Ordinary Differential Equations 1. Introduction …………………………………………… 1 2. Ordinary Differential Equations – The Basics ……… 1 3. 1st Order, Separable …………………………………… 3 4. Linearity ………………………………………………. 6 5. 2nd Order, Linear ODEs with Constant Coefficients ….. 8 a) Homogeneous Equations ……………………. 9 b) Non-Homogeneous Equations ………………. 19 Tutorial Exercises ……………………………………… 25 Answers to Tutorial Exercises ………………................ 27

SCHOOL OF ENGINEERING & BUILT ENVIRONMENT

Mathematics

Dr Derek Hodson

1

ODES_Rev/D Hodson Ordinary Differential Equations 1) Introduction A differential equation is an equation that contains derivatives of a function. For example

12 −= xdxdy [1]

0=− ydxdy [2]

02

2

=++ xcdtdxb

dtxda [3]

are all differential equations. Technically they are ordinary differential equations (ODEs) since they contain ordinary derivatives as opposed to partial derivatives. An equation that contains partial derivatives is called a partial differential equation (PDE). The equation

2

22

2

2

xuc

tu

∂∂=

∂∂ [4]

is an example of a partial differential equation. In this module we will only be dealing with ordinary differential equations. Differential equations are important as they can describe mathematically the behaviour of physical processes such as population dynamics, oscillatory phenomena in electrical and mechanical systems, heat transfer within materials, as well as many others. 2) Ordinary Differential Equations – The Basics

Given an ODE in terms of dxdy , where y is called the dependent variable and x is the

independent variable, our aim is to determine y in terms of x , without any derivatives. Consider the simple ODE

12 −= xdxdy . [5]

The solution we desire will be of the form )( xfy = . In this case, the task is simple. We seek the function that, when differentiated, gives us 12 −x .

2

Hence

∫ −= dxxy )1( 2

and so

Cxxy +−= 3

31 . [6]

Notice that an arbitrary constant has been generated. This means that expression [6] does not represent just a single solution of ODE [5], but infinitely many (one for each possible value of C). Expression [6] is called the GENERAL SOLUTION of [5] since it covers all possible solutions of the given ODE. Solutions with arbitrary constants are characteristic of ODEs. The ORDER of an ODE is given by the highest derivative appearing in it. For example

12 −= xdxdy

is a 1st order ODE,

02

2

=++ xcdtdxb

dtxda

is 2nd order, and so on. The order of an ODE determines the number of arbitrary constants appearing in its general solution: 1st order - 1 arbitrary constant 2nd order - 2 arbitrary constants. We will not be considering ODEs of order higher that 2. Once the general solution of an ODE has been found, a PARTICULAR SOLUTION may be determined by applying additional constraints called boundary conditions or initial/final conditions. For the simple 1st order ODE above we might be asked to solve

12 −= xdxdy

subject to 1)0( =y . That is, from the infinity of solutions of the ODE we are looking for the one that will give 1=y when 0=x . The general solution is

Cxxy +−= 3

31 . [7]

3

To determine the particular solution that satisfies the given condition, we determine a value for C by substituting 1=y and 0=x into the general solution and solving the resulting equation : C+−= 001 1=C , giving

131 3 +−= xxy , [8]

which satisfies both the ODE and the condition. In general, we require one condition per arbitrary constant to determine a particular solution. So the number of conditions required equals the order of the ODE. In the simple case above we were able to determine the general solution of the ODE by a single, direct integration. Other ODEs may require different methods. We shall now consider some special types of ODEs. 3) 1st Order, Separable A separable ODE (in terms of y and x ) is one where we can algebraically move everything

involving x , including the bottom bit of dxdy , to one side of the equation, and everything

involving y to the other side, next to the top bit of dxdy . That is, the ODE can be rewritten as

dxxgdyyf )()( = . Integrating both sides we obtain

∫∫ = dxxgdyyf )()( .

Providing we can perform the two integrations, we can obtain the functional relationship between y and x . Examples / Over the page . . .

4

Examples In the following examples, Examples (1) – (5) illustrate the basic processes in solving separable ODEs and tidying up the solution. Example (6) illustrates how a boundary or initial condition can be applied as soon as the constant of integration appears. Example (7) illustrates that some solutions cannot be manipulated into the explicit form )( xfy = , and so we might have to be content with an implicit form of solution.

(1) ODE 12 += xdxdy

Separate dxxdy )1( 2 +=

Form integrals ∫∫ += dxxdy )1(1 2

Integrate Cxxy ++= 3

31

[ Note that we only need one constant of integration. ]

(2) ODE yx

dxdy =

Separate dxxdyy =

Form integrals ∫∫ = dxxdyy

Integrate Cxy ~2

212

21 +=

Tidy up )~2(,22 CCCxy =+= Cxy +±= 2

5

(3) ODE )31( 2xydxdy −=

Separate dxxy

dy )31( 2−=

Form integrals ∫∫ −= dxxy

dy )31( 2

Integrate Cxxy +−= 3ln Tidy up Cxxey +−=

3

Cxx eey

3−= Cxx eAeAy == − ,

3

(4) ODE 0=− ydxdy

Re-arrange ydxdy =

Separate dxy

dy =

Form integrals ∫∫ = dxy

dy 1

Integrate Cxy +=ln Tidy up Cxey += Cx eey = Cx eAeAy == ,

6

(5) ODE + condition 1)2(,)31( 2 =−= yxydxdy

From Example (3): Cxxy +−= 3ln . Apply condition: 1=y when 2=x C+−= 3221ln C+−= 60 6=C . Substitute value: 6ln 3 +−= xxy . Tidy up: 63 +−= xxey . 4) Linearity In the Section (5) we will be moving on to special types of 2nd order ODEs. The following definitions will prove useful for the understanding of parts of that section. Definition (1): An ODE is said to be non-linear if it contains powers and/or products of the

dependent variable or its derivatives, and linear if otherwise. For example

)()( xqyxpdxdy =+

is linear, whereas

13 =+ ydxdy

is non-linear. The remaining types of ODEs that we shall consider are all linear and we shall exploit the special properties of linear equations in determining their solutions. Also, we shall switch to variables x and t since many applications of these type of ODEs are time dependent. Definition (2): / Over the page . . .

7

Definition (2): If we have two functions, )(1 tx and )(2 tx , then a linear combination of these functions takes the form )()()( 2211 txatxatx += where 1a and 2a are constants. A non-linear combination would contain

powers and/or products of the yi's. Definition (3): Two functions, )(1 tx and )(2 tx are said to be linearly independent if one

function is not a constant multiple of the other, i.e. )()( 21 txktx ≠ . 5) 1st Order, Linear ODEs For the moment, we shall restrict ourselves to the basic form:

0=+ xkdtdx

where k is a constant. We can solve this equation using separation of variables:

.

.

ln

)(

tk

Ctk

Ctk

eAx

eex

ex

Ctkx

dtkx

dx

dtkx

dx

xkdtdx

−

−

+−

=

=

=

+−=

−=

−=

−=

∫∫

Notice the exponential function appearing in the final form of the solution and how the k in the original ODE appears in that exponential. Exponential solutions also crop up in 2nd order, linear ODEs.

8

6) 2nd Order, Linear ODEs with Constant Coefficients These type of ODEs come in two forms:

02

2

=++ xcdtdxb

dtxda (Homogeneous) [9]

)(2

2

tfxcdtdxb

dtxda =++ (Non-homogeneous) [10]

where a , b and c are constants. Equations of the above form are extremely useful for modelling certain types of physical systems. Consider a simple suspension system linking a mass to some fixture via a spring and dashpot:

O

Equilibrium state Disturbed state

x

m

m

kk

cc

In this system: mass=m stiffness spring=k tcoefficien damping=c mass ofnt displaceme=x

mass ofvelocity =dtdx mass ofon accelerati2

2

=dt

xd

xk−=spring todue masson Force

dtdxc−=dashpot todue masson Force .

9

Newton's 2nd law of motion states that Mass × Acceleration = Force, so

dtdxcxk

dtxdm −−=2

2

or

02

2

=++ xkdtdxc

dtxdm . (c.f. equation [9])

If there are other forces acting on the mass (such as a forced vibration) then we obtain

)(2

2

tfxkdtdxc

dtxdm =++ (c.f. equation [10])

where )( tf is called the forcing term. Similar ODEs can be derived that model the behaviour of current or voltage in electronic circuits. These type of equations are applicable to many areas where oscillations or vibrations are present. This will become apparent when we look at the solutions to the ODEs. a) Homogeneous Equations

02

2

=++ xcdtdxb

dtxda [11]

Rather than solving equations of this type by a step-by-step integration process, as we did for 1st order equations, we are going to construct solutions based on inherent properties of linear, homogeneous equations and certain mathematical functions. Consider a function x of the form tex λ= [12] where λ is unspecified. If we differentiate this function twice we obtain

tedtdx λλ=

and

tedt

xd λλ22

2

= .

10

Notice that x and its derivatives have a common factor of te λ . If we substitute these expressions into ODE [11] we obtain 0)( 2 =++ tecba λλλ or 02 =++ cba λλ , [13] since 0≠te λ . If we choose λ to satisfy this quadratic equation then, automatically, the function [12] will be a particular solution of the ODE. Equation [13] is called the auxiliary equation or characteristic polynomial of the ODE. Solving it allows us to construct particular solutions and, hence, the general solution of the ODE. Since we have to solve a quadratic equation, we have to deal separately with the different types of solution that can occur. Recall that

a

cabb2

42 −±−=λ .

Case (i) 042 >− cab When we have 042 >− cab we will end up with two values of λ ; call these 1λλ = and 2λλ = . So tex 1λ= will be one particular solution of the ODE and tex 2λ= will be another. These solutions are linearly independent. Because of the homogeneity of the ODE (i.e. zero RHS) and its linearity, any linear combination of the two particular solutions will also be a solution of the ODE. That is, any composite function of the form tt eBeAx 21 λλ += [14] will be a solution. In fact expression [14] covers all possible solutions of the ODE (when

042 >− cab ) and is therefore the general solution of the ODE. Note the two arbitrary constants. We have constructed the general solution without the need to integrate.

11

Case (ii) 042 <− cab In this case we will end up with two complex roots of the auxiliary equation: βαλλ i+== 1 and βαλλ i−== 2 . Just as in case (i) the general solution can be written down as tt eBeAx 21

~~ λλ += or tjtj eBeAx )()( ~~ βαβα −+ += or tjttjt eeBeeAx βαβα −+= ~~ or [ ]tjtjt eBeAex ββα −+= ~~ . Using a result from complex number theory we can write )(sin)(cos tjte tj βββ += and )(sin)(cos tjte tj βββ −=− . Therefore the general solution of the ODE can be rewritten as [ ])(sin)(cos tBtAex t ββα += [15] where )~~( BAA += and )~~( BAjB −= . In general, the constants A and B can be complex, but we shall only deal with real values. Note: Real numbers are a subset of complex numbers. Equation [15] provides us with a template solution when the auxiliary equation yields complex roots.

12

Case (iii) 042 =− cab For this third case we will only have one root to play around with: 0λλ = . This doesn't seem enough to give us a general solution with two arbitrary constants. However in this special case it can be shown that tex 0λ= [16] and tetx 0λ= [17] are both particular solutions of the ODE. They are also linearly independent, so taking a linear combination of [16] and [17] gives us the general solution tt etBeAx 00 λλ += [18] or tetBAx 0)( λ+= . [19] Summary To solve the ODE

02

2

=++ xcdtdxb

dtxda :

I) form the auxiliary equation 02 =++ cba λλ ; II) solve the auxiliary equation; III) write down the general solution of the ODE, depending on the nature of the solution of the

quadratic :– Two real roots: tt eBeAx 21 λλ += Two complex roots: [ ])(sin)(cos tBtAex t ββα += One real root: tetBAx 0)( λ+= .

13

Examples

(6) ODE 022

2

=−− xdtdx

dtxd

Auxiliary equation: 022 =−− λλ 0)1()2( =+− λλ 1,2 −== λλ Two real and distinct roots: tt eBeAx −+= 2

(7) ODE 02

2

=++ xdtdx

dtxd

Auxiliary equation: 012 =++ λλ

23

21

231

j±−=

−±−=λ

Complex roots: 2

321 , =−= βα

( ) ( ) ]sincos[ 2

3232/1 tBtAex t += −

14

(8) ODE 022

2

=++ xdtdx

dtxd

Auxiliary equation: 0122 =++ λλ 0)1( 2 =+λ 1−=λ One root only: tetBAx −+= )( Since the general solution of a 2nd order ODE has two arbitrary constants, we shall require two boundary or initial conditions to evaluate a particular solution. These conditions may be given in the form 0)( cax = 1)( cbx = , where they are both applied to the general solution to give (in general) two simultaneous equations for the constants A and B . Alternatively, they may be given in the form

0)( cax = 1)( cadtdx = ,

where the 2nd condition is applied to the derivative of the general solution. Again, two simultaneous equations for A and B will (in general) result. Example

(9) Solve 0652

2

=++ x dt

xd dt

xd subject to 2)0( =x , 1)0( =dtdx .

(a) Auxiliary equation: 0652 =++ λλ 0)2)(3( =++ λλ 3,2 −=−= λλ General solution: tt Be Aex 32 −− += Apply 2)0( =x : 03022 ×−×− += eBeA 2 B A =+

15

Before we can apply the second condition, we must differentiate the general solution: tt Be Aex 32 −− +=

tt Be Aedtdx 32 32 −− −−= .

Now apply 1)0( =dtdx : 0302 321 ×−×− −−= Be Ae

132 BA =−− The two underlined equations solve simultaneously to give 7=A and 5−=B . Substitute these values into the general solution to give the particular solution tt e ex 32 57 −− −= (10) Consider the spring-mass-dashpot system at the start of the section. Set 1=m , 1=c and

1=k . Determine the motion of the mass if it is initially displaced by an amount h and then released from rest.

ODE 02

2

=++ xdtdx

dtxd

Initial displacement hx =)0(

Initial velocity 0)0( =dtdx

The ODE is just that from Example (7). So

( ) ( ) ]sincos[ 23

232

1

tBtAext

+=−

. Apply first condition, hx =)0( :

( ) ( ) ]0sin0cos[ 23

230

21

BAeh +=×−

Ah = hA = .

16

Substitute:

( ) ( ) ]sincos[ 23

232

1

tBthext

+=−

. Before we can apply the second condition, we must determine the derivative of x. Applying

the Product Rule gives

( ) ( )

( ) ( ) .]cossin[

]sincos[

23

23

23

232

1

23

232

1

21

tBthe

tBthedtdx

t

t

+−+

+−=

−

−

Now apply the second condition, 0)0( =dtdx :

( ) ( )

( ) ( ) ]0cos0sin[

]0sin0cos[0

23

23

23

230

21

23

230

21

21

Bhe

Bhe

+−+

+−=×−

×−

Bh 2

3210 +−=

3

hB =

Substitute this to give the final form for the displacement x:

( ) ( ) ]sin3

cos[ 23

232

1

ththext

+=−

17

(11) Solve 042

2

=+ xdt

xd subject to

(a) 1)0( =x 2)( 8 −=πx

(b) 1)0( =x 4)0( =dtdx .

(a) Auxiliary equation: 042 =+λ 42 −=λ j20 ±=λ 0=α , 2=β General solution: [ ])(sin)(cos tBtAex t ββα += [ ])2(sin)2(cos0 tBtAex t += × )2(sin)2(cos tBtAx += Apply 1)0( =x : )2(sin)2(cos tBtAx += )02(sin)02(cos1 ×+×= BA A=1 1=A Apply 2)( 8 −=πx : )2(sin)2(cos tBtAx += )(sin)(cos2 44

ππ BA +=−

2

12

12 BA +=−

2−=+ BA The two underlined equations solve simultaneously to give 1=A and 3−=B . Substitute these value into the general solution to give the particular solution )2(sin3)2(cos ttx −=

18

(b) Everything starts exactly as in part (a): General solution: )2(sin)2(cos tBtAx += Apply 1)0( =x : )2(sin)2(cos tBtAx += ( same as before) )02(sin)02(cos1 ×+×= BA A=1 1=A Before we can apply the second condition, we must differentiate the general solution: )2(sin)2(cos tBtAx +=

)2(cos2)2(sin2 tBtAdtdx +−= .

Now apply 4)0( =dtdx : )2(cos2)2(sin2 tBtA

dtdx +−=

)02(cos2)02(sin24 ×+×−= BA B204 += 2=B Substituting the two underlined values into the general solution gives the particular solution )2(sin2)2(cos ttx +=

19

b) Non-Homogeneous Equations

)(2

2

xfycdxdyb

dxyda =++ [24]

The general solution of a non-homogeneous ODE of the form [24] is found as follows: STEP I Determine the general solution of the corresponding homogeneous ODE (i.e.

initially set the RHS to zero). In this context, we call this the complementary function and denote it by )( xyc .

STEP II Determine one particular solution of the full, non-homogeneous ODE by the

method of undetermined coefficients (see below). We denote this by )( xy p . STEP III The general solution of the full, non-homogeneous equation is then given by )()( xyxyy pc += . The Method of Undetermined Coefficients This method is best illustrated by example. Consider the ODE

42 22

2

+=−− xydxdy

dxyd .

First we determine the complementary function: Aux. Eq.: 022 =−− λλ 0)1()2( =+− λλ 12 −== λλ . Two real roots, so xx

c eBeAxy −+= 2)( .

20

Next, the aim is to determine one particular solution of the full, non-homogeneous equation. Its format depends on the type of function on the RHS of the ODE. Looking at the RHS above we see (in this case) a quadratic. We look for a particular solution of a similar form, namely a quadratic. We write down the most general form that this can take: cxbxay p ++= 2 , [26] where a , b and c are the, as yet, undetermined coefficients. Our task is now to determine the values of a , b and c that make [26] a solution of ODE [25]. We do this by substituting the general form [26] into the LHS of ODE [25] and force this new LHS to equal the given RHS. To do this we form the 1st and 2nd derivatives of [26]:

bxadx

dy p += 2 [27]

adx

yd p 22

2

= . [28]

Substituting [26], [27] and [28] into the LHS of [25] we get 4][2]2[]2[ 22 +=++−+− xcxbxabxaa . Collecting together the powers of x on the LHS gives us 42)22()22( 22 +=−+−−− xxaxbacba . [29] If [26] is to be a solution of [25] then we must determine a , b and c so that the LHS of [29] equals the RHS. Equating the coefficients of the powers of x on both sides yields three equations in a , b and c : coefficients of x2 : 12 =− a coefficients of x : 0)22( =+− ba constant term : 422 =−− cba . Solving these equations gives 4

1121

21 −==−= cba .

Our particular solution is therefore 4

11212

21)( −+−= xxxy p .

21

We can now write down the general solution of ODE [25] by adding together the complementary function and the particular solution: 4

11212

212)( −+−+= − xxeBeAxy xx .

For other forms of RHS functions )( xf we look for functions which are similar in form. There are one or two exceptions so the table on the next page gives the starting choice for )( xy p corresponding to various RHSs. If )( xf is a sum of different terms then we add together corresponding forms of yp's , using different undetermined coefficients. Examples See Examples (12) – (13) after the table of yp starting forms.

22

Table of Starting Forms for Particular Solutions of Non-homogeneous ODEs

RHS function )( xf Starting form for )( xy p

Any constant term ( e.g. 4 , −3 )

a

Any linear term ( e.g. 2 x , 5 x − 8 )

a x + b

Any quadratic term ( e.g. 3 x2 + 2 x − 6 )

a x2 + b x + c

RHS contains an exponential of the form xke

(where k is not a root of the AE)

xkea

RHS contains an exponential of the form xke (where k is one of two distinct roots of the

AE)

xkexa

RHS contains an exponential of the form xke

(where k is a repeated root of the AE)

xkexa 2

RHS contains )(cos xk or )(sin xk or both

)(sin)(cos xkbxka +

or ])(sin)(cos[ xkbxkax + ∗

∗ This only occurs with ODEs of the form

bothor )(sinor)(cos containing RHS22

2

xkxkykdx

yd =+ .

23

Examples

(12) ODE 4232

2

+=++ xydxdy

dxyd

Aux. Equation 0232 =++ λλ 0)2()1( =++ λλ 2,1 −=−= λλ Comp. Function xx

c eBeAy 2−− += Since the RHS of the ODE is a linear function we select a particular solution of the

form bxay p += Differentiate: ay p =′ 0=′′py Substitute into ODE: 4)(230 +=+++ xbxaa Collect like terms: 4)23(2 +=++ xbaxa Equate coefficients: x terms: 12 =a constant terms: 423 =+ ba Solve simultaneous equations and substitute into py : 2

1=a 4

5=b 4

521 += xy p

Combine cy and py to give the general solution of the full ODE: 4

5212 +++= −− xeBeAy xx

24

(13) ODE )2(sin2

2

xydx

yd =+

Aux. Equation 012 =+λ 12 −=λ

j

j±=

±=0

λ

Comp. Function xBxAyc sincos += Since the RHS of the ODE is a trig function whose coefficient is 2 , we select a

particular solution of the form )2(sin)2(cos xbxay p += Differentiate: )2(cos2)2(sin2 xbxay p +−=′ )2(sin4)2(cos4 xbxay p −−=′′ Substitute into ODE: )2(sin])2(sin)2(cos[])2(sin4)2(cos4[ xxbxaxbxa =++−− Collect like terms: )2(sin)2(sin3)2(cos3 xxbxa =−− Equate coefficients: cosine terms: 03 =− a → 0=a sine terms: 13 =− b → 3

1−=b Substitute into py : )2(sin3

1 xy p −= Combine cy and py to give the general solution of the full ODE: )2(sinsincos 3

1 xxBxAy −+=

25

Tutorial Exercises 1st Order ODEs – Separable (1) Determine the general solution of each of the following ODEs:

(i) 02 =− ydxdy (ii) 04 2 =+ y

dxdy

(iii) 42 =− ydxdy (iv) 0=+ y

dxdyx

(v) 02 =+ yedxdy x (vi) 12 1 −= − yx

dxdy

(2) Determine the particular solution of each of the following ODEs that also satisfies the given

boundary or initial condition:

(i) ydxdyx 2)1( =+ , 1)0( =y

(ii) 022 =+ yedxdy x , 1)0( −=y

(iii) xdtdxt 32 = , 4)1( =x

2nd Order ODEs – Linear, Constant Coefficients (Homogeneous) (3) Determine the general solution of each of the following ODEs:

(i) 062

2

=−− ydxdy

dxyd (ii) 0422

2

=−+ ydxdy

dxyd

(iii) 0962

2

=+− ydxdy

dxyd (iv) 022

2

=++ ydxdy

dxyd

(v) 0422

2

=+− ydxdy

dxyd (vi) 0222

2

=++ ydxdy

dxyd

(vii) 092

2

=− ydx

yd (viii) 092

2

=+ ydx

yd

26

(4) Determine the particular solution of each of the following ODEs that also satisfies the given boundary or initial conditions:

(i) 0822

2

=−− ydxdy

dxyd , 6)0(0)0( =′= yy

(ii) 01362

2

=++ ydxdy

dxyd , 0)0(2)0( =′= yy

(iii) 0442

2

=++ ydxdy

dxyd , 4)0(1)0( =′= yy

(iv) 0442

2

=++ ydxdy

dxyd , 0)1(1)0( == yy

(v) 042

2

=+ ydx

yd , 3)(1)0( 4 == πyy

2nd Order ODEs – Linear, Constant Coefficients (Non-Homogeneous) (5) Determine the general solution of each of the following ODEs:

(i) 8016102

2

=++ ydxdy

dxyd (ii) 33652

2

+=+− xydxdy

dxyd

(iii) 68432

2

−−=−+ xydxdy

dxyd (iv) xey

dxdy

dxyd 22

2

223 =++

(v) 24103 32

2

+=−+ xeydxdy

dxyd (vi) )3sin(42

2

xydx

yd =+

27

Answers

(1) (i) xeAy 2= (ii) )4(

1Cx

y−

=

(iii) 22 −= xeAy (iv) xAy =

(v) Ce

y x −= 1 (vi) 1]ln[ 2 ++= Cxy

(2) (i) 2)1( += xy

(ii) 3

22 −

= xey

(iii) 32 16 tx = (3) (i) xx eBeAy 23 −+= (ii) xx eBeAy 67 += − (iii) xexBAy 3)( += (iv) xexBAy −+= )( (v) ])3(sin)3(cos[ xBxAey x += (vi) ])(sin)(cos[ xBxAey x += − (vii) xx eBeAy 33 −+= (viii) )3(sin)3(cos xBxAy +=

28

(4) (i) xx eey 24 −−= (ii) ])2(sin3)2(cos2[3 xxey x += − (iii) xexy 2)61( −+=

(iv) xexy 2)1( −−= (v) )2(sin3)2(cos xxy += (5) (i) 528 ++= −− xx eBeAy

(ii) 1211

2132 +++= xeBeAy xx

(iii) 324 +++= − xeBeAy xx (iv) xxx eeBeAy 2

612 ++= −−

(v) 513

2125 −++= − xxx eeBeAy

(vi) )3(sin)2(sin)2(cos 5

1 xxBxAy −+=