Solutions of Diff Eqn

8/9/2019 Solutions of Diff Eqn

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F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S

FLAP M6.3 Solving second-order differential equationsCOPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

Module M6.3 Solving second order differential equations1 Opening items

1.1 Module introduction1.2 Fast track questions1.3 Ready to study?

2 Methods of solution for various second-order differential equations

2.1 Classifying second-order differential equations2.2 Equations of the form d 2 y / dt 2 = f (t

);

direct integration2.3 The equation for simple harmonic motion:

d 2 y / dt 2 + 2

y = 0

2.4 Equations of the form d 2 y / dt 2 2 y = 0

2.5 The equation for damped harmonic m a (d 2 y / dt 2) + b(

dy / dt ) + cy = 02.6 Equations of the form

a (d 2 y / dt 2) + b(

dy / dt ) + cy = f (t )

2.7 A worked example:damped, driven harmonic motion3 Closing items

3.1 Module summary

3.2 Achievements3.3 Exit test


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1 Opening items

1.1 Module introductionSuppose that we are interested in the behaviour of a mass m, free to bob up and down along the y-of a spring of force constant k , and also subject to a damping (or resistive) force the magnitude

proportional to the speed of the mass, as well as to an externally imposed periodic force in the y-by F y = F 0 1 sin 1 ( t ) where F 0 is a positive constant. ( F y is commonly abbreviated to F to avoconfusion with the subscripts.) The equation of motion of the mass is obtained by applying Nelaw . If y is the displacement of the mass from equilibrium, then we have

m d 2

ydt 2 = ky b dydx + F 0 sin ( t ) (1)


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All the forces acting on the mass have been added together on the right-hand side of Equation 1,

m d 2 ydt 2

= ky b dydx

+ F 0 sin ( t ) (Eqn 1)

and are as follows:ky (the restoring force exerted by the spring);

b dydt

, where b is a positive constant (the damping force);

F 0 1 sin 1 ( t ) (the periodic applied force).

Equation 1 is an example of a second-order linear differential equation . It belongs to a categdifferential equations known as linear equations with constant coefcients , where the dependent vaderivatives are multiplied by constants (not by functions of the independent variable ).


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The most general second-order equation of this sort can be written as

a d 2 ydt 2

+ b dydt

+ cy = f ( t ) (2)

where a , b , c are constants and a 0; it is equations of this type that will be discussed in this module.

The simplest of this type of equation is one for which b and c are zero, so that the equation becomes

a d 2 ydt 2

= f ( t ) (3)

This can be solved by direct integration , as will be explained in Subsection 2.1. Another relatively siarises when we have zero on the right-hand side of Equation 2, instead of a function of t , so that

a d 2 ydt 2

+ b dydt

+ cy = 0 , where a , b , c are constants and a 0 (4)

Equations of this type (known as linear homogenous equations ) have innumerable applications in parise, for example, in the discussion of simple harmonic motion and the analysis of a.c. circuits .module (Subsections 2.2 to 2.5) is therefore devoted to nding solutions to various forms of Equat
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Subsection 2.6 tells you how to solve some equations of the type

a d 2

ydt 2 + b dydt + cy = f ( t ) (Eqn 2)

for some different forms of f 1 (t ).

The method explained there will then be used in Subsection 2.7 to nd the general solution of equations type

m d 2 ydt 2

= ky b dydx

+ F 0 sin ( t ) (Eqn 1)

Study comment Having read the introduction you may feel that you are already familiar with the material coveredmodule and that you do not need to study it. If so, try the Fast track questions given in Subsection 1.2. Idirectly to Ready to study? in Subsection 1.3.


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1.2 Fast track questions

Study comment Can you answer the following Fast track questions ?. If you answer the questions successfuonly glance through the module before looking at the Module summary (Subsection 3.1) and the AchieveSubsection 3.2. If you are sure that you can meet each of these achievements, try the Exit test in Subsection difculty with only one or two of the questions you should follow the guidance given in the answers and read the reparts of the module. However, if you have difculty with more than two of the Exit questions you are strongly astudy the whole module .


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Question F1

The angular displacement of the bob of a simple pendulum satises the equation of motiona

d dt

b d dt

c2

2 0

+ + =

(a) Find the general solution of this equation if a = 0.1 1 kg, b = 0.2 1 kg 1 s1 and c = 1.0 1 kg 1 s2.

(b) Find the particular solution if the bob is hit when in its rest position = 0 such that it is givespeedd dt

t = =0 3 01. rad s at

(c) Sketch this solution as a function of t .

Question F2

Find the general solution to the equationd 2 xdt 2

+ 5 dxdt

+ 6 x = 2 t 2 3
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Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the nthrough the module and to proceed directly to Ready to study? in Subsection 1.3.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly Closing items .


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1.3 Ready to study?Study comment In order to study this module you will need to be familiar with the following term(of a polynomial ), exponential function , general solution , initial condition , linear differential

particular solution . You will also need to be familiar with various trigonometric identities (although we w

here); to be able to solve rst-order differential equations by direct integration (which requires a fair integration methods ); to know how to check a proposed solution to a differential equation by substitutiongood differentiation skills); and to be able to use initial conditions to obtain a particular solution from a To appreciate the physical signicance of some of the equations appearing in this module, you should know how

Newtons second law to write down a differential equation describing the motion of an object if you are given info

about the forces acting on it. Some familiarity with simple harmonic motion (SHM) would be partiIf you are uncertain about any of these terms, you can review them now by reference to the Glossary ,indicate where in FLAP they are developed. The following Ready to study questions will allow you to establisneed to review some of these topics before embarking on this module.
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Question R1

Find the general solution to the differential equationdydx

= x1 + x2

Question R2

If y = 4e x /2 cos 1 2 x, calculate dy / dx and d 1 2 y / dx 2 .
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Question R3

Show by substitution that y = 14 (2 x 3) + Ae x, where A is an arbitrary constant, is a solution tod 2 ydx 2

+ 3 dydx

+ 2 y = x

Is it a general solution ?

Question R4

In answering this question, you should make use only of trigonometric identities , you shouldcalculator .

(a) If cos 1 = 1/3, what is the value of sin 1 , given 0 < < /2 ?

(b) If cos1

= 1/3 and sin1

= 1/2, what is the value of cos1

( + ), given 0 < < /2 , < < 3 /
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Question R5

In each of the two following expressions for y , use the given initial conditions to calculate varbitrary constants A and B.(a) y = ( At + B)e2 t ; where y = 1 and dy / dt = 3 at t = 0.(b) y = A 1 cos 1 2 x + B 1 sin 1 2 x; where y = 4 and dy / dx = 2 at x = 0.

Question R6

An object of mass m is free to move along a line; its displacement from the origin is denoted by

by three forces: a restoring force, the magnitude of which is proportional to the magnitude of(or damping) force, the magnitude of which is proportional to the cube of the objects speed; and a coforce of magnitude F 0 acting in the negative x-direction. Use Newtons second law to write doworder differential equation that x must satisfy.


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2 Methods of solution for various second-order differential equations

Notation 3 Very often in the problems that arise in physics the independent variable is time, and sodiscussing a general type of differential equation, the independent variable is denoted by t andvariable by y. However, in some of the examples, drawn from specic problems in physics, we will notation for the variables that is appropriate to the situation. Bear in mind that the quantity being different

will always be the dependent variable.


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2.1 Classifying second-order differential equations

There is such a wide variety of second-order differential equations that it is useful to divide them into vacategories, and in this subsection we will introduce some terminology that arises when we do so. Yoprobably already familiar with the distinction between linear and non-linear differential equatiogeneral linear second-order differential equation is of the form

a ( t ) d 2 y

dt 2 + b( t ) dydt + c( t ) y = f ( t )

where a (t ), b(t ), c(t ) and f 1 (t ) are functions of t .

However, in many of the second-order linear equations that you will encounter in your study of physidependent variable and its derivatives appear multiplied only by constants rather than functionsSuch an equation is known as a linear differential equation with constant coefcients and has t

a d 2 ydt 2 + b

dydt + cy = f ( t ) (5)
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Equation 5

a d 2 ydt 2

+ b dydt

+ cy = f ( t ) (Eqn 5)

is easiest to solve when its right-hand side is zero, i.e. when

a d 2 ydt 2

+ b dydt

+ cy = 0 (6)

As you see, all the terms in Equation 6 contain the dependent variable y or a derivative of y; for equation of this form is called a linear homogeneous differential equation . We will discuss equsort in Subsections 2.3, 2.4 and 2.5. If f 1 (t ) in Equation 5 is not zero, then the equation ilinear inhomogeneous differential equation (as there is now a term in the equation which does not idependent variable y). Equation 6 is relatively easy to solve, whereas the ease with which we are able

Equation 5 is crucially dependent on the form of the function f 1 (t ).
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Question T1

State whether the following differential equations are linear with constant coefcients and, if so, whetheare homogeneous or inhomogeneous:

(a)d 2 ydx 2

= x 3 y3

(b) x d 2 ydx 2

+ 2 dydx

+ y = 03

(c)d 2 xdt 2

5 dxdt

= 6 x3

We will rst dispose of the easiest of all second-order differential equations 1 1 those that may direct integration . This is the subject of the next subsection.


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2.2 Equations of the form d 2 y / dt 2 = f ( t); direct integration

You should already know how to solve a rst -order differential equatio n of the form

dy

dt = f ( t ) (7)

where the derivative of the dependent variable is equal to a function of the independent variable. This equmay be integrated directly, to give

y = f ( t ) dt

(8)

The indenite integral will involve one arbitrary constant of integration, this indicates that Equation 8 givthe general solution to Equation 7.
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It is just as straightforward to solve a second-order differential equation of the form

d 2 ydt 2

= f ( t ) (9)

where the second derivative of the dependent variable is equal to a function of the independent vHere, we simply integrate directly twice . Recalling that d 1 2 y / dt 2 is the derivative of dy / dt , weintegrate both sides of Equation 9, we obtain

dydt

= f ( t ) dt = F ( t ) + A (10)where F (t ) is an indenite integra l of f 1 (t ) (i.e. F (t ) is a function such that F 1 (t ) = dF (t )/ dt = fconstant of integration. If we integrate Equation 10 again we obtain:

y = F ( t ) + A( ) dt = G ( t ) + At + B (11)where G(t ) is an indenite integral of F (t ), and B is another constant of integration. We see that Econtains two arbitrary constants, A and B; this indicates that it is the general solution to Equation 9.


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Equations which can be directly integrated often arise when considering objects that are moving along a

For instance, an object moving along the x-axis, with position coordinate x and acceleration function of time, satises the equation

d xdt

a t x2

2 = ( ) (12)

The following example illustrates this.
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Example 1 A car is travelling in the x-direction along a straight road. At time t = 0, it starts tofor the next 8 1 s, its acceleration a x(t ) is given by a x(t ) = a + bt . If a = 2.00 1 m 1 s2, b = 0.50 1 m 1 s3 and

velocity is 5.00 1 m 1 s1 at t = 0, how far does the car travel during the 8 1 s period?Solution We must rst solve the differential equation

d x

dt

a bt 2

2 = +

where x = x(t ) is the position of the car and we take the origin to be the position of the car at t = at t = 0. Integrating once gives us

dx

dt = (a + bt )

dt = at + b t

2

2

+ A = at + b

2t 2 + A

where A is an arbitrary constant. On substituting dx / dt = 5.00 1 m 1 s1 at t = 0 into this equation, we nd A = 5 1 m 1 s1.
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Integrating dx / dt to nd x gives us

x at b t A dt at bt At B= + + = + +2 2 62

2 3

where B is an arbitrary constant. If we substitute x = 0 at t = 0, we nd B = 0.

So x t t

t = + + ( . ) ( . ) ( . )0 50 6 2 00 2 5 0033

2

2

1m s m s m s

Therefore at t = 8 1 s, we nd that the car has travelled 146.6 1 m. 3


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Sometimes, instead of being given information directly about the acceleration of the object, you may be

the force F x(t ) acting on it as a function of time. However, since F x(t ) = ma x(t ), where m is thobject, it is very easy to write down a differential equation of the form given in Equation 12.

d xdt

a t x2

2 = ( ) (Eqn 12)

Such a differential equation, arising from an application of Newtons second law , is ofequation of motion .

Try the following question, which illustrates another example.

Question T2

An object of mass 2 1 kg, which is constrained to move along the x -axis, is subjec

F x(t ) = F 0 1 cos 1 ( t ), acting along the x-axis (in the direction of increasing x), where F 0 = 4 1 NAt time t = 0, the objects displacement is +3 1 m and its velocity is +1 1 m 1 s1. Find its displacement of time. 3
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2.3 The equation for simple harmonic motion:

d 2 y / dt 2 + w

1 1

1 1

2 y = 0We will now make a start at nding solutions of Equation 6,

a d 2 y

dt 2 + b dy

dt

+ cy = 0 (Eqn 6)

In this subsection (and the next), we will look at the simplied case where b = 0, so that Equatioform

a d 2 y

dt 2 + cy = 0 (13)

In Equation 13, a 0 (so the equation is of second order) and c 0. (If c were zero, we could solveby direct integration.) Since a 0, we can divide both sides of the equation by a , and write h = c / a

d 2 ydt 2

+ hy = 0 (14)


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d 2 ydt 2 + hy = 0 (14)

You will see shortly that the form of solution of Equation 14 depends on whether the constantnegative. We will consider these two possibilities separately, starting with the case of posit

(The case of negative h will be discussed in Subsection 2.4.) If h is positive, we can make it clear thby writing h = 02 , so that Equation 14 becomes:

d 2 y

dt 2 +

0

2 y = 0 (SHM equation ) (15)

Before discussing the solutions to Equation 15, we will emphasize the importance of this equation in phIt is the equation of motion of any object undergoing simple harmonic motion (SHM) 1 1 mo

force acting on the object is proportional to the objects displacement from some point and acts in the oppdirection to the displacement. For this reason, Equation 15 is often known as the SHM equation .
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You may already know that an object undergoing SHM executes oscillations and we will see sh period of the oscillations depends on the parameter

0 which is known as the angular frequency

of Equation 15.

d 2 ydt 2

+ 02 y = 0 (Eqn 15)

So it is important that whenever you encounter a case of SHM, you should be able to rewrite the differequation describing the physical situation in the form of Equation 15, and so nd an expression forof the parameters appearing in the problem. The following question gives you practice in this.
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L

C

Figure 1 3 See Question T3(a).

l

Figure 2 3 See

Question T3Rewrite each of the following equations

in the form of Equation 15,d 2 ydt 2

+ 02 y = 0 (Eqn 15)and so nd an expression for 0 :

(a) L d 2 I dt 2

+ I C

= 0 3

This equation describes the way in whichthe current I varies with time in a circuitcontaining an inductor L and a chargedcapacitor C , as shown in Figure 1.

(b) ml d 2

dt 2 = mg 3

This equation describes the motion of a simple pendulum 1 1 a mass m suspended from a xed poinof length l, as shown in Figure 2. 3


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It is not too difcult to guess the general solution of Equation 15.

d 2 ydt 2 + 0

2 y = 0 (Eqn 15)

Any solution of this equation must be a function such that when we differentiate it twice, we recovfunction itself, multiplied by 02 . This should remind you of a cosine or a sine function. In fact, bothand sin 1 ( 0 t ) have precisely this property, as does any function of the form B 1 cos 1 ( 0 t ) + C 1 sin 1 ( 0C are arbitrary constants.

Show that y = B 1 cos 1 ( 0 t ) + C 1 sin 1 ( 0 t ) is a solution to Equation 15.

You have now shown that

the rst form of the solution to the SHM equation (Equation 15) is:

y(t ) = B1

cos1

( 0 t ) + C 1

sin1

( 0 t ) (16)

Since it contains two arbitrary constants B and C , it is the general solution.


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Question T4

Write down the general solution of the following equations:

(a)d ydx

2

2 + 25 y = 0 3 (b) 9

2

2

d Qdx

= 4Q 3

It is easy to see from Equation 16

y(t ) = B 1 cos 1 ( 0 t ) + C 1 sin 1 ( 0 t ) (Eqn 16)

that y(t ) is a periodic function of t 1 1 that is, a function that repeats itself each time t increaamount, known as the period of the function. As you know, the value of a sine or cosine unchanged if the argument of the function is increased by 2 , or an integer multiple of 2 . Thusvalue if t increases by an amount T = 2 / 0.

The quantity T =2

0

is therefore the period of the function given in Equation 16.
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The solution to Equation 15 in a different form

In order to see some other features of the dependence of y on t , it is helpful to write the generaEquation 15

d 2 ydt 2

+ 02 y = 0 (Eqn 15)

in a different form. This is done by expressing the arbitrary constants B and C that appear in Equati


in terms of two other arbitrary constants, A and , as follows:

B = A 1 sin 1 (17a)

and C = A 1 cos 1 (17b)


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With these expressions for B and C , Equation 16

y(t ) = B 1 cos 1 ( 0 t ) + C 1 sin 1 ( 0 t ) (Eqn 16)becomes

y = A 1 sin 1 1 cos 1 ( 0 t ) + A 1 cos 1 1 sin 1 ( 0t ) = A[sin 1 1 cos 1 ( 0 t ) + cos 1 1 sin 1 ( 0 t )] (18)

We may now use the trigonometric identity

sin 1 ( + ) = sin 1 1 cos 1 + cos 1 1 sin 1

to rewrite the right-hand side of Equation 18 and make it look a lot neater.


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d 2 ydt 2

+ 02 y = 0 (Eqn 15)

Thus:

the second form of the solution to the SHM equation (Equation 15) is:

y(t ) = A 1 sin 1 ( 0 t + ) (19)

Equation 19 shows us that, for any particular choice of the constants A and , the graph of thEquation 15 always has the shape of a sine curve (though it may not pass through the origin). Such a cucalled sinusoidal ; an example is shown in Figure 3 (next page). We can also deduce from Equation since the maximum value attained by a sine function is 1, the maximum value of y is equal to ththerefore A is the amplitude of the oscillations. The other constant is known as the phaseinitial phase of the oscillations. Clearly the value of y at t = 0 depends on both A and sinc

A 1 sin 1 . Thus, if we know the value of 0, and we can discover the values of A and (so that wwith a particular solution of Equation 15) we can easily use Equation 19 to construct the graph of the sol

y /m
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2

2

0

12

y

The example shown in Figure 3corresponds to the values A = 2 1 m, = /6 and 0 = 2 1 s1 , so that

y = (2 1 m) 1 sin 1 [(2 1 s1)t + /6].

(Notice that y = 0 when t = ( /12) 1 s,and y = 1 1 m when t = 0.)

Figure 3 3 A sinusoidal curve with period 1 s1, amplitude 2 1 m and phase constant /6.


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Converting from one form of the solution into the other

It is clearly important to be able to switch between the two forms of the solution to the SHM equation givEquations 16 and 19.


y(t ) = A 1 sin 1 ( 0 t + ) (Eqn 19)From Equations 17a and b we can easily calculate B and C if we are given values for A and ; calculate A and from known values of B and C ? To see how to do this, note that if

B = A1

sin1

and C = A1

cos1

(Eqns 17a, b)then

B2 + C 1 2 = A2 (sin 2 + 1 cos 1 2 ) = A2

so that

A2 = B2 + C 1 2


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By convention, A is always taken to be positive, so we have the following two equations for AEquation 21 is obtained from Equations 17a and b

B = A 1 sin 1 and C = A 1 cos 1 (Eqns 17a, b)

by dividing the expression for B by the expression for C .

Equations converting from the rst form of the solution of the SHM equation into the second form:

A B C = +2 2 (positive square root) (20) BC

A A= =

sincos tan

i.e. = arctan

1

( B / C ) but with 0 < 2 (21)

By convention, is always chosen to lie within the range 0 < 2 in this context. You may be won

it cannot be chosen to lie within the range /2 < /21

1

the standard range for the inverse tan funreason is that although the function tan 1 repeats itself as is increased by , sin 1 and cos 1 dorange /2 < /2 would not cover all possible values (both positive and negative) of the constants


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In fact, using Equation 17,

B = A 1 sin 1 and C = A 1 cos 1 (Eqns 17a, b)

and our knowledge of the properties of sin 1 and cos 1 , we can lay down some rules for the rangwithin which must lie, depending on whether B and C are positive or negative:

B 0 and C 0 0 2 (22a)

B 0 and C < 0 2 < (22b) B < 0 and C < 0 < < 3 2 (22c)

B < 0 and C 0 3 2 < 2 (22d)You can now use Equations 17, 20, 21 and 22

A B C = +2 2 (positive square root) (Eqn 20) BC

A A

= =sincos

tan

i.e. = arctan 1 ( B / C ) but with 0 < 2 (Eqn 21)

to answer the following questions:


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Question T5

Write the particular solution y = 61

sin1

(3 t + /3) in the form y = B1

cos1

( 0t ) + C 1

sin1

( 0 t ).3

Question T6

Write the two following particular solutions in the form y = A 1 sin 1 ( 0 t + ):(a) y = 4 1 cos 1 (2 t ) + 3 1 sin 1 (2 t )(b) y = 5 1 sin 1 (4 t ) 12 1 cos 1 (4 t ) 3


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2.4 Equations of the form d 2 y / dt 2 - l 2 y = 0We will now return to the case of negative h in Equation 14.

d 2 ydt 2

+ hy = 0 (Eqn 14)

If h is negative, we can make it clear that this is so by writing h = 2 , where (by convention)Equation 14 becomes

d 2 ydt 2

2 y = 0 (23)

You may not have yet encountered any physical situations that could be described by an equation of thisHowever, it does have important applications in quantum physics and elsewhere. So it is well worth your to learn how to solve this equation. Moreover, the solution is very easy to nd.


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To solve Equation 23,

d 2

ydt 2

2 y = 0 (Eqn 23)

let us proceed using the same sort of informed guesswork that we used to solve Equation 15.

d 2 ydt 2 + 0

2 y = 0 (Eqn 15)

In the case of Equation 15, we wanted to nd a function such that its second derivative was equal to the funitself multiplied by a negative constant. Looking at Equation 23, we see that this time we want a (or functions) the second derivative of which is equal to the function itself multiplied by a po

Exponential functions have the property that all their derivatives are proportional to the function itself.

So let us see if an exponential function, of the form

y = Be pt (24)

where B is an arbitrary constant, is a solution of Equation 23.
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If we differentiate Equation 24 once,

y = Be pt (Eqn 24)

we nd dy / dt = pBe pt = py.

When we differentiate again, we nd

d 2 ydt 2

= p dydt

= p2 y

and on substituting this result into Equation 23,

d 2 ydt 2 2 y = 0 (Eqn 23)

we obtain

p2 y 2 y = 0

which is an identity provided that p2 = 2 .

Thus Equation 24 is a solution to Equation 23 provided p = + or p = .


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It follows that any function of the form

y = Be t (25a)

or y = C e t (25b)

(We have replaced the arbitrary constant B in Equation 25a by the arbitrary constant C in Equation

is a solution to Equation 23.

d 2 ydt 2

2 y = 0 (Eqn 23)

Neither of these equations can themselves be the general solution of Equation 23, as neither containarbitrary constants. But perhaps their sum , which does contain two arbitrary constants, is the general so

Show that y = Be t + C e t is a solution to Equation 23.


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d 2 ydt 2

2 y = 0 (Eqn 23)

So you have shown that:

The general solution of Equation 23 is

y(t ) = Be t + C e t (26)

Since Equation 26 contains two arbitrary constants, it is the general solution. You may wonder what this sollooks like graphically. If either B or C is zero, the curve is exponential. Otherwise, the solutionEquation 26 behaves like y = B 1 exp 1 ( t ) when t is large and positive (as then exp 1 ( t ) is very sm

y = C 1 exp 1 ( t ) when t is large and negative (as then exp 1 ( t ) is very small). Figure 4 (next page) shogeneral shapes of curve you can expect, depending on the different signs of B and C .

y y y y B > 0, C > 0 B < 0, C < 0 B > 0, C < 0


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t 0 t 0 t 0

(a) (b) (c) (d)

, , ,

Figure 4 3 Solutions to Equation 23. General solution is y(t ) = Be t + C e t (Equation 26)

Q ti T7


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Question T7

Find the general solution of

d 2 ydx 2

4 y = 0

and the particular solution if y = 6 and dy / dx = 0 at x = 0. 3

We obtained the general solution given in Equation 26

y(t ) = Be t + C e t (Eqn 26)

by taking the sum of the two solutions given in Equation 25.

y = Be t (25a) or y = C e t (Eqn 25b)

In fact, it can be proved that, for any linear homogeneous equation (not necessarily one withcoefcients) the sum of two solutions (or more than two, if the equation is of higher order than the secoalways also a solution. We will not give the proof here (although the last marginal note should prto how the proof might go), but you should bear this useful result in mind; we will make use of it in thesubsection.


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2.5 The equation for damped harmonic motion:

a ( d 2 y / dt

2) + b( dy / dt ) + cy = 0

In Subsection 2.3, we mentioned that an equation of the form

a d 2 ydt 2

+ cy = 0 3 where c / a = h > 0

has many different applications in physics, describing, as it does, the behaviour of an object undergoing sharmonic motion. Its solutions are sinusoidal oscillations of constant amplitude. However, you knowexperience that the vibrations of an oscillating object always die away with time (pendulum clocks run dmasses bobbing up and down on springs come to rest, and so on). This is due to the effect ofdamping forces, which oppose the motion of the oscillating object. How can we modify the SHM eq(Equation 15).

d 2 ydt 2

+ 02 y = 0 (Eqn 15)

to take account of these forces? Let us return to the example mentioned in Subsection 1.1, a massup and down on a spring of force constant k .
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If the restoring force of the spring, ky, is the only force acting on the mass, then, according to Nelaw , its displacement y(t ) must satisfy the equation

m d 2 ydt 2

= ky

To incorporate the effects of a damping force, we must add a term on the right-hand side which always act

direction opposite to the direction of the velocity of the mass (the bob). A simple way of doing this is to that the damping force can be written in the form b(dy / dt ), where b is a positive constant, so that of motion of the mass becomes

m d 2 y

dt 2 = ky b dy

dt i.e. m

d 2 ydt 2

+ b dydt

+ ky = 0 (27)

LAn equation of the form given in Equation 27
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L

R

Figure 5 3 A circuit cinductance L, a resistocapacitor C , connecte

An equation of the form given in Equation 27

m d 2 ydt 2 + b

dydt + ky = 0 (Eqn 27)

also arises in the theory of a.c. circuits.

The current I in the circuit shown in Figure 5, containing an inductance L,

a resistor R and a capacitor C , obeys the differential equation L

d 2 I dt 2

+ R dI dt

+ I C

= 0 (28)
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To predict the behaviour of the mass on the spring, or the current in the circuit, you need to be able to differential equations of the type

a d 2 ydt 2

+ b dydt

+ cy = 0 3 where a > 0, b > 0 and c > 0 (29)

We need the ratio b / a to be positive if the coefcient of dy / dt is to represent a resistive forcevelocity of the object; and we need the ratio c / a to be positive so that if b is zero, we recover the S(Equation 15).

d 2 y

dt 2 + 02

y = 0 (Eqn 15)This condition is most simply achieved by restricting all three constants to positive values. (In fact, the soluwe obtain to Equation 29 will apply equally well if any of a , b , or c is negative.)


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Let us now try to nd solutions to Equation 29.

a d 2

ydt 2 + b dydt + cy = 03 where a > 0, b > 0 and c > 0 (Eqn 29)

We know that if b = 0, then the solution is a sinusoidal function. However, we do not want a pusolution if b > 0 (we want y to tend to zero as t becomes large and positive, due to the damping); m

you try a solution of the form y = sin1

( pt ) or cos

1

( pt ) in Equation 29, you will quickly nd that it doe(unless b = 0, of course). But perhaps an exponential function would work, just as it did for Equation 23?

d 2 ydt 2

2 y = 0 (Eqn 23)

We have nothing to lose by trying it, so let us substitute

y = Be pt (30)

into Equation 29.

2


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The rst derivative of y is equal to py, and the second derivative is equal to p2 y; so we nd, on subs

ap 2 y + bpy + cy = 0which is an identity provided that p satises the equation

ap 2 + bp + c = 0 ( auxilliary equation ) (31)

This quadratic equation in p is known as the auxiliary equation of Equation 29.

a d 2 y

dt 2 + b dy

dt + cy = 0 3 where a > 0, b > 0 and c > 0 (Eqn 29)

As it is a quadratic, the roots of Equation 31 are given by

p

b b ac

a p

b b ac

a1

2

2

242

42=

+ =

and (32)

This formula will give us two real values for p1 and p2 provided that b2 > 4 ac . (We will deal wwhere b2 < 4 ac or b2 = 4ac shortly; but for the moment, let us assume that the two roots are real.)
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R l t f th ili ti (h d i g)


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Real roots of the auxiliary equation (heavy damping)

We can write the two solutions of Equation 29a

d 2 ydt 2

+ b dydt

+ cy = 0 3 where a > 0, b > 0 and c > 0 (Eqn 29)

we have just found as

y = B 1 exp 1 ( 1 p1 t ) 3 and 3 y = C 1 exp 1 ( 1 p2 t )

As we mentioned at the end of Subsection 2.4, we can obtain the general solution to Equation 29 by adding two solutions together:

The general solution of Equation 29 in the case b2 > 4 ac is

y(t ) = B 1 exp 1 ( 1 p1t ) + C 1 exp 1 ( 1 p2t ) (33)

where 3 p b b aca

12 4

2= + 3 and 3 p b b ac

a2

2 42

=

or, written out in full,


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or, written out in full,

y = B exp b + b2 4 ac

2 a t

+ C exp

b b2 4 ac2 a t

(34)

There is no need to memorize this unpleasant-looking equation 1 1 but you should remember the meused to derive it, and be able to apply it to a given differential equation, as in the following example.

Example 2 Find the general solution of the differential equation

d 2 ydt 2

+ 5 dydt

+ 6 y = 0

Solution First write down the auxiliary equation,

p2 + 5 p + 6 = 0

This equation has two real roots:

p = 52 12 25 24 = 2 or 3 3

Thus the general solution is y = Be 2 t + C e3 t . 3

Question T8


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Find the general solution of the differential equation

6 d 2 ydt 2

+ 17 dydt

+ 12 y = 0 3

yYou can see that if a , b and c are all positive quantities, the positive square root


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0

Figure 6 3 Hmotion.

b 2 4 ac must be less than b, and therefore both roots of the auxiliary equationare negative. Thus the solution to Equation 29

a d 2 ydt 2

+ b dydt


that is given in Equation 34

y = B exp b + b2 4 ac

2 at

+ C exp b b

2 4 ac2 a

t

(Eqn 34)

is the sum of two decreasing exponentials, and as t becomes very large and positive, yAn example of this is shown in Figure 6.

yRemember that in physical situations b gives an indication of the magnitude of the


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0

Figure 6 3 Hmotion.

damping force. If b = 0 the motion is undamped and we have oscillations

corresponding to SHM (as in Equation 15),d 2 ydt 2

+ 02 y = 0 (Eqn 15)

but if b is so large that b2 > 4 ac , then there are no oscillations (as in Figure 6).In such a case, the motion is said to be heavily damped . You can probably guesswhat will happen if we choose a value for b somewhere between these twoextremes. Nonetheless, we will now analyse such problems systematically.

Complex roots of the auxiliary equation (light damping)
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Complex roots of the auxiliary equation (light damping)

We will now solve Equation 29

a d 2 ydt 2

+ b dydt


for the case b2 < 4 ac . We will give two alternative treatments of this important case.The rst treatment depends on a result that comes from the study of complex numbers , and ssolution in this case is merely an extension of the previous result.

The second treatment does not require as much knowledge of complex numbers, we will just give yo

solution of the differential equation and ask you to verify that what we say is correct.

The rst treatment requires the following result from the theory of complex numbers
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ei = cos 1 + i1 sin 1 (35)

If you are familiar with this result continue reading; if not go straight to Question T9 .

If b2 < 4 ac in Equation 29,

a d 2 y

dt 2 + b dy

dt + cy = 0 3 where a > 0, b > 0 and c > 0 (Eqn 29)

then the two roots of Equation 31

ap2 + bp + c = 0 (auxilliary equation) (Eqn 31)

are complex and can be written (using Equation 32) p

b b aca

p b b ac

a1

2

2

242

42

= + = and (Eqn 32)

as p i p i1 22 2

= + = + and

where = b / a 3 and 3 = 4 ac b2 /(2 a ) 3 (Notice that, since b2 < 4 ac , we can be sure that is a real quantity.)

Thus, the general solution to Equation 29
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us, t e ge e a so ut o to quat o 9

a d 2 ydt 2 + b

dydt + cy = 0

3 where a > 0, b > 0 and c > 0 (Eqn 29)

when b2 < 4 ac is similar to that given in Equation 33

y(t ) = B 1 exp 1 ( 1 p1t ) + C 1 exp 1 ( 1 p2t ) (Eqn 33)

and we may write it as

y = Rexp 2

+i t

+ S exp 2

i t

= exp ( t 2) Rexp( i t )[ ]+ S exp( i t )[ ]

where R and S are arbitrary constants.

It looks as though this solution will give complex values to y. However, since the case b2 < 4 ac fr


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g g p y ,in physics problems, it must be possible to arrange Equation 36

y = Rexp 2

+i t

+ S exp 2

i t


in a form which need not involve any complex quantities. We do this by employing Equation 35,

ei = cos 1 + i1 sin 1 (Eqn 35)

to write

exp 1 (i 1 t ) = cos 1 ( 1 t ) + i 1 sin 1 ( 1 t ) 3 and 3 exp 1 (i 1 t ) = cos 1 ( 1 t ) i 1 sin 1 ( 1 t )

if we substitute these results into Equation 36, we nd

y = exp 1 ( 1 t /2)[( R + S ) 1 cos 1 ( 1 t ) + i( R S ) 1 sin 1 ( 1 t )]

We can now dene new arbitrary constants B = R + S and C = i( R S ), and rewrite the solution as y = exp 1 ( 1 t /2)[ B 1 cos 1 ( 1 t ) + C 1 sin 1 ( 1 t )]

You may think that this rewriting has achieved nothing since C may be complex, so we have still


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y g g y pthat the solution y(t ) is a real quantity. However, at this point it is important to realize that the constantsin Equation 36

y = Rexp 2

+i t

+ S exp 2

i t


are quite arbitrary, they did not even have to be real. Equation 36 is a perfectly valid solution to Equat

a d 2 ydt 2

+ b dydt


even if R and S are complex. In fact, if R and S are complex conjugates , i.e. complex numbers of the

R = a + ib 3 and 3 S = a ib 3 where a and b are real constants

then B = R + S = 2 a 3 will be real

and C = i( R S ) = i(2 ib) = 2b3

will be realMoreover, the values of a and b are unrelated, so we are free to choose arbitrary values for the constants

just as we were for R and S.


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Question T9


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Show by substitution that, for arbitrary constants B and C ,

y(t ) = exp 1 ( 1 t /2)[ B 1 cos 1 ( 1 t ) + C 1 sin 1 ( 1 t )]

where

= b / a 3 and 3 = 4 ac b2 /(2 a )is a solution to Equation 29.

a d 2 ydt 2

+ b dydt


(Notice that is a real quantity if b2 < 4 ac .) 3

The combination [ B 1 cos 1 ( 1 t ) + C 1 sin 1 ( 1 t )] appears in Equation 37,


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y(t ) = exp1

( 1

t /2)[( B

1

cos 1

( 1

t ) + C 1

sin 1

( 1

t )] (Eqn 37)and it may be convenient to write this instead in the form A 1 sin 1 (

1

t + ), as we did in Subsection 2.3,

A B C = +2 2 (Eqn 20)and = arctan 1 ( B / C ) 3 but with 0 < 2 (Eqn 21)

This gives:

The alternative form of the solution of Equation 29 in the case b2

< 4 ac y(t ) = A 1 exp 1 ( 1 t /2) 1 sin 1 (

1

t + ) (39)

with = b /2 3 and 3 = 4 ac b2 /(2 a ) (Eqn 38a and b)

From Equation 39 it is easy to determine the behaviour of the function y(t ). Since b / a is pospositive 1 1 so y (t ) is given by a sinusoidal function multiplied by an exponential function that decbecomes large and positive.

yFigure 7 shows the form of the solution in this case.You can see that the effect of multiplying the


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0

2

Figure 7 3 Lightly damped oscillations.

You can see that the effect of multiplying thesinusoidal function by the exponential is to produceoscillations with amplitudes that decrease with time.

This is just the sort of behaviour we expect of asystem where the resistive forces are not toolarge 1 1 remember that b2 must be less than 4acfor the solution in Equation 39 to apply.

y(t ) = A 1 exp 1 ( 1 t /2) 1 sin 1 ( 1

t + ) (Eqn 39)

In this case, the system is said to be lightly damped .Note that the values of t for which y = 0 are stillequally spaced, with separation, t say, given by

1

t = , or t = / . Successive maxima andminima are also equally spaced, by an interval in t

equal to 2 / (though they no longer occur exactlyhalf-way between the points where y = 0).For these reasons, we still speak of the quantity2 / as the period of the oscillations.

Equal roots of the auxiliary equation (critical damping)
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Equal roots of the auxiliary equation (critical damping)

Finally, we will consider the case where b2 = 4ac which (although it is somewhat articial in that it rarin practice) is of interest because it marks the transition from light to heavy damping. In this caseEquation 31

ap2 + bp + c = 0 (auxilliary equation) (Eqn 31)has only one root; from Equation 32,

p b b ac

a p

b b ac

a1

2

2

24

2

4

2= + = and (Eqn 32)

we see that p 1 = p 2 = b /(2 a ). We deduce that y = B 1 exp 1 [bt /(2 a )] is a solution of the differenEquation 29.

a d 2 y

dt 2 + b dydt + cy = 03 where a > 0, b > 0 and c > 0 (Eqn 29)

However, it cannot be the general solution, since it only contains one arbitrary constant. The other t dd t it i t t ll b i ill j t t ll th d l t h k


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we must add to it is not at all obvious, so we will just tell you the answer, and leave you to check

substitution.

a d 2 ydt 2

+ b dydt


It turns out

The general solution of Equation 29 in the case b2 = 4ac is

y(t ) = ( B + Ct ) 1 exp 1 [bt /(2 a )] (40)

Question T10

Show that Equation 40 is a general solution of Equation 29 in the case b2 = 4 ac .3

This has been a long and quite complicated subsection, which is perhaps misleading since in practisolution of Equation 29 is fairly straightforward and just a matter of knowing how to deal with three


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solution of Equation 29 is fairly straightforward, and just a matter of knowing how to deal with three

Here then is a summary of the steps you need to follow in order to solve the equation of damped harmotion:

a d 2 ydt 2

+ b dydt

+ cy = 0 (Eqn 29)

Step 1 Evaluate the quantity b2 4ac .Step 2o If b2 > 4 ac , nd the roots p1 and p2 of the auxiliary equation, Equation 31.

ap2 + bp + c = 0 (auxilliary equation) (Eqn 31)The general solution is then given by Equation 33.


o If b2 < 4 ac , nd the quantities and , using Equation 38.


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with = b / a (Eqn 38a)and = 4 ac b2 /(2 a ) (Eqn 38b)The general solution is then given by Equation 37

y(t ) = exp1

( 1

t /2)[( B 1

cos 1

( 1

t ) + C 1

sin 1

( 1

t )] (Eqn 37)(and then use Equations 20 and 21

A B C = +2 2 (Eqn 20)

and = arctan 1 ( B / C ) 3 but with 0 < 2 (Eqn 21)if you want the solution in the form of Equation 39).

y(t ) = A 1 exp 1 ( 1 t /2) 1 sin 1 ( 1

t + ) (Eqn 39)o If b2 = 4 ac , the general solution is given by Equation 40.

y(t ) = ( B + Ct ) 1 exp 1 [bt /(2 a )] (Eqn 40)

Now practise these steps by trying the following question. You must make sure that you have masteretechniques required in these exercises. You will not be able to make progress with the next section unles


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techniques required in these exercises. You will not be able to make progress with the next section unles

have done so.

Question T11

Find the general solution of each of the following differential equations:

(a)d 2 xdt 2

+ 6 dxdt

+ 10 x = 0 (c) 5 d 2 y

dt 2 + 6 dy

dt + 2 y = 0

(b) 2 d 2 y

dx 2 + 5 dy

dx+ 3 y = 0 (d) d

2 y

dt 2 + 2 dy

dt + y = 0

3

The method of comparing b2 with 4 ac and then selecting the form of the solution will of course wob = 0, that is, for the differential equations discussed in Subsections 2.3 and 2.4. Consider, for example,


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b 0, that is, for the differential equations discussed in Subsections 2.3 and 2.4. Consider, for example,

d 2 ydt 2

+ 0 2 y = 0 (Eqn 15)

Here, b = 0, a = 1 and c = 0 2 . So b2 < 4 ac , and Equation 38

= b / a (Eqn 38a)

= 4 ac b2 /(2 a ) (Eqn 38b)tells us that = 0, = 0 . Thus using Equation 37,

y(t ) = exp 1 ( 1 t /2)[( B 1 cos 1 ( 1 t ) + C 1 sin 1 ( 1 t )] (Eqn 37)

we see that the general solution is y(t ) = B 1 cos 1 ( 0 t) + C 1 sin 1 ( 0t ), which is just what we founEquation 16.


2.6 Equations of the form a ( d 2 y / dt 2) + b( dy / dt ) + cy = f ( t)


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We are now in a position to set about nding solutions to the general second-order linear inhomogeequation with constant coefcients:

a d 2 ydt 2

+ b dydt

+ cy = f ( t ) (Eqn 5)

We will rst of all show that if we can nd any one particular solution to Equation 5, then we canthe general solution. We will then show a method that can sometimes be used to nd particular solutions.

Let us suppose that we have somehow managed to nd a particular solution to Equation 5;

d 2 d


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a d 2 ydt 2 + b

dydt + cy = f ( t ) (Eqn 5)

we will call it yp(t ). We now assert that the general solution is obtained by taking the sum ofgeneral solution to the homogeneous equation which is obtained by setting f (t ) = 0 in Equation 5:

a d 2 y

dt 2 + b dydt + cy = 0 (41)

In this context, the general solution to the homogeneous equation (Equation 14)

d 2 ydt 2 + hy = 0 (Eqn 14)

is called the complementary function ; we will denote it by yc(t ). Thus the claim here is that the geneto Equation 5 is

y(t ) = yp(t ) + yc(t ) (42)

This result is easy to prove. We simply substitute Equation 42


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y(t ) = yp(t ) + yc(t ) (Eqn 42)into Equation 5.

a d 2 ydt 2

+ b dydt

+ cy = f ( t ) (Eqn 5)

The left-hand side becomes

a d 2

dt 2 ( yp + yc ) + b d

dt ( yp + yc ) + c( yp + yc )

which is equal to

ad 2 ypdt 2

+ b dypdt

+ cyp + a d 2 yc

dt 2 + b dyc

dt + cyc (43)

but by assumption, yp is a solution to Equation 5; that is

ad 2 ypdt 2

+ b dypdt

+ cyp = f ( t )

and yc is the general solution to Equation 41,

a d 2 y +b dy + cy = 0 (Eqn 41)


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adt 2

+ bdt

+ cy = 0 (Eqn 41)

so that

a d 2 yc

dt 2 + b dyc

dt + cyc = 0

So we see from Equation 43

ad 2 ypdt 2

+ b dypdt

+ cyp + a d 2 yc

dt 2 + b dyc

dt + cyc (Eqn 43)

that, with the substitution y = yp + yc, the left-hand side of Equation 5

a d 2 ydt 2

+ b dydt

+ cy = f ( t ) (Eqn 5)

is equal to f 1 (t ) + 0, i.e. equal to f 1 (t ). Thus Equation 42

y(t ) = yp(t ) + yc(t ) (Eqn 42)

is a solution to Equation 5; it is the general solution since yc contains two arbitrary constants.


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Solution To nd the general solution, we must add the complementary function to the particular solution.The complementary function is the general solution of the homogeneous equation that we obtain by setting right hand side of Equation 44 to zero


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right-hand side of Equation 44 to zero,

d ydt

y t 2

2 4+ = (Eqn 44)

d 2 y

dt 2 + 4 y = 0

This equation is of the form given in Equation 15,

d 2 ydt 2

+ 02 y = 0 (Eqn 15)

with 0 = 2. Its general solution is given by Equation 16


to be B1

cos1

(2 t ) + C 1

sin1

(2 t ). So the general solution to Equation 44is y = t /4 + B 1 cos 1 (2 t ) + C 1 sin 1 (2 t ). 3

Question T12Find the general solution to the differential equation

2


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d 2

ydt 2 + 3 dydt 4 y = 2e

3 t

given that y = e3t

2 is a particular solution. 3

Finding a particular solution

So we now know how to nd a general solution to a linear inhomogeneous equation if we are giv


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So we now know how to nd a general solution to a linear inhomogeneous equation if we are givsolution. But how can we nd a particular solution? There are several methods for doing this, andexplain here only the simplest. It is sometimes known as the method of undetermined coefcientswill see, it consists of little more than intelligent guesswork. The idea is that, guided by the form ofshould assume a particular form for yp(t ) which contains some undetermined constants, and simply

this into the differential equation. If the form we have chosen is correct, we will be able to determiconstants appearing in our trial solution from the requirement that we must obtain an identitysubstitution. If we have not chosen the correct form, then we will nd that it is not a solution, and we must again! The method can best be explained by an example.

Example 4 Find a particular solution to the differential equation

d 2 y + 4 dy + 5 y = t + 2
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dt 2 dt y

Solution There is a function of the form Ht + K on the right-hand side of this equation. A function of (a polynomial ) has the property that when it is differentiated any number of times, no new functionin its derivatives 1 indeed, the results are just constants (or zero). This suggests that the differential equati

might be satised by a function of this sort, with H and K suitably chosen. We may as well try it, anywIf we put y = Ht + K into the equation, we obtain

4 H + 5( Ht + K ) = t + 2

i.e. 5 Ht + (4 H + 5 K ) = t + 2This equation must be an identity if y = Ht + K is a solution. Thus the coefcients of t on both sidessame, as must the constant terms. This gives us two equations to be solved for H and K :

5 H = 13

and3

4 H + 5 K = 2These are easily solved, to give H = 1/5, K = 6/25. So we have found a particular solution; it is

y = t /5 + 6/25. 4

If we were to try to generalize the reasoning in Example 4, it would go something like this. The partsolution must be such that when we substitute it into the left-hand side of the equation, f 1 (t


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If f 1

(t ) is a function whose derivatives are all of the same form as f 1

(t ) itself (such as a polynomial, anfunction or a sinusoidal function), we may be able to achieve this by trying as a particular solution a funwhich is also of the form of f 1 (t ), but contains some as yet unknown constants (the undetermined coefthe values of which we will determine when we make the substitution. This leads to the following rulnding particular solutions, for certain forms of f 1 (t ):

Rules for nding particular solutions, for certain forms of f

1 1

1 1 ( t):1 If f 1 (t) is a polynomial of degree m : try a particular solution that is also a polynomial of degree


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1 If f (t ) is a polynomial of degree m : try a particular solution that is also a polynomial of degree

yp(t ) = Ht m + Kt m1 + + N but containing undetermined coefcients H , K N

2 If f 1 (t ) is an exponential, C ekt : try a particular solution that is also an exponential,

yp(t ) = H ekt

where H is an undetermined coefcient. 3 If f 1 (t ) is a sinusoidal function, f 1 (t ) = C 1 sin 1 (kt ) + D 1 cos 1 (kt ): try a particular solution that is also

function yp(t ) = H 1 sin 1 (kt ) + K 1 cos 1 (kt )

where H , K are undetermined coefcients.

You can now use Rules 13 to answer the following question.

Question T13

Find a particular solution to each of the following equations


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(a) d 2 y

dx 2 + 2 dy

dx+ 3 y = 2e x

(b) d y

dx

dy

dx

y x x2

2 4 4 2 11+ + = cos sin 3

The method of undetermined coefcients will work for some more complicated forms of f 1 (t ), buenough here for you to be able to nd particular solutions to many second-order inhomogeneous equatio

physical interest. (It is worth pointing out, though, that if f 1 (t ) is given by a sum of two or more funsorts mentioned in Rules 13, then the particular solution is simply the sum of the particular solcorresponding to each of these functions.) At the end of the next subsection, in Question T14, you catogether the methods you have learnt so far to nd general solutions to inhomogeneous equFirst, however, we will consider an example of great importance in physics.

2.7 A worked example: damped driven harmonic motion
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We are now in a position to solve Equation 1, introduced in Subsection 1.1 ;

m d 2 ydt 2

= ky b dydx

+ F 0 sin ( t ) (Eqn 1)

rewritten slightly, it is

m d 2 ydt 2

+ b dydt

+ ky = F 0 sin ( t ) (45)

An equation of this sort applies whenever an oscillating object is subjected to a damping f

as well as an external driving force, F 0 1 sin 1 ( 1 t ) with angular frequency .It is said to describe forced , damped oscillations or damped, driven oscillations .In this subsection, we will nd the general solution to Equation 45 for the case b2 < 4 mk .
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which can be rewritten in the form

[( k m 2 ) H b K ]cos( t ) [( k m 2 )K b H ]sin( t ) F sin ( t )


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(

) +

( ) +

(

)

( ) = 0

( )

If this is to be an identity, the coefcients of cos 1 ( 1 t ) and sin 1 ( 1 t ) on each side of the equation must

So we obtain two equations (for H and K )

( k m 2 ) H + b K = 0 3 and 3 ( k m 2 ) K b H = F 0On solving these equations for H and K , we nd (after some algebra! )

H = ( b ) F 0

( k m 2 )2 + ( b )23

and3

K = ( k m 2 ) F 0

( k m 2 )2 + ( b )2

Thus the particular solution is the rather complicated looking expression

yp (t ) = (b )F 0

(k m 2 )2 + (b )2 cos( t ) + (k m 2 )F 0

(k m 2 )2 + (b )2 sin ( t ) (47)

We can simplify this greatly if we use Equations 20 and 21

A B C 2 2 (positive square root) (Eqn 20)
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= + BC

A A

= =sincos

tan


to write yp in the form

yp(t ) = A 1 sin 1 ( 1 t + )

Applying Equations 20 and 21, we nd (after some more algebra)

A F

k m b= +02 2 2( ) ( ) (48a)

and = arctan b k m 2

3 but with 0 2 (48b)

The general solution

Thus the general solution of Equation 45,


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m d 2 ydt 2

+ b dydt

+ ky = F 0 sin ( t ) (Eqn 45

given by the sum of the complementary function (Equation 46)

y t bt m B t C t c exp[( ) ( )][ cos ( ) sin ( )]= +2 (Eqn 46and the particular solution (Equation 47),

yp (t ) = (b )F 0(k m 2 )2 + (b )2 cos( t ) +

(k m 2 )F 0(k m 2 )2 + (b )2 sin ( t ) (Eqn 47

is

( ) [ b (2 )][ B ( ) C i ( )] A i ( ) (49)


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y(t ) = exp[ bt (2 m)][ Bcos( t ) + C sin( t )] + Asin( t + ) (49)where = 4 22mk b m( )

and A and are given in Equations 48a and b.

A F

k m b= +

02 2 2( ) ( )

(Eqn 48a)


3 but with 0 2 (Eqn 48b)

Interpretation

We can see from the form of the complementary function in Equation 46


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y t bt m B t C t c exp[( ) ( )][ cos ( ) sin ( )]= +2 (Eqn 46(or Equation 49)

y(t ) = exp[ bt (2 m)][ Bcos( t ) + C sin( t )] + Asin( t + ) (Eqn 49that, whatever the values of the arbitrary constants B and C , this part of the solution will eventually t(because of the presence of the decaying exponential factor, exp 1 [bt /(2 m)]). All that remains aftertherefore, is the particular solution, which represents the response of the oscillating mass to the applied

So it is this part of the solution which is of greatest interest to us. Equation 47

yp (t ) = (b )F 0(k m 2 )2 + (b )2 cos( t ) +

(k m 2 )F 0(k m 2 )2 + (b )2 sin ( t ) (Eqn 47

shows that it corresponds to sinusoidal oscillations, of the same angular frequency as the applied

However, there is a constant phase difference between the oscillations of the mass and oscillatidriving force. Mathematically, represents the amount by which the movement of the mass lead

h i ll it k t th t th ti l g b hi d th f b


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physically it makes more sense to say that the motion lags behind the force by an amouNotice (from Equations 48a and b)

A F

k m b= +

02 2 2( ) ( )

(Eqn 48a)


3 but with 0 2 (Eqn 48b)

that the amplitude A of the oscillations depends not only on F 0, but also on ; in other words, appdifferent frequency may invoke oscillations of the mass of very different magnitude.

0.05Figure 8 shows A as a function of , for acertain choice of the parameters m, k , b and F 0 .You can see that it has a pronouncedmaximum at a value of close to the


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0.04

0.03

0.02

0.01

00 5 10 15 20 25 30

/s 1

A / m

Figure 8 3 Graph of A as a function of , in the caseb = 1.0 1 N 1 s 1 m1, k = 80 1 N 1 m1, F 0 = 0.8 1 N.

maximum at a value of close to thenatural angular frequency 0 = k m .This large response for a particular value of

the angular frequency of the applied force isthe well-known phenomenon of resonance .

The above example involved some rather unpleasant algebra (because we wished to discuss a general however, nding the general solution to a specic inhomogeneous equation is generally straightfo

enough1 1

provided of course that f1

(t) is one of the functions mentioned in Rules 1 3 at the end of S
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enough provided, of course, that f (t ) is one of the functions mentioned in Rules 13 at the end of S2.6 1 1 the method is summarized here:

1 Find the complementary function, using the methods of Subsection 2.5

2 Find the particular solution, using one of Rules 13 in Subsection 2.6

3 Add the two together.

You can practise these steps by trying the following question.

Question T14

Find the general solution to the equation

d 2 ydt 2

+ 6 dydt

+ 9 y = 1 + 3 t 3

3 Closing items

3 1 Module summary


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3.1 Module summary1 A linear differential equation with constant coefcients has the form:

a d 2 ydt 2

+ b dydt

+ cy = f ( t ) (Eqn 5)

If f 1 (t ) in Equation 5 is set to zero it is a linear homogeneous differential equation . If f 1 (t Equation 5 it is a linear inhomogeneous differential equation

2 Equations of the form

d 2 y

dt 2 = f 1 (t ) (Eqn 9)

can be solved by direct integration .

3 The general solution of the SHM (simple harmonic motion ) equation

d 2 y

dt 2 + 02

y =0 (Eqn 15)


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dt + y = 0 (Eqn 15)can be written either as

y(t ) = B 1 cos 1 ( 0t ) + C 1 sin 1 ( 0 t ) (Eqn 16)

or y(t ) = A 1 sin 1 ( 0 t + ) (Eqn 19)corresponding to sinusoidal oscillations of period 2 / 0 . The values A and are determined f

B and C (or vice versa) by Equations 17, 20 and 21.

B = A1

sin1

and C = A1

cos1

(Eqns 17a, b) A B C = +2 2 (positive square root) (Eqn 20) B

C

A

A

= =sin

cos

tan


4 The general solution of an equation of the form

d 2 y

dt 2 2 y = 0 (Eqn 23)


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dt

is y(t ) = Be t + C e t (Eqn 26)

5 The nature of the solution of the ( damped harmonic motion ) equation

a d 2 ydt 2

+ b dydt

+ cy = 0 (Eqn 29)

depends upon the sign of b2 4ac.

The case where b2

> 4 ac corresponds to heavy damping , as illustrated in Figure 6 (with the solution given by Equation 33).


where 3 p b b ac

a12 4

2= +

3 and 3 p b b ac

a22 4

2=

The case where b2 < 4 ac corresponds to light damping , as illustrated in Figure 7 (with the solution given by Equation 37, or alternatively Equation 39).

y(t) = exp 1 ( 1 t /2)[( B 1 cos 1 ( 1 t) + C 1 sin 1 ( 1 t)] (Eqn 37)


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y(t ) exp ( t /2)[( B cos ( t ) + C sin ( t )] (Eqn 37)

y(t ) = A 1 exp 1 ( 1 t /2) 1 sin 1 ( 1

t + ) (Eqn 39)

The case where b2 = 4 ac corresponds to critical damping (with the solution given by Equation 40

y(t ) = ( B + Ct ) 1 exp 1 [bt /(2 a )] (Eqn 40)

6 To solve the (damped, driven harmonic motion ) equation

a

d 2 y

dt 2 +b dy

dt + c = f ( t )


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a dt + b dt + c = f ( t )we rst obtain the complementary function by nding a general solution of the equation

a d 2 ydt 2

+ b dydt

+ c = 0

The general solution of a d 2 ydt 2

+ b dydt

+ c = f ( t ) can then be found, if we know one partiby adding the complementary function to that particular solution.

Particular solutions can be found by the method of undetermined coefcients if f 1

(t ) is a exponential function, or a sinusoidal function.In the case of a sinusoidal function f 1 (t ), the amplitude of the solution (for large values ofupon the angular frequency of f . In some cases this amplitude may be very large for a particular the applied angular frequency, leading to the phenomenon of resonance .

3.2 AchievementsHaving completed this module, you should be able to:A1 Dene the terms that are emboldened and agged in the margins of the module


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A1 Dene the terms that are emboldened and agged in the margins of the module.A2 Recognize linear differential equations with constant coefcients and distinguish between homo

inhomogeneous linear differential equations.

A3 Solve equations of the form d 2

ydt 2 = f ( t ) by direct integration.A4 Recognize the SHM equation whenever it arises, identify the parameter that determines

oscillations and solve the equation.

A5 Convert between a particular solution of the form y = Bcos( t ) + C sin( t ) and one y = Asin( t + ) and vice versa.


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3.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions, each of wh


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one or more of the Achievements .

Question E1

( A2 and A3 ) 3 The differential equationd 2 xdt 2

= ge kt

where g is the magnitude of the acceleration due to gravity and k is a positive constant, may be used tmotion of a parachutist falling under gravity . The x-axis is chosen to be the downward vertical, when t = 0.

Is this equation linear, linear with constant coefcients, or homogeneous?

Find the particular solution if the parachutist starts from rest.

Question E2

( A4) 3 If the Earth is represented by a sphere of uniform density, the gravitational force on a part

inside the Earth at a distance r from the centre is of magnitude mgr / R, and is directed towards the Earth, where R = 6400 km is the (approximate) radius of the Earth and g is the magnitude of the a
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Earth, where R 6400 1 km is the (approximate) radius of the Earth and g is the magnitude of the ato gravity at the Earths surface. If it were possible to pass a straight tube through the centre of the Earththat a particle placed in the tube would execute simple harmonic motion , and nd the period (Take g as 10 1 m 1 s2.)
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Question E4

( A2 , A4 and A6 ) 3 In a certain medium , the z-component of the electric eld , E z, varies with x, the

the boundary of the medium, according to the equationd 2 E 2


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FLAP M6.3 Solving second-order differential equationsCOPYRIGHT 1998

Solutions of Diff Eqn

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