of too many terms will give a correct result, with superfluous

coefficients coming out zero.


18/20


Ex.1 ~ 4.

(1) r(x) = 8x2 yp(x) = k2x2 + k1x+ k0.

If you choose yp(x) = k2x2 would fail, why ?

y + 4y= 8x2

2k2 + 4k2x2 = 8x2If you choose yp(x) = k2x

2 +k1x+ k02k2 + 4k2x

2 + 4k1x+ 4k0 = 8x2

(2) r(x) = ex, yh(x) = c ex yp(x) = cxe

x.

(3) r(x) = ex+x, yh(x) = (c1 + c2x) ex yp(x) = cx

2 ex+ k1x+ ko .

(4) r(x) = 16ex+ sin 2 x, yh(x) = ex(A cos2x+ B sin2x)

yp(x) = kex+ Mcos2x+ Nsin2x.



2.10 Solution by variation of parameters

A general method to solve nonhomogeneous DE without special right

side r(x). For example

For the DE y +p(x) y + q(x) y= r(x). The method variation of

parameters gives a particular solution yp in the form

where y1 and y2 form a basis of solutions of the corresponding equation

y +p(x) y + q(x) y= 0

and w= y1y2 y1y2 is the Wronskian of y1 and y2.

xexyyy 2 23

=+

,)( 122

1 dxw

ryydx

w

ryyxyp +=


19/20


Ex.1. Solving y+ y= secx.

y1 = cosx, y2 = sinx, and W[y1,y2] = 1.

( ) ( ) .sincoscosln

sincoslncossincos

sincoslncos

sin)cos

sco(cos

cos

cos

sincos

sin

cos

seccossinsecsincos

21

21

xxcxxc

xxxxxcxcyyy

xxxx

xxdxx

xx

dxx

x

xdxx

x

x

dxxxxdxxxxy

ph

p

+++=

+++=+=

+=

+

=

+=

+=


How to get ?

For a nonhomogeneous DEand the corresponding homogeneous DE:

y +p(x) y + q(x) y= r(x) .. (1)

and y +p(x) y + q(x) y= 0.

Ifp and q are continuous in an open interval I, the homogeneous DE

has a general solution

yh(x) = c1y1(x) + c2y2(x) on I.

Now we replace c1 and c2 by functions u(x) and v(x) and to determine

u(x) and v(x) functions such that

yp(x) = u(x)y1(x) + v(x)y2(x) . (2)

Differentiating Eq.(2) to obtain

yp(x) = uy1 + uy1 + vy2 + vy2 .... (3)

We need two conditions to determine u(x) and v(x); one condition is that

yp(x) satisfies Eq(1). Now we assume another condition

( ) dxw

ryydx

w

ryyxyp +=

12

21

.021 =+ yvyu


20/20


This reduce yp(x) to the form yp(x) = uy1 + vy2

Differentiating the equation, we have

Substituting Eqs.(2), (3) and (4) into Eq.(1), we obtain

)4(............................................."''"''" 2211 vyyvuyyuyp +++=

( )

.],[

'and],[

'

''''

0''

'''')'"('"

''"''"''

21

1

21

2

21

21

21222111

21212211

yyW

yrv

yyW

yru

ryvyu

yvyu

ryvyuqypyyvqypyyu

rqvyquypvypuyvyyvuyyu

=

=

=+

=+

=+++++++

=+++++++

ryvyu =+ '''' 21


.

and

0],[Since

12

21

12

21

+=

==

dxw

yrydx

w

yryy

dxw

yrvdx

w

yru

yyW

p


em702p second-order linear ordinary differential equations

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