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Chapter 3

Second Order Linear Differential Equations

Sec 3.1 : Homogeneous equation

A linear second order differential equations is written as

When d(x) = 0, the equation is called homogeneous, otherwise it is called nonhomogeneous.

,

For the study of these equations sometimes we will consider the standard forms given by

)()(')('' xgyxqyxpy

0)(')('' yxqyxpy

Linear differential operators(ways to write the linear 2nd ODE):

In mathematics a function that ``transforms a function into a different function'' is called an operator.

(1) L -Operator

Let y'' + p(x)y' + q(x)y = g(x)

be a second order linear differential equation. Then we call the operator

L(y) = y'' + p(x)y' + q(x)y

the corresponding linear operator(Differential operator). Thus in this chapter we want to find solutions to

the equation

L(y) = g(x) y(x0) = y0 y'(x0) = y'0

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Proof:

(1)

(2) the proof of (2) is exercise #28 sec 4.2

Note: Any operator that satisfies (1) & (2) is called linear operator. If (1) & (2) fails to hold, the operator is

nonlinear

(2) D- Operator

Recall from calculus that

dx

dyDy ,

2

22

dy

ydyD , and in general

n

nn

dy

ydyD . Using this notations, we can express L as

])[(][ 22 yqpDDqypDyyDyL

for example, if

,3'4''][ yyyyL

then we can write

Linearity of the differential operator L :

Let L(y) = y'' + p(x)y' + q(x)y. if y , 1y and 2y are any twice-differeniable functions on the interval I

and if c is any constant, then

(1) )()()( 2121 yLyLyyL ,

(2) )()( ycLcyL

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])[34(34][ 22yDDyDyyDyL

when p and q are constants we can treat qypDyyD 2 as polynomial , so for the above example

we can write

yDDyDDyDyyDyL )3)(1(])[34(34][ 22

Exercises:

# 3.1.1 Let yyyL ''][2 compute

(a) ][sin2 xL (b) ][2

2 xL (c) ][2r

xL (d) ]2[2

2x

eL

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Sec 3.2: Solution of Homogeneous Linear Second Order Differential

Equations

Theory of Solutions

Here we will investigate solutions to homogeneous differential equations. Consider the homogeneous linear

differential equation

We have the following results:

(1)

The above theorem makes clear the importance of being able to

decide whether or not a set of solutions )(1 xy and )(2 xy of

Are linearly independent.

Methods to check if two functions are Linearly independent:

(1) Linear Dependence and Indepedence using Definition :

Definition:(Linear dependence)

Two functions 22 yandy are said to be linearly independent on the interval I, if

there exist nonzero constants 22 candc such that for all x in I

02211 ycyc

General Solution for a Second-Order Homogeneous Linear Equation: Theorem: The 2nd order homogeneous equation

0)(')('')( yxcyxbyxa (1)

always has exactly two linearly independent solutions )(1 xy and )(2 xy . The general

solution to equation (1) is given by the formula

2211)( ycycxy

where the functions 21, yy is the pair of linearly independent solutions to

equation (1).

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f 22 yandy are linearly

independent then 021 cc is the only solution for

02211 ycyc

Another way of stating this definition is ,

ratio test as follows:

Two functions 22 yandy are linearly dependent in I if cy

y

2

1

Two functions 22 yandy are linearly independent in I if cy

y

2

1

Remark: Proportionality of two functions is equivalent to their linear dependence.

(2) Linear Dependence and Independence using Wronskian:

It is not easy to use the above definition for higher order, Instead we can use the wronskian.

The Wronskian : for two functions

Let and be two differentiable functions. and are linearly dependent(proportional) if and only if

cy

y

2

1 . an equivalent criteria for linearly dependent can be derived from the followingt:

and are linearly dependent

So cy

y

2

1

Differentiate both sides 02

1

y

y

.

is called the Wronskian of of and , that is

Wronskian of and is

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07

Therefore, we have the following:

Now we have the following conclusion:

Let and be two differentiable functions. The Wronskian , associated to and , is the

function

We have the following important properties:

(1) If and are two solutions of the equation y'' + p(x)y' + q(x)y = 0, then

In this case, we say that and are linearly independent.

(2) If and are two linearly independent solutions of the equation y'' + p(x)y' + q(x)y = 0, then

any solution y is given by

for some constant and . In this case, the set is called the fundamental set of solutions.

Exercises:

# 3.1.3 Determine whether the given solutions of the following DE are linearly independent

(a) xyxyyy 3cos3sin09'' 21

(b) xx

eyeyyy 20'' 21

(c) xx

eyeyyyy2

21 302''

There is a fascinating relationship between second order linear differential equations and the

Wronskian. This relationship is stated below.

Two functions 1y and 2y are linearly dependent if 0),( 21 yyW

0),( 21 yyWif dependentinlinearly are 2yand 1yTwo functions

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Theorem: Abel's Theorem

Let y1 and y2 be solutions on the differential equation

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

where p and q are continuous on [a,b]. Then the Wronskian is given by

dxxp

CexyyW)(

21 ))(,(

where C is a constant depending on only y1 and y2, but not on x. The Wronskian is

either zero for all x in [a,b] or always positive, or always negative.

Proof :

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Methods for solving second order Linear

Homogeneous DE:

Section 3.3: Method (1) )(For Constant coefficients)

Characteristic equation (Auxiliary equation)

A second order homogeneous equation with constant coefficients is written as

where a, b and c are constant. This type of equation is very useful in many applied problems (physics,

electrical engineering, etc..). Let us summarize the steps to follow in order to find the general solution:

if 0a , we get the first order equation of the same family

0' cyby

and this is first order seperable differential equation which has a solution of the form

bcAeyx

/,

In the same way we can think that , x

ey will be a solution of the second order equation

so, by substituting x

ey ,'

xey ,

2 xey in the left-hand side of differential equation,

we get

02 xxx

ceebea

now, dividing both sides by ,x

ey , we obtain

the characteristic equation (Auxiliary equation)

02 cba

This is a quadratic equation. Let 1 and 2 be its roots we have

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a

acbb

2

4,

2

21

The actual form of the solution is strongly dependent on whether the roots are real versus

complex or distinct versus repeated. In fact three special cases can be identified based on

whether the term inside the radical is positive, negative, or zero. There are three cases as follows

Three Special Cases

Case I. Real Distinct Roots:

If 1 and 2 are distinct real numbers (this happens if ), then the general solution is

xxececy 21

21

Exercises 3.3.2 Find the general solution of 04'3'' yyy

Case II. Repeated Roots:

If 21 (which happens if ), then the general solution is

xxxececy 11

21

Exercises 3.3.4 Find the general solution of 0'4''4 yyy

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Case III. Complex Conjugate Roots: (section 2.3)

If 1 and 2 are complex numbers (which happens if ),

Then a

baci

a

b

2

4

2,

2

21

And the general solution is

where

Example: Find the general solution 2y''+2y'+3y=0

Exercises 3.3.12 Solve 016'' yy y(0)=1 y’(0)=0

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02

Justification of Case 3:

xixi ececy )(

2

)(

1

using Euler identity: xixeix sincos

the solution can be written as

)]sin(cos)sin(cos[ 21 xixcxixcey x

]sin)(cos)[( 2121 xiccxcce x

)sincos( 21 xCxCe x where 211 ccC and iccC )( 212

Proof of Euler identity:

From Maclaurin series expansion

.....!4

)(

!3

)(

!2

)(1

432

ixixix

ixeix

.....!5!4!3!2

15432

ixxixxix

....!5!3

sin53

xx

xx

...!6!4!2

1cos642

xxxx

From the above expansions , we conclude the following

xixeix sincos

which is Euler identit

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Method (2): Euler Equations

An Euler-Cauchy equation is of the form

(EC) 0'''2 cyxyyax x>0

where a, b and c are constant numbers. Let us consider the change of variable

x = et.

Then we have

xdt

dy

dx

dt

dt

dy

dx

dy 1 and

222

2

22

2

2

2 1111

xdt

dy

xdt

yd

xdt

dy

xdx

dt

dt

yd

dx

yd

)(1

2

2

2 dt

dy

dt

yd

x

Substitute in (EC)

The equation (EC) reduces to the new equation

0)(2

2

cydt

dyab

dt

yda

We recognize a second order differential equation with constant coefficients. Therefore, we use the previous

sections to solve it. We summarize below all the cases:

(1)

Write down the characteristic equation

0)(2 caba

(2)

If the roots r1 and r2 are distinct real numbers, then the general solution of (EC) is given by

21

21)(

xcxcxy

(3)

If the roots 1 and 2 are equal ( 21 ), then the general solution of (EC) is

xxcxcxy ln)( 11

21

02

02

(4)

If the roots 1 and 2 are complex numbers, then the general solution of (EC) is

xxcxxcy lnsinlncos 21

Short cut for using Euler method:

We assume a solution of the form

xy

Therefore,

1'

xy and 2

)1('' xy

Upon substitution into the original ODE, we get

0))1(( xcba

and dividing by

x gives the characteristic equation,

0)(2 caba .

Which is similar to equation in step (1) above, so we can continue finding the general solution as

in steps 2,3,4.

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Example: Find the general solution to 0)(6)('2)(''2 xyxxyxyx x>0

Solution: First we recognize that the equation is an Euler equation, with a=1, b=2 and c=-6.

Exercise # 3.3.14 solve 04'''2 yxyyx x>0

Example: Find the general solution of

0)(14)(')1(10)('')1(2 xyxyxxyx x>-1, Ans:

72

21 )1()1( xcxc

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Method(3): For variable coefficients

integration using one known solution(Reduction of order method)

____________________________________________________________________________________

This technique is very important since it helps one to find a second solution independent from a known one.

Therefore, according to the previous section, in order to find the general solution to y'' + p(x)y' + q(x)y = 0,

we need only to find one (non-zero) solution, .

Let be a non-zero solution of

Then, a second solution independent of can be found as

,

Development of the Method

Given a valid solution to a homogeneous linear 2nd order differential equation, one can obtain a

second linearly independent solution by letting

)()()( 12 xyxuxy

where y1(x) is known and u(x) is to be determined. This technique applied to a 2nd order system

is as follows:

Given the original differential equation

with known solution y1(x), we let

dxxy

exyxy

dxxp

2

1

)(

12)]([

)()(

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or

Now, substitution into the original equation gives

Or

Letting z = u' and noting that the third term vanishes from definition of the original ODE, we

have

Now, separating variables and integrating gives

Or

Finally, since du/dx = z, we have

Example 22 )(;0,06'6'' xxfxyxyyx .Ans:

3 xy

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Sec 3.4 Nonhomogeneous Second Order Linear

Equations

So far we have considered equation of the form

)(''' xgcybyay for the case where 0)( xg ,

When 0)( xg , then 02 cbrar giving 1rr and 2rr and the solution is in general

xrxr

h ececy 21

21 .

In the equation )(''' xgcybyay , the substitution xrxr

h ececy 21

21 would make the left-

hand side zero. Therefore , there must be a further term in the solution which will make the L.H.S equal to

)(xg and not zero.

The complete solution will therefore be of the form

p

xxyececy 21

21

where

py is the extra function yet to be found.

xx

h ececy 21

21

is called the complementary solution or homogeneous solution.

)(xyyp

is called the particular solution.

And General solution = homogeneous solution + particular solution

In the previous sections we discussed how to find . In this section we will discuss two techniques to find

:

(1) Method of Principle of superposition:

(2) Method of variation of parameters

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Method (1) : Principle of superposition:

Let y be a solution of the differential equation

)(''' 1 xfcybyay

and let y

be a solution of

)(''' 2 xfcybyay

Then for any constants

21,cc , the function ycyc

21

is a solution of the differential equation

)()(''' 2211 xfcxfccybyay

Example:

Given that tty cos)(1 is a solution to tyyy sin'''

And 3/)( 2

2

tety is a solution to

teyyy

2'''

Use the superposition principle to find solution to the following differential equations

(a) tyyy sin5'''

(b) t

etyyy23sin'''

(c) t

etyyy218sin4'''

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Method (2) : Variation of Parameters

This method has no prior conditions to be satisfied. Therefore, it may sound more general than the previous

method. We will see that this method depends on integration while the previous one is purely algebraic

which, for some at least, is an advantage. It is easily extended to higher order linear ODEs.

Consider the equation )()(')('' xgyxqyxpy

In order to use the method of variation of parameters we need to know that 21, yy is a set of

fundamental solutions of the associated homogeneous equation y'' + p(x)y' + q(x)y = 0. We know that, in this

case, the general solution of the associated homogeneous equation is . The idea behind

the method of variation of parameters is to look for a particular solution such as

where and are functions. From this, the method got its name.

The functions and can be calculated as follows,

Let’s start with the general 2nd order equation written in standard form,

)(')('' xgyyxpy

with homogeneous solution

2211)( ycycxyh

Now assume yp of the form

2211 )()()( yxuyxuxyp (1)

which gives,

At this point we recognize that an addition constraint, beyond the original balance equation (1),

will be needed because we are trying to find two unknowns, u1 and u2. For simplicity, let’s

choose the following relationship,

0'' 2211 yuyu (2)

which greatly simplifies the above equation for yp’, giving

2211 ''' yuyuy p

22221111 ''''' yuyuyuyuy p

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Now the second derivative becomes

22221111 '''''''''' yuyuyuyuy p

Substitution of these expressions into the original differential equation gives

)('''''''''' 2211221122221111 xgyquyquypuypuyuyuyuyu

Rearranging gives

)('''')'''()'''( 221122221111 xgyuyuqypyyuqypyyu

Therefore, since both the first and second terms vanish (because y1 and y2 are solutions to the

homogeneous equation), we have

)('''' 2211 xgyuyu (3)

The functions and are solutions to the system of equations (2) and (3)

which can be written in matrix form (two equations with two unknowns) gives

gu

u

yy

yy 0

'

'

'' 2

1

21

21

Using Cramer’s rule the solution to this system can be written as

gW

y

y

yy

yy

yg

y

u2

2

21

21

2

2

1

1

0

''

'

0

' and gW

y

y

yy

yy

gy

y

u1'

0

''

'

0

'1

1

21

21

1

1

1

Finally, integrating these latter expressions to give u1 and u2 allows one to write the desired

expression for yp(x) as

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This result is consistent with the general relationships given above.

Exercise 3.4.6 find the general solution of xyy sec'' .

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Exercise # 3.4.10 Solve using variation of parameters

12

3'5'' xyxyyx