linear differential equations of second and higher orders

8/7/2019 Linear Differential Equations of Second and Higher Orders

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Chapter 2

Linear Differential Equations of Second and Higher Orders

Recall, Chapter 1, a linear differential equation of 1st-order can be written in the form:

)()()(01

xfyxayxa

A 2nd

-order linear differential equation is of the form)012 ()()()( xfyxayxayxa

In general, a linear differntial equation of order n can be written in the form

)()()()(...)()( 012)1(

1)(

xfyxayxayxayxayxa nnn

n

If ,0)( xan we obtain the standard form of a linear nth

-order differential equation

)()()()(... 012)1(

1

)(

xfyxayxayxayayn

n

n

Examples: The following are linear differential equations

xyxyxorderst 2

sectancos:1

0:22 yyxyxordernd

xxeyyorderth cos:4)4(

If 0)( xf in the equations above, the differential equation is called homogeneous.

Otherwise, it is nonhomogeneous.

For instance, the 2nd

order differntial equation above is homogeneous, but the 1st- and 4

th-

orders above are both nonhomogeneous.


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Contents of Chapter 2:

Section 2.1: Homogeneous Linear DE of 2nd

-order

Section 2.2: Second-Order Homogeneous Differential Equations With Real

Constant Coefficients

Section 2.3: The Complex Case for the Second-Order Homogeneous Differential

Equations With Real Constant Coefficients

Section 2.6: Cauchy-Euler Equations

Section 2.7: Existence and Uniqueness Theorem, Wronskain

Section 2.8: NonHomogeneous Equations

Section 2.9: Undetermined Coefficients Solution Method

Section 2.10 Solution by Variation of Parameters Methods

Section 2.12 Modeling of Electric Ciruits

Section 2.13: Higher-Order Linear Differential Equations

Section 2.14: Higehr-Order Linear Homogeneous Differential Equations

With Constant Coefficients

Section 2.15: Higher-Order Nonhomogeneous Linear Differential Equations


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Section 2.1: Homogeneous Linear DE of 2nd

-orderThe standard form for homogeneous linear DE of 2

nd-order is

0)()( yxqyxpy (1)

A homogeneous solution, hy , of (1) on an interval bxa

, is any function )(xhy

defined and continuous on the interval ),( baI and satisfies the eqn (1) above for all

Ix , i.e.,0)()()()()( xhxqxhxpxh

Example 1: Each of the functions xey 1 andx

ey2 satisfies the homogeneous

differential equation

0 yy (2)

So, both of 1y and 2y are homogeneous solutions.

Note also that xxxx eeee 42&4,2 are all homogeneous solutions to the DE in (2).

Actually, for any arbitrary constants 21 & cc , the linear combinationxx

ecec 21 ofx

e

and xe is a homogeneous soluion to the DE in (2).

Theorem 1 (Homogeneous Eqn (vs) Linearity and Superposition)For the homogeneous linear DE (1), any linear combination 2211 ycyc of two solutions

1y and 2y on an open interval Iis also a solution of the DE. So, sums and constant

multiples of solutions are also solutions.

Proof: Substituting 2211 ycyc into eqn (1) above:

))(())(()( 221122112211 ycycxqycycxpycyc

))()(())()(( 22221111 yxqyxpycyxqyxpyc

0)0()0( 21 cc Done

This is the concept of linearity and superposition in case of homogeneaous solutions ofthe homogeneous DEs.

This is not true for non linear or non homogeneous as in the following examples.


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Example 2: Both xy cos11 and xy sin12 are solutions to the linear nonhomogeneou

DE: 1 yy . But, 12y and 21 yy are not solutions.

Example 3:2

1 xy and 12 y are both solutions to the nonlinearhomogeneous DE0 yxyy . But, neither 1y nor 21 yy is a solution.

General and Particular Solutions:

For the homogeneous linear DE (1), the general solution, denoted gy , is any solution in

the form 2211 ycycyg , where 1y and 2y are two solutions and 1c and 2c are any arbitrary

constants. This general solution gy is also called homogeneous solution .hy

If 1cand 2c

are specified values, say 2 & 4 for instance, the solution 21 42 yy is then

called a particular solution, .py

Initial-Value Problem (IVP):

If initial conditions, say 1000 )(&)( yxyyxy , at some point 0x are given then the eqn

1000 )(&)(0)()( , yxyyxyyxqyxpy is called an initial-value problem (IVP).

Example 4: The initial-vlaue problem (IVP) 2)0(,4)0(,0 yyyy

has the general solution xxhg ececyy 21 .

To find the values of 1c and 2c , we substitute the initial conditions in the solution gy as

follow: 214)0( ccy and 212)0( ccy . From these equations we compute

3&1 21 cc . Thusxx

p eey 3 is the only particular solution that satisfies the IVP.

Basis Solution:

Definition Two functions f & g are said to be linearly independent on a set S if0)()( 21 xgcxfc is satisfied (for all x in S) only if 021 cc . Otherwise, f & g are

linearly dependent.

For example, if 0)( xf then f& g are linearly dependent for any function g and all x in

domain g since choosing 0&0 21 cc , we obtain 00)0()(.0)(. 11 cxgxfc .


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On the other hand, xxf cos)( and xxg sin)( are linearly independent on any set since if

01 c then 0sin.cos. 21 xcxc implies that xc

cx sin.cos

1

2 , i.e., xcos is a constant

multiple of xsin , which is not true.

So, two functions f& g are linearly dependent on a set S if one of them is a constant

multiple of the other one.

Definition The set }{ 21, yy of solutions of the DE 0)()( yxqyxpy is said to be abasis solution of the DE on a set S if 1y and 2y are linearly inedependent on S.

Example 5: The set }{ , xx ee

is a basis solution to the DE 0 yy since each of xx ee

&

a solution and xx ee & are linearly independet ( cee

e xx

x

2 for any constant c).

Example 6: Since xx sin&cos are linearly independet and since each of them is a

solution to the DE 0 yy , then }{ sin,cos xx is a basis solution to the given DE.

Remark: A basis solution of a homogeneous differential eqn is not unique. For if the set

}{ 21, yyof solutions of the DE 0)()( yxqyxpy

is a basis of the DE then }{ 2211 , ykyk

is also a basis solution to the DE for any nonzero constants21 & kk .

How do we find a basis solution to the homogeneous 2nd

-order differential equation

0)()( yxqyxpy ?

For now, we assume we have one solution 1y . We now use Reduction of order Method

obtain a second solution 2y that is linearly independent from 1y .

Reduction of order method:

Find a function u such that 12 uyy is a solution to the DE above. Since 112 yuyuy and1112 2 yuyuyuy we substitute into the equation we obtain:

0)())(()2( 11111 uyxqyuyuxpyuyuyu

0)2)(())()(( 111111 yuyyxpuyxqyxpyu

0)/)(2( 11 uyyxpu

Let uv . Then the eqn above reduces to the separable 1st-order differential eqn:


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)/2)(( 11 yyxpv

v

Integrating both sides with respect ot x, we obtain:

1ln2)(ln ydxxpv

dxxpey

vu )(2

1

1

Hence dxey

udxxp

)(

2

1

1and dxe

yyuyy

dxxp

)(

2

1

112

1.

Since .constant1 )(2

11

2

dxeyy

y dxxp , then }{ 21, yy is a basis solution and so the general

solution is then 2211 ycycyg .

Example 7: Find a basis solution to the following DE with xy 1 is one solution:

02 yyxyx

Solution: Write the DE in standard form: 0)/1()/1( 2 yxyxy

.0,)/1()( xxee dxxdxxp Hence, xdxx

xu ln

12

, xxuyy ln12 and

)ln(ln 1111 xccxxxcxcyy gh

Problems:

(6) Solve xxyxyyyx /sin,02 1

Answer: xxcxcyxxuyy h /)cossin(/cos 2112

(4) Solve yyx 32 Hint: Substiute yz , then yz and the eqution then becomes: zzx 32

Answer: 22/5

1 cxcy

(5) Solve 2)(2 yyy

Hint: Substiuteyz

, thenz

dy

dz

dx

dy

dy

ydy

and the eqution then becomes:2

2zdy

dzyz

Answer:21

1

cxcy


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Section 2.2: Second-Order Homogeneous Differential Equations With Real

Constant Coefficients

Consider the 2nd

-order linear homogeneous Differential Equation

0 cyybya (1)

Where a, b and c are real constants.

Idea of Solution:Recall a first-orderlinear homogeneous DE

0 kyy has the exponential solution kxey , where kis constant.

So, to find the solution of (1), lets try rxey , where ris constant to be determined.

Substituting rxey into (1), we get the auxiliary (or characteristic) quadratic equation

02

cbrar (2)The roots of this quadratic equation are:

acbb

aracbb

ar 4

2

2

1&4

2

2

121

And we have two solutionsxr

ey 11 andxr

ey 22 (3)From these two solutions, we obtain a basis solution and hence the general solution .gy

Types of Solutions:

There are three types of solutions depending on status of the the roots 21 & rr . We have

three cases of the roots:

(i) Case 1: 21 & rr are real and distinct(ii) Case 2: 21 rr and real(iii) Case 3: 21 & rr are complex and distinct

We now discuss the general solution in each case.

Case 1: Distinct Real Roots

Rrr 21

This case occurs if the discriminant 042 acb , hence the roots 21 & rr are real and distin

and the two solutions xrey 11 andxr

ey 22 are linearly independent

( tconseyy xrr tan/ )(2121 ).

So, in this case, the basis is },{ 21xrxr

ee and the general soultion is xrxr ececg

y 21 21 .


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Example 1: Solve the IVP: .1)0(,1)0(,034 yyyyy Solution: Let rxey . Then we hav 0342 rr The roots are 21 13 rr which are real, hence

xxecec

gy

23

1

Substituting 1)0(,1)0( yy , we obtain 2121 31&1 cccc

And the values of 21 &cc are 121 &0 cc , so the solution isx

ey

Case 2: Double Real Roots Rrr 21

This case occurs if the discriminant 042 acb , hence we have double real roots:

a

brr

221 .

Hence, the two solutions abxeyy 2/21 are dependent.

Question: How do we obtain a basis of soltuion in this case?

Answer: We apply the Rduction-of-Order method (discussed in Section 2.1) on the DE0)/()/( yacyaby

with a solution abxey 2/1 is given. Let 12 uyy . Then dxe

yu

dxxp

)(

21

1, where

abxp /)( . So, xdxee

u abxabx

/

/

1, 12 xyy and the general solution is then

xrg exccy

1)( 21

Example 2: Solve the IVP: .1)0(,1)0(,096 yyyyy Solution: Let rxey . Then, the auxiliary (characteristic) equation is 0962 rr .The roots are 21 3 rr , a real double root. Hence, by case 2 above,

xxxg exccxececy

321

32

31 )(

Substituting 1)0(,1)0( yy , we obtain 211 31&1 ccc

And from which we have 22 c , so the solution isxexy 3)21(


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Section 2.3: The Complex Case for the Second-Order Homogeneous Differential

Equations With Real Constant Coefficients

We are solving the 2nd

-order linear homogeneous Differential Equation

0 cyybya (1)

Where a, b and c are real constants.

Substituting rxey into (1), we get

acbb

aracbb

ar 4

2

2

1&4

2

2

121

Case 3: Complex Conjugates Roots Crr 12

This case occurs if the discriminant 042 acb . Definea

b

2 and abac 2/4 2 .

Since the coefficients a, b and c are real constants then the roots are complex conjugates:

irrir 121 & , where1

i and the two solutions ,xix

eey

1 andxix

eey 2 are linearly independent ( tcons

xieyy tan2

/ 21 ) . But they are complex.

Question: How do we obtain a basis of real soltuions in this case?

Answer:Using Eulers Identities: xixeix sincos and xixeix

sincos

Hence, xee

ixix

cos2

and xi

eeixix

sin2

To find a real basis, note that )21

(xixix

g ececey

Choosing 2

121 cc , we obtain xey xp cos1

.

Choosingi

ci

c2

1&

2

121 , then xey

xp

sin

2

.

Note that21

&pp

yy are real, linearly independent and homogeneous solutions of (1). So,

the basis is }sin,cos{ xx

exx

e

and the general soultion is

)sin2

cos1

( xcxcx

eyg

Example 3: Solve the IVP: .1)0(,1)0(,052 yyyyy

Solution: Letrx

ey . Then we hav 0522

rr The roots are irir 21

2&21

1 . So, 1 and 2 , then

)2sin2

2cos1

( xcxceg

y x

Substituting 1)0(,1)0( yy , we obtain 211 21&1 ccc . Hence, 21 1 cc , so the

solution is )2sin2(cos xxey x .


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Section 2.7 Existence and Uniqueness Theorem, Wronskain

In this section, we give a general theory for the homogeneous linear 2nd

-order differential

equation of the form

0)()(01

yxayxay (1)

where )(&)( 10 xaxa are continuous functions on a given open interval I.

Consider solving the initial-value prorblem:

1001 )0(&)0(,0)()( yyyyyxayxay (2)

Theorem 1: (Existence and Uniqueness Theorem)

If )(&)( 10 xaxa are continuous functions on some open interval Iand if Ix 0 , then the

initial-value problem (2) has a unique solution )(xy on the interval I.

Linear Independence of Solutions,WronskianThe Wronskian ),( 21 yyW of two solutions )(&)( 21 xyxy of (1) is the determinant

2121

21

21

21 det),( yyyyyy

yyyyW

Theorem 2: (Linear Independence and Dependence of Solutions)

Suppose )(&)( 10 xaxa are continuous functions on some open interval I. Then two solutions

)(&)( 21 xyxy of (1) on Iare linearly dependent on Iif their Wronskian is zero at some poin

Ix 0 . Furthermore, if 0W at some Ix 0 , then 0W on I; hence if there is Ix 1 at

which 0W , then )(&)( 21 xyxy are linearly independent on I.

Example 1: xxyxxy sin)(&cos)( 21 are solutions of 02 yy , with 0 , on any

interval. Their wronskian is 0cossin

sincosdet)sin,(cos

xx

xxxxW

So, by Theorem above, xx sin&cos are linearly independent on all ofR. Hence, the

general solution to 02 yy is xcxc sincos 21 , where 21 & cc are arbitrary constants.

Example (2):xx

xexyexy )(&)( 21 are solutions of 02 yyy on any interval.

Their wronskian is 0)1(

det),(2

x

xx

xx

xxe

exe

xeexeeW

So, xx xee & are linearly independent on all ofR , and the general solution to 02 yyy

is xexcc )( 21 where 21 & cc are arbitrary constants.


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Theorem 3: (Existence of a General Solution)

If )(&)( 10 xaxa are continuous functions on some open interval I, then the differential

equation 0)()( 01 yxayxay has a general solution on I.

Theorem 4: (General Solution)Suppose that equation (1) has continuous coefficients )(&)( 10 xaxa on some open interval .

Thenevery solution of (1) on Iis of the form )()()( 2211 xycxycxyg , where )(&)( 21 xyxy

form a basis of solutions of (1) on Iand 21 & cc are arbitrary constants.

Hence, (1) does not have singular solutions.


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Section 2.13 Higher-Order Linear Differential Equations

In this section, the concepts and definitions introduced to the linear differential equations

order 2 in Sections (2.1 & 2.7) are generalized to the Linear Differential Equations of

higher-orders (n > 2):

)()()()( 01)1(

1)( ... xfyxayxayxay nn

n (1)If 0)( xf , we obtain the homogeneous differential equation in (2). Otherwise, it is

nonhomogeneous.

0)()()( 01)1(

1)( ... yxaxayxay ynnn (2)

A homogeneous solutionhy of (2) on an interval bxa , is a function )(xhy , defined

and continuous on the interval ),( baI , that satisfies eqn (2) above for all Ix .

Example 1: Each of the functions xey 1 ,x

ey2 and

xxey 3 satisfies the following

3rd-order linear and homogeneous differential equation: 0 yyyy

So, all of 1y , 2y

and 3y are homogeneous solutions.

Note also that xxxx xeeee 42&4,2 are also homogeneous solutions to the above DE

Theorem 1 (Superposition and Linearity Principles)For the homogeneous linear DE (2) , sums and constant multiples of solutions on an open

interval Iare again solutions of (2) on I.

Proof: It is a generalization of the proof in Sec. 2.1 Done

Again, this is not true for non linear or non homogeneous.

Linear Independence and Dependence Definition

The n functions )(...,),(),( 21 xyxyxy n are called linearly independent on an interval Iif

0)(...)(11 xycxyc nn (3)

For all x in Iimplies that all nccc ...,,, 21are zero. Otherwise, they are linearly dependent.

For example, if 0)(1 xy then )(...,),(),( 21 xyxyxy n

are linearly dependent by putting

0...,1 21 nccc Also, )(...,),(),( 21 xyxyxy n are linearly dependent on I iffone of the functions can be writte

as a linear combination of the other (n-1) functions.

For instance, ))(...)((1

)( 221

1 xycxycc

xy nn

iff 01 c

iff )(...,),(),( 21 xyxyxy n are linearly dependent.


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Example 2: The functions 2321 )(&2)(,)( xxyxxyxxy are linearly dependent since

)(.0)(2

1)( 321 xyxyxy

Example 3: Show that the functions3

3

2

21 )(&)(,)( xxyxxyxxy are linearlyindependent on any interval, say ]3,1[I

Solution:

Equation (3) is: 0332

21 xcxcxc Putting 3&2,1 xxx , into (3) we obtain the following system of linear equations:

02793

0842

0

321

321

321

ccc

ccc

ccc

Solving these equations we obtain 0321 ccc . Hence32

&, xxx are linearly independen

on the interval ]3,1[ .

General Solution, Basis and Particular Solutions:

A basis of solution of (2) on an open interval Iis a set }...,,,{ 21 nyyy ofn linearly

independent solutions.

A general solution of (2), denoted ,gy on an open interval Iis any solution of the form

nng ycycycy ...2211 (4)

where }...,,,{ 21 nyyy is a basis (fundamental system) of solutions of (2) and nn cccc &...,,, 121 are any arbitrary constants. This gy is also called homogeneous solution .hy If nn cccc &...,,, 121

in (4) are specified values, the solution is then called a particular

solution py .

Example 4: The functions 32 &, xxx in Example 3 above are linearly independent solution

to the 3rd

-order linear and homogeneous differential equation 0663 23 yyxyxyxon

any interval does not contain 0.

So, a general solution to the above equation is3

3

2

21 xcxcxcyg and (for example)32

523 xxxy p is a particular solution to the DE in Example 4 above.

Initial-Value Problem, Existence and Uniqueness

Consider the IVP:10

)1(1000

01)1(

1)(

)(...,,)(,)(

0)()()( ...

nn

nn

n

yxyyxyyxy

yxayxayxay(5)


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Theorem 2: (Existence and Uniqueness Theorem)

If )(...,,)( 10 xaxa n are continuous functions on some open interval Iand if Ix 0 , then the

initial-value problem (5) has a unique solution )(xy on the interval I.

Example (5) Solve the IVP

4)1(,1)1(,2)1(

066323

yyy

yyxyxyx

Solution: Note that this equation is not in standard form. Its standard form is

066332

yx

yx

yx

y

So, the coefficeints are continuous on any interval not containing 0. Since ,010 x then b

Theorem 2 above, a unique solution exists on the interval where .0x

In Example 4,we saw that the general solution to this eqn. is3

32

21 xcxcxcyg Substituting the given initial conditions, we obtain

32

321

321

624)1(

321)1(

2)1(

ccy

cccy

cccy

Solving for the constants, we obtain 322 xxxy p as the unique solution with x > 0.

Linear Independence of Solutions. Wronskain

The Wronskian ),...,( 1 nyyW ofn solutions )(,...,)(1 xyxy n of (2) is the nth-orderdeterminant

)1()1(1

21

21

1

........

.

.

.......

......

det)...,,(

n

n

n

n

n

n

yy

yyy

yyy

yyW

Theorem 3: (Linear Independence and Dependence of Solutions)

Suppose )(...,,)( 10 xaxa n

are continuous functions on some open interval I. Then n solutio

)(),...,(1 xyxy n of the differentia equation

0)()(...)( 01)1(

1)( yxayxayxay

nn

n (2)

on Iare linearly dependent on Iif their Wronskian is zero at some point Ix 0 .

Furthermore, if 0W at some Ix 0 , then 0W on I; hence if there is Ix 1 at which W

, then )(,...,)(1 xyxy n are linearly independent on I.


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Example 6: 32

32

322

620

321det),,( x

x

xx

xxx

xxxW

By Example 5, the functions

32&,xxx are solutions to the 3

rd

-order linear and homogeneodifferential equation 0663 23 yyxyxyx on any interval not containing 0.

Since, 02 3 xW for any ,0x then the solution set },,{ 32 xxx is liearly independent on

),0( I and hence a basis of solutions of the equation above.

Example 7: Since 04

)2(

)1(det),,(

x

xxx

xxx

xxx

xxx e

exee

exee

xeee

xeeeW

And sincexxx

xeee &,

are solutions to 0 yyyy for all x, then by Theorem above,},,{

xxxxeee is a basis of solutions to the equation.

Theorem 4: (Existence of a General Solution)

If the coefficients )(...,,)( 10 xaxa n of (2) are continuous functions on some open interval I,

then (2) has a general solution on I.

Theorem 5: (General Solution)

Suppose that (2) has continuous coefficients )(...,,)( 10 xaxa n on some open interval . Then

every solution of (2) on Iis of the form )(...)()()( 2211 xycxycxycxy nng where )(,...,)(1 xyxy n form a basis of solutions of (2) on Iand 21 & cc are arbitrary constant

Hence, (2) does not have singular solutions.


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Section 2.14: Higehr-Order Linear Homogeneous Differential Equations

With Constant CoefficientsIn this section, we discuss the solution to the Higehr-Order Homogeneous Differential

Equations with real constant coefficients 001)1(

1)( ... yayayay nnn (1)

Since the coeffecients of (1) are all consatnts, and hence contintuous on R, then,

by Theorem 4 of Section 2.13, a general solution always exists on all ofR.

To find the solution of (1), we imitate the method of in Sections 2.2 (the 2nd

-order

homogeneous equation with constant coefficients). So, we assume an exponential

solution rxey , and we substitue it in eq. (1) we obtain the auxiliary equation:

0011

1 ... ararar nnn (2)

Let nrr ...,,1 be the n roots of (2). Then we obtain the solutionsxr

n

xr n

eyey ...,,1

1 To find a basis of solutions, we look for n linearly independent solutions of the form ey

Remark:Before discussing the cases of the roots, we need the following Vandermond Determinant

of order n

:

distinctarerallifnonzero

jisomeforrrif

rr

rrr

rrr

V

i

ji

ji

nji

nn

n

n

nn

n

,

,,0

)()1(

.

.

1.......11

det

1

2/)1(

11

2

1

1

21

Cases of the Roots:As in Section 2.2, we have three cases to consider:

(iv) Case 1: all roots are real and distinct(v) Case 2: some complex roots(vi) Case 3: some repeated real roots.


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Case 1: Distinct Real Roots jiRrr ji ,

In this case, the n solutions are xrnxr neyey ...,,11 . Using Vandermond determinnat, the

Wronskian of these solutions is nonzero:

xrnn

xrnxrn

xrn

xrxr

xrxrxr

xrxr

n

n

n

n

ererer

ererer

eee

eeW

11

2

1

1

21

21

21

21

1

.

.

.......

det),...,(

0

)()1(

.

.

1.......11

det

1

2/)1()...(

11

2

1

1

21

)...(

21

21

ji

nji

nnxrrr

n

n

nn

n

xrrr

rre

rrr

rrr

e

n

n

So the solutions above form a basis and a general solution in this case is nxr

g ececy ...1

1

for all real x.

Theorem 1: (Basis)

n solutionsxr

nxr n

eyey ...,,1

1 of (1) form a basis of solutions of (1) iff all the n roots

nrr ...,,1 of (2) are distinct.

Theorem 2 (Linear Independence)

Any number of solutions xrkxr keyey ...,,11 of (1) are linearly independent on any open

interval I iff krr ...,,1 are distinct.

Example 1: Solve the differential Equation022 yyyy

Solution: rxey 022 23 rrr The roots of this auxiliary equation are: 2&1,1 and the general solution is then

xxxg ecececy

2321


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Remark To look for the roots, we use the following helpful steps:

- By rational zero test, the rational zeros are obtained from the set }1

2,1{

where 1 &

2 are the divisors of 2 (the constatnt term) and 1 is the divisor of 1 (the leading

coefficient)

- By Descartes rule of sign, there are 2 or no positive real roots and exactly onenegative root.

- Use synthetic division by the possible rational zeros to obtain the roots.

Case 2: Some Complex Conjugate Roots Crr ji Note: If the coefficients 01 ...,, aan are all real, then if there is a complex root, say

ir , then its conjugate ir is also a root, i.e., complex roots occurd as pairs.

As in Section 2.3, the two real solutions that correspond to the complex conjugate roots

ir are xe x cos and xe x sin .

Example 2: Solve the IVP: 299)0(&11)0(,4)0(,0100100 yyyyyyy

Solution: rxeyThe auxiliary equation is 010010023 rrr

-The rational zeros are divisors of 100.

- There are 3 or 1 positive root

- There are no negative roots

- By synthetic division by 1, we obtain a quadratic quation:

010001

10001

100100111

So, 11 r is a real root. The other two roots are the roots of the quadratic equation

01002 r which are the complex conjugate roots ir 103,2

So, the general solution is xcxcecy xg 10sin10cos 321

We now substitute the initial values

21

21

21

100299)0(

1011)0(

4)0(

ccy

ccy

ccy

Solving for the constants 321 &, ccc we obtain xxeyx

g 10sin10cos3


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9

Case 3 Some Repeated Real Roots mrrr ....21

Ifa real root 1r is repeated m-times ( 1r is a root of multiplicity m), say mrrr ....21 ,

then (as a generalization of the 2nd

-order in this case) we obtain m independent solutions

as follow: xrmxrxr exmyxeyey 111

11 ...,,2

,

Example 3: Solve the DE: 033 yyivyvy

Solution: The auxiliary equation is 033 2345 rrrr

The roots are: 0, 0, 1,1 ,1 and the general solution is xexdxddxccy )()( 221010

Ifa complex root, say

ir and hence its conjugate ir are repeated, say

mrrr ....21 and mrrr ....21 then indepndent solutions are obtained by multiplying

the solutios xe x cos and xe x sin by x for each time the roots are repeated, so we

obtain the solutions,sin...,,sin,sin

cos...,,cos,cos

1

1

xexxxexe

xexxxexe

xmxx

xmxx

For example, if ir 211 is a root of multiplicity 3 of the auxiliary equation of a

homogeneous DE with real constant coefficients, then its conjugate ir 211 is also a

repeated root of multiplicity 3. Hence, the solutions corresponding to these roots are

)2sin)(2cos)((2

2102

210 xxBxBBxxAxAAex


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21

Section 2.6: Cauchy-Euler Equations

Consider the 2nd

order Cauchy-Euler DE: 02 cyybxyax (1)

where a , b and c are real constants. Writing this equation in standard form, we obtain

02

y

ax

cy

axby

(2)

By Theorem 4 of Section 2.13, since the coeffecients of (2) are all contintuous on }0/{R ,

then a general solution always exists on any interval not containing 0.

So, a solution of (2)( hence of (1)), exists on ),0( or )0,( .

Idea of Solution:

We look for a function fsuch that fxfx 2& are both constant multiples off. For

instance, rxxf )( satifies frrfxrffx )1(& 2 .

So we assume the solution of the formr

xy where r is a constant to be determined.Substituting rxy into (1), we get the characteristic (auxiliary) quadratic equation

0)1( cbrrar (3)

Or 0)(2 crabar (4)

The roots 21 &rr of this quadratic equation are:

acababa

r 4)()(2

1

12 and acabab

ar 4)()(

2

1

22

And we have two solutions 11r

xy and 22r

xy (5)

From these two solutions, we obtain a basis solution and hence the general solution .gy

Types of Solutions:

There are three types of solutions depending on the the roots 21 &rr . We have three cases

of the roots:

Case 1: 21 & rr are distinct and real

Case 2:21rr and real

Case 3: 21 & rr are distinct but complex conjugates.

Note that if 21 &rr are distinct, then the two solutions:1

1

r

xy and2

2

r

xy areindependent since consatntx

y

y rr 12

1

2 (

(The Wronskian is 0)( 11212

11

21

21

21

rr

rr

rr

xrrxrxr

xxW (all 0x )).


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2

Case 1: Distinct Real Roots Rrr 21

In this case, the two solutions 11r

xy and 22r

xy are independent (by the Wronskian) and

hence the general solution is 21 21rr

g xcxcy

Example 1: Solve 025.22 yyxyx

Solution: Let rxy . Then we hav 025.32 rr The roots are 21 2/14 rr which are real, hence

2/141 2

xcxcg

y

Case 2: Double Real Roots Rrr 21

This case occurs if the discriminant 04)( 2 acab , hence we have double real roots:

a

abrr

2

)(21

. Hence, the two solutions are dependent ( aabxyy 2/)(21

).

To obtain a basis of solution in this case, we apply the Rduction-of-Order method(Section 2.1) on the DE 0)/()/( 2 yaxcyaxby , where a solution aabxy 2/)(1

is give

Let 12 uyy . Then xdey

udxxp

)(

21

1, where )/()( axbxp .

So, ,ln/)(

/

xdxx

xu

aab

ab

xyy ln12 and the general solution is then1)ln( 21r

g xxccy

In this case, the Wronskian of the solutions is then 0//ln

ln 21

111

11

xyxyxyy

xyy

Example 2: Solve 0432 yyxyx

Solution: Let rxy . The auxiliary equation is 0442 rr . The roots are 21 2 rr , a

real double root. So, 221 )ln( xxccyg .

Case 3: Complex Conjugates Roots Crr 12

This case occurs if the discriminant 04)( 2 acab .

Definea

ab

2

and aabac 2/)(4 2 , then the roots are 0& ,21 irir

Note that the roots 221 rrr are complex conjugates and the two solutions1

1r

xy &

22

r

xy are linearly independent since 02),( 1221 xiyyW or ( tconsxyy i tan/ 221

).

But, they are complex: ixxy 1 and i

xxy2 .

To obtain a basis of real soltuions in this case, note that

)lnsin()lncos(&)lnsin()lncos(lnln xixexxixex xiixii


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22

Then: )lncos()(21 xxx ii and )lnsin()(

2

1xxx ii

i

So, )lncos(1 xxy and )lnsin(2 xxy

are two real and linearly independent

solutions of (1) when the roots are complex conjugates )tan)lncot(/( 21 tconsxyy

Hence, the general soultion is ))lnsin()lncos(( 21 xcxcxyg

Example 3: Solve 01372 yyxyx

Solution: Let rxy . Then we hav 01362 rr The roots are irir 23&23 21 , so, 3 and 2 and

))ln2sin()ln2cos(( 213 xcxcxyg

Example 4: Boundary Value Problem (BVP)Electric Field Between two Concentric Spheres

Find the electrostatic potential )(rvv between two concentric spheres of radiicmrcmr 8&4 21 kept at potentials voltsvvoltv 0&110 21 , respectively.

Solution: Physical InformationThe potential )(rvv satisfies 02 vvr .

This is Cauchy-Euler equation, So, let mrv .

Then we have 02 mm . The roots are .1&0 m

So, the potential is rccrv /)( 21 . Substituting the boundary values:

8/21

0)8(

4/

21

110)4(

ccv

ccv

The solution is then rrv /880110)(


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23

Section 2.8 NonHomogeneous Equations

In this section, we give a general theory for the non homogeneous linear 2nd

-order

differential equation of the form

)()()(01

xfyxayxay

(1)

where )(&)( 10 xaxa are continuous functions on a given open interval I.

Consider also the homogeneous linear 2nd

-order differential equation of the form:

0)()( 01 yxayxay (2)

Theorem 1: Relations Between Solutions of (1) & (2)

a- If 21 & yy are two solutions of (1) on some open interval I, then 21 yy is a solutionof (2) on Ialso.

b- If py is a solution of (1) and hy is a solution of (2) on some open interval I, thenhp yy is a solution of (1) on I.

Proof: Define yxayxayyL )()(][ 01 .

Note that ][yL is a linear operator, i.e., ][][][ 22112211 yLcyLcycycL a- If 21 & yy are solutions of (1), then )(][][ 21 xfyLyL

Hence, 0)()(][][][ 2121 xfxfyLyLyyL , so 21 yy is a solution of (2) on Ialso

b- If py is a solution of (1) on some open interval Iand hy is a solution of (2) on Ialso, then 0][&)(][

hp

yLxfyL . Hence )(0)(][][][ xfxfyLyLyyLhphp

Definitions: General Solution of the nonhomogeneous eqn (1)

If )()( 2211 xycxycyh is a solution of (2) and if )(xy p is a solution of (1), then

)()()( 2211 xyxycxycyyy pphg (3)

is a general solution of (1).

A Particular Solution of the nonhomogeneous eqn (1) is a solution of the form

)()()( 2211 xyxycxyc p where the constants 21 & cc are specified numbers.

General Solution of (1) Includes All Solutions

Theorem 2: Suppose the coefficients )(),( 10 xaxa and the function )(xf of (1) are all

continuous functions on some open interval I. Thena general solution of (1) exists on I

and every solution of (1) is obtained by assigning suitable values to the arbitrary

constants in the genearl solution (3).


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24

How to Solve (1) or IVP for (1)?

We find the homogeneous solution hyof equation (2) and find the particular solution py

of (1).

Example 1: Find the general solution of .454 2xeyyy

Solution:

Step 1: Find the homogeneous solution of 054 yyy

Since the coefficients are constants, then the solution is of the form rxh ey .

Substitutinginto the homogeneous equation above , we get the auxiliary

equation 0542 rr . We obtain irir 2&2 11 and so

)sincos( 212 xcxcey xh

.Step 2: We now find the particular solution py of .454

2xeyyy Since xexf 24)( and since the coefficients of the equation are conatant,

then py and its derivatives should be in the form of .2xAe So, let

xp Aey

2 ,

then we have xxxx eAeeAeA 2222 45)2(4)4( .Solving for A, we obtain 17/4A

Step 3: So, the general solution is then xxg excxcey2

212

)17/4()sincos( .

Example 2: Solve the initial value problem IVP

.1)0(&0)0(,454 2 yyeyyy x

Solution:

Step1: xxh ececy5

21 is the homogeneous solution of 054 yyy

Step 2: Again, as in Example (1), we assume the particular solution as xp Aey

2 .

Then we have, xxxx eAeeAeA 2222 145)2(4)4(

Solving for A, we obtain .2A

Step 3: xxxg eececy25

21 2

Step 4: Find the coefficients 21 &cc such that the solution satisfies the initial values.

With ,0)0( y then 20 21 cc And ,1)0( y then 451 21 cc

Solving these two equations for 21 &cc , we obatin .2

1&

2

521 cc

Hence the unique particular solution to the IVP is xxxp eeey25

22

1

2

5 .


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25

Section 2.9 Undetermined Coefficients Solution MethodIn this section, we solve the nonhomogeneous eqn

)()()( 01 xfyxayxay (1)

by the Undetermined Coefficients Solution Method. This method requires eqn (1) to havetwo properties:

1- The coefficients are constants, i.e., eqn (1) becomes:)(012 xfyayaya (2)

In this case, the homogeneous solution hy is found as in Sections 2.2 & 2.3 by

assuming .rxh ey

2- The function )(xf is a linear combination of bxorandbxbxbxex axr cosh/sinh,cos,sin,,Steps of Solution

Step 1: As in Sec 2.2 & 2.3 find the homogeneous solution )()( 2211 xycxycyh of

0012 yayaya (3)

Step 2: Find the particular solution py of (2) according to the following rules:

(a)Basic Rule:Choose *py

in the same form as )(xf but with undetermined coefficients as in the

following table:

)(xf Assumed *py

nkx 0

11 ... AxAxA

nn

nn

rxke rxAe

bxkcos or bxksin bxBbxA sincos nrxxke rxnn eAxA )...( 0

bxkerx

cos or bxkerx sin rxebxBbxA )sincos(

bxkxn

cos or bxkxn sin bxAxA nn cos)...( 0 bxBxBn

n sin)...( 0

bxekxrxn

cos or bxekx rxn sin bxAxAe nnrx cos)...( 0 bxBxBe

nn

rxsin)...( 0

(b) Modification Rule:


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26

If the assumed *py or any of its derivatives is already a homogeneous solution of (3) then

multiply *py byrxxx ...,,,

2 where ris the least positive integer such that *pryx is not a

homogeneous solution. The particular is then *pr

p yxy .

(c) Sum Rule (Superposition)If )(...)()()( 21 xfxfxfxf m , then for each )(xfk , we assume a particular solution kpy ( modified if needed) and the particular solution will be the superposition ppp yyy ...1

Example 1: Solve 284 xyy

Solution:

Step1: The homogeneous solution of 04 yy is xcxcyh 2sin2cos 21

Step 2: Since 2)( xxf , then the assumed paticular solution is CBxAxy p 2* .

No modification is needed, so *pp

yy .

To find A, B& C, substitute py in the nonhomogeneous eqn we obtain22

8)(42 xCBxAxA and from which we have 042&04,84 22 CABxxAx

So, 1&0,2 CBA and then )12(2sin2cos 221 xxcxcyg

Example 2: Solve xeyyy 2423

Solution:

Step1: xxh ececy2

21

Step 2: Since xexf 24)( , we assume xp

Aey2* which is a homogeneous solution. So

a modification is needed by multiplying by x: xpp Axexyy2* .

To find A, substitute py in the nonhomogeneous eqn we obtainx

p Axey2 xp Axey

222

xp exAy

2)12(

xp exAy

2)36(3

xp exAy

2)44(

xp exAy

2)44(

We add and obtain: 44 22 AeAe xx . So,xxx

g xeececy22

21 4

Example 3: Solve the IVP 1)0(,0)0(,62 yyeyyy x

Solution:

Step1: xxh xececy 21

Step 2: : Since xexf 6)( , we assume xp Aey* . Again, a modification is needed

since xAe is a homogeneous solution. To modify it, multiply by x . Again


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27

xAxe

is a homogeneous solution, so multiply by x again and this timex

pp eAxyxy 22 is not a homogeneous solution.

To find A, substitute py in the nonhomogeneous eqn we obtain

xp eAxy 2 xp eAxy 2

xp exxAy

)2( 2

xp exxAy

)42(2 2

xp exxAy

)24( 2

xp exxAy

)24( 2

We add and obtain: 362 AA , so, xg exxccy

)3( 221

Step 3: Find the coefficients 21 &cc

10)0( cy and 211)0( ccy , sox

g exxy )3( 2

Example 4: Solve the IVP 1)0(,0)0(,2cos452 yyxxeyyy x

Solution:

Step1: )2sin2cos( 21 xcxceyx

h

Step 2: xDCxexBAxey xxp 2sin)(2cos)( .

Note that part of *py is a homogeneous solution, namely

)2sin2cos( xDxBe x . So, modify *py by multiplying it by x to obtain

xDxCxexBxAxexyy xxpp 2sin)(2cos)(22

Step 3: Find the constants A, B, C & D by substituting py in the nonhomogeneouseqn. Then ]2sin)(2cos)[( 2

21

2 xcDxCxxcBxAxey xg


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28

Section 2.15 (a) Undetermined Coefficients Solution Method for Higher Order DEsWe now apply Undetermined Coefficients Solution Method to Higher Order

nonhomogeneous differential equations with the same properties on the coefficients and

f(x).

)(...0

)1(

1

)( xfyayaya n

n

n

n

(4)where )(xf is a linear combination of bxorandbxbxbxex axr cosh/sinh,cos,sin,,

Remark: Let yayayayL nnn

n 0)1(

1)(

...][

So eqn (4) becomes)(][ xfyL (5)

We give example using Superposition Principle:

If )(...)()()( 21 xfxfxfxf m , then for each )(xfk , we assume a particular solution kpy ( modified if needed) and the particular solution will be the superposition

mpppyyy ...1

Example 1: Without evaluating the undetermined coefficients, write the general solution o

xeexyyy xxiv sin27854223

Solution:

Step 1: Soving 054 yyy iv , we obtain )sincos()( 322

10 xcxcexccyx

h Step 2: To find the form of py , we find the particular solution for each )(xfk .

)(xf Assumed: py Modified: py

38x DCxBxAx 23 )( 232 DCxBxAxx

xe2

7 xEe2 xEe2

xexsin2

2 )cossin(2 xGxFe x )cossin(2 xGxFxe x

So, )cossin()( 22232 xGxFxeEeDCxBxAxxy xxp and phg yyy .

Example 2: Suppose the homogeneous solution and )(xfof (2) are given, write the

general sloution in each of the following cases:

(i) xh ecxxccxxccy 254321 2sin)(2cos)( and xx exexxf 22 2sin102cos)( Then: xxp EexDxCexBxAy

22)2sin2cos()2sin2cos(

Modification is needed for the first and third terms of py The modified py is then

xxp ExexDxCexBxAxy

222)2sin2cos()2sin2cos(


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29

(ii) xxh HeeGFxExxDxCBAxy 22

)()sincos()(

and xx eexxxxf 52cos)( 223

Then: xxp JeIHxGxeFExDxCxxBxAy22223

)()()sincos(

Modification is needed for all the terms

The modified py

is thenxx

p JxeIHxGxexFExDxCxxxBxAxy2223232

)()()sincos(


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31

Section 2.10 Solution by Variation of Parameters Methods

In this section, we use the Variation of Parameters Method to solve the

nonhomogeneous DE

)()()(01

xfyxayxay (1)

where )(&)(),( 10 xfxaxa are continuous functions on a given open interval I.

Contrary to the method of Undetermined Coefficients, this method requires no specific

conditions on the coefficients )(),( 10 xaxa nor on the function )(xf other than being

continuous on an open interval I.

How does this method (Variation of Parameters ) work?

Steps of the Solution Method:

Step 1: Find the homogeneous solution )()( 2211 xycxycyh of the homogeneous

equation: 0)()( 01 yxayxay (2)

Step 2: Find the particular solution py as follow:

2211 )()( yxuyxuy p (3)

where 21 &uu are variables (functions of x) to be determined.

The general solution is then phg yyy

Remark(1): This way of finding py resembles the method of finding a 2nd

- order

linearlyindependent solution 12 uyy from a given one 1y .

Computation of 21 &uu :

2211 yuyuyp

)( 22112211 yuyuyuyuyp

In order not to obtain 2nd

- order differential equation in the new variables 21 &uu , we

require0

2211

yuyu (4)Then 22112211 yuyuyuyuyp

Substituting ppp yyy &, in equation (1) as a particular solution and simplifying we obtain:

)()())()((

))()(()()(

2211202122

10111101

xfyuyuyxayxayu

yxayxayuyxayxay ppp

(5)

Since 21& yy are homogeneous solution satisfying (2), then (5) simplifies to


31/40

3

)(2211 xfyuyu

(6)

We now have two equations relating 21 &uu , namely (4) and (6):

02211 yuyu

)(2211 xfyuyu

Using Cramers Rule, we solve for 21 &uu and obtain:

W

Wu

W

Wu 22

11 & , where 1

1

1

22

2

2

1 )()(

0,)(

)(

0yxf

xfy

yWyxf

yxf

yW

and

21

21

yy

yyW The wronkain of 21 & yy .

Integrating both 21 & uu and substituting 21 &uu in (3), we obtain dxuydxuyyp2

211

Remark(2) Note that equation (1) is put in standard form

)()()( 01 xfyxayxay

Example (1) Solve xyy sec

Solution:

Step1: xcxcyh sincos 21 and hence 1cossin

sincos 1

xx

xxW

Step 2: xuxuyp sincos 21 .

Then xxxyxfxx

xW tansinsec)(

cossec

sin021

and 1cossec)(secsin

0cos12

xxyxf

xx

xW

So, xxdxuxu coslntantan 11 and xdxuu 22 1 And the general solution is xxxxxcxcyg sincos)cos(ln)sincos( 21


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32

Section 2.15(b) Solution of Nonhomogeneous Higher Order Differential Equations

using Variation of Parameters Method

In this section, we generalize the Method Variation of Parameters to solve the higher

order nonhomogeneous eqn

)()()(...)( 01)1(

1)(

xfyxayxayxayn

nn (1)

Steps of the Solution Method

Step 1: Find the homogeneous solution )(...)(11 xycxycy nnh of the homogeneous

equation

0)()(...)( 01)1(

1)( yxayxayxay

nn

n

(2)

Step 2: Find the particular solutionpy as follow:

nnp yxuyxuy )(...)( 11

(3)

Where nuuu ...,,, 21 are variables (functions ofx) to be determined.

The general solution is then phg yyy

Computation of nuu ...,,1 Given nnp yuyuy ...11 )...(... 1111 nnnnp yuyuyuyuy

In order not to obtain higher order differential equations in the new variables nuu ...,,1 ,

we require 0...11 nn yuyu (4)

We continue computing )1(...,, npp yy and in each step we set:

1...,,1,0...)1()1(

11

nkyuyuk

nnk (5)

Then compute)(n

py and substitue each of)()1(

&...,,,,n

p

n

pppp yyyyy in equation (1), we get

)(

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

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

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

)1()1(

1101

202122

1011110)(

xf

yuyuyxayxayu

yxayxayu

yxayxayuyxay

n

nn

n

nnnnn

n

nppn

(6)

Using the fact that each of nyyy ...,, 21 is a homogeneous solution satisfying equation (2),

we then obtain )(... )1()1(22)1(

11 xfyuyuyun

nnnn (7)


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33

So, from equations (4), (5) and (7) we obtain the following sytem of linear equations in

nuu ...,,1 :

0...2211 nn yuyuyu 0...2211 nnyuyuyu

.

.

.

)(...)1(

22

)1(

11

)1(

xfyuyuyun

nnn

n

To solve for each of nuu ...,,1 , we apply Cramers Rule:

W

Wu kk , nk ...,,2,1 (8)

Where Wis the Wronskian of )(),...,(1 xyxy n , namley

)1()1(2

)1(

1

21

21

..

...

...

...

...

det

n

n

nn

n

n

yyy

yyy

yyy

W and

...).(..

.0.

..

0

0

det

)1()1(

1

1

1

n

n

n

n

n

k

yxfy

yy

yy

W

Where the column

)(

.

.

.0

0

xf

replaces the kth column of the matrix of the wronskian W.

Integrating each of nuu &...,1 obtained from (8) and substituting nuu ...,,1 in (3), we obtain

dxuydxuyyn

np ...11

Remark(2) Note that equation (1) is put in standard form.


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34

Example (1) Solve xxyyxyxyx ln663 423

Solution: This is a 3rd

- order Cauchy-Euler differential equation as seen in Section 2.6.

Step 1: Solve the homogeneous equation 0663 23 yyxyxyx

We substitut rxy and obtain the auxiliary equation 06116 23 rrr

Its roots are 1, 2 & 3 and the homogeneous solution is then 33221 xcxcxcyh

Step 2: Compute: 32

32

2

620

321 x

x

xx

xxx

W

Step3: Compute 333

21 xuxuxuyp

Write the equation in standard form: )(ln661

332

xfxxyx

yx

yx

y

Then compute: xx

xxx

xx

xx

W ln

62ln

320

0

52

32

1 , xx

xxx

x

xx

W ln2

6ln0

301

0

42

3

2 , and

xx

xx

xx

xxx

W ln

ln20

32132

32

3

So, 312

3

51

118

)1ln3(

2

ln

2

lnx

xu

xx

x

xx

W

Wu

223

42

24

)1ln2(

2

ln

2

lnx

xu

xx

x

xx

W

Wu

And xx

ux

x

xx

W

Wu

2

)1(ln

2

ln

2

ln33

33

3

So the particular solution is then:

36

)11ln6(

2

)1(ln

4

)1ln2(

18

)1ln3(4444

332211

xxxxxxxxyuyuyuy p

And the general solution is )11ln6(36

)(

43

32

21 xx

xcxcxcyg


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35

Example (2) Solve 211

43343 xyyxyx

Solution: Again this is 3rd

-order Cauchy-Euler differential equation

Step 1: Solve the homogeneous equation 0334 3 yyxyx

We substitut rxy and obtain the characterestic equation 0311124 23 rrr

Its roots are 1,2

1 &2

3 and the homogeneous solution is then 21

321 cxcxcyh

Step 2: Compute:4

1

4

3

4

10

2

3

2

11

2

1

2

3

2

1

2

1

2

3

2

1

xx

xx

xxx

W

Step 3: To obtain 23

2

1

321 xuxuxuyp , write the equation in standard form:

)(4

3

4

32

5

32xfxy

xy

xy

Then compute:

27

2

1

2

3

2

5

2

1

2

1

2

3

2

1

1

4

3

4

1

2

3

2

10

0

x

xxx

xx

xx

W

2

7

11 4x

W

Wu

2

9

19

8xu

42

2

1

4

30

2

301

0

2

1

2

5

2

1

2

3

x

xx

x

xx

W

422 2x

W

Wu 52

5

2xu

and

33

2

1

4

30

02

11

0

2

5

2

1

2

1

2

1

x

xx

x

xx

W

333 2xW

Wu

432

1xu


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36

So the particular solution is then: 211

2

3

42

1

52

9

90

1.

2

1.

5

2.

9

8xxxxxxxyp

And the general solution is 211

321

90

12

3

2

1

xxcxcxcyg


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37

Section 2.12 Modeling of Electric Ciruits

In this section we study an RLCseries circuit, where, as in Sectin 1.7, R representsresistance, L represents inductance and Crepresents capacitance.

RLC- Series CircuitFind the current in the given RLC- series circuit for a periodic electromotive force

wtEV sin0

Solution: By Kirchhoffs voltage law 0v , we have:

wtEtVdttI

C

RI

dt

dIL sin)()(

10 or

L

wtwEI

LCdt

dI

L

R

dt

Id cos1 02

2

(1)

The general solution to this equation is: phg III (2)

To find hI , we solve the homogeneous differential equation

01

2

2

ILCdt

dI

L

R

dt

Id(3)

With R, L and Cbeing constants, we use undetermined coefficients method. So assume

rt

h eI and find ras the root of the auxiliary equation 0

12

LCrL

R

r

The roots are:LCL

R

L

Rr

1

22

2

2,1


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38

Define

01

,1

01

,1

&2

22

22

LCif

LC

LCif

LC

L

R

Cases of the roots The homogeneous solution

1- Two distinct reals tth ececI )(2)(1 2- One repeated real: th etccI )( 21 3- Two complex roots )sincos( 21 tctceI th

Recall: From Section 1.7: An electrical system is said to be in steady-state if the

variables describing the behaviour are either periodic or constant. Otherwise, transient

Remark If ,0R then 0 and 0lim

t

te . Hence the homogeneous current tends to

zero and the steady state current tend to the particular current .pI

To find pI , we use undetermined coefficients method.

SinceL

wtwE

dt

tdV

L

cos)(1 0 is a cosine function and assuming jwr 2,1 , we let

wtBwtAIp sincos .

Then wtwAwtwBIp sincos

And wtBwwtAwIp sincos 22

Substituting these into equation (1), we obtain

L

wtwEwt

LC

B

L

RwABwwt

LC

A

L

RwBAw

cossin)(cos)( 0

22

From this we have the two linear equations in the unknowns A and B:

0)1(

)1

(

2

02

BC

LwAwR

wEBwRAC

Lw

Using Cramers Rule, we solve forA & B and obtain


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39

22

0

221

10

2212

10

2

2

2

2

0

)(

)(

))((

)(

1

1

10

RS

SE

RwC

wL

wCwLE

RwC

wLw

wCwLEw

CLwwR

wR

C

Lw

CLw

wRwE

A

WherewC

wLS 1 is called reactance. Similarly,

22

0

2

2

02

1

1

0

1

RS

RE

CLwwR

wR

C

Lw

wR

wEC

Lw

B

With this we have: )sincos(222222

0 wtRS

Rwt

RS

S

RS

EIp

Note that

B

A

R

S and 22

22

0 BA

RS

E

.

To write the solution in amplitude/phase form )sin(0 wtIIp , note that the two terms

2222&

RS

R

RS

S

can be represented as the sine and cosine, respectively of an angle

in a right triangle. Then we have:

2222cos,sin

RS

R

RS

S

and pI becomes

)sin()sincoscossin( 022

0

wtIwtwtRS

EIp

WhereR

S

RS

EI

tan&

22

00

The quantity 22 RS is called the impedence and it is equal to the ratio .0

0

I

E


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ExampleFind the current in an RLC- series circuit with R= 8 ohms, L=2 henrys, C=0.1 farad

and tV 5sin160 .

Solution ,011

&2

2

2

LCL

R Hence the roots are complex conjugate

jr 22,1

and the homogeneous solution is then )sincos( 212

tctceI th

To find the steady state current (the particular current) we copmute:

82101

wC

wLS , 1tan&2106464

160

22

00

R

S

RS

EI

So, )5sin(2104

tIp and the general solution is then

)5sin(210)sincos(421

2 ttctceI t

linear differential equations of second and higher orders

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