Differential Equations for Engineers and Scientists

By C.G. Lambe and C.J. Tranter

This concise applications-oriented text is intended for undergraduate students in engineering, mathematics, and other areas of science. The first chapters focus on solutions of first order equations, linear equations with constant coefficients, and simultaneous equations and reducible equations. Subsequent chapters explore the method of solution by infinite series and the more important special functions of mathematical physics.
The treatment examines the solution of partial differential equations as well as numerical methods of solution, including that of relaxation. Readers also receive an introduction to the theory of nonlinear differential equations. Nearly 900 worked examples and exercises include complete solutions, making this volume ideal for self-study as well as an excellent classroom text.

• Language: English
• Publisher: Dover Publications
• Release date: Jun 13, 2018
• ISBN: 9780486831343

Book preview

INDEX

CHAPTER 1

PRELIMINARY IDEAS AND DIRECT METHODS

1.1 Introduction

Most problems in science and engineering have to be idealized before their solution can be attempted. This idealization is generally necessary to bring the problem into a form capable of solution by known mathematical techniques, but it is essential, of course, that the actual and idealized problems should bear a close resemblance to one another. The solution then normally starts from well-established physical laws which are of two main types:

(a) statements which can usually be stated as mathematical equations, and

(b) assertions, not necessarily of a mathematical character, which provide a set of rules for the selection of physically acceptable solutions.

Typical of the first type of physical law is the knowledge that the drag of the air on an aeroplane (over a certain range of speeds) is proportional to the square of its velocity. As a mathematical equation this law can be expressed in the form D = kυ², where D is the drag, υ the velocity, and k is a constant whose value can be determined experimentally. An example of a law of the type (b) is the principle that mechanical systems cannot exist in which energy is created or destroyed. Such principles are useful in choosing an appropriate solution to a physical problem, since the mathematician can often produce a whole set of solutions each of which is correct in the sense that no mathematical error has been perpetrated.

The essentials in specifying an idealized problem are therefore provided by a number of physical laws and some rules which exclude unsuitable solutions. The only other requirement is a principle for the incorporation of these essentials into a mathematical equation. This is provided by the so-called doctrine of determinism. Put simply, this doctrine asserts that the state of a system at any instant is uniquely linked to, or determined by, the succession of states which precedes it. It is generally only possible to apply this principle to neighbouring states of the system, that is, those which differ infinitesimally in respect to time and space. Hence differential coefficients are usually involved and the resulting equations are known as differential equations. These equations have to be solved (or integrated) to give expressions relating states of the system separated by finite intervals of time and space.

There are two main classes of differential equations, ordinary and partial. An ordinary differential equation is one in which there is only one independent variable. For example, if an aeroplane of mass m is flying horizontally with propeller thrust P and drag D, its acceleration (which is the rate of change of its velocity υ at time t) is given by the application of Newton’s law and the relation D = kυ² by the equation

The equation

is a partial differential equation in which the unknown function V depends on the time t and the coordinates (x, y, z) of a representative point. It arises, for instance, in problems in the conduction of heat in a solid body.

The differential equations arising in physical problems are statements of physical laws and indicate features which may well be common to several problems. Besides appearing in heat conduction problems, equation (2) occurs in problems in the diffusion of matter in still air and in the drag of a surface in contact with a moving fluid. It is necessary therefore to possess additional information to specify any particular problem.

This information is given by initial and boundary conditions. Suppose, for example, a rectangular block of metal is heated on one face. The subsequent temperature in the block is determined by equation (2) in conjunction with a knowledge of the initial temperature of the block and the conditions on its other faces. The differential equation expresses the way in which the heat flows in any solid and the initial and boundary conditions specify the particular case under consideration.

Some differential equations can be solved quickly and easily by comparatively elementary methods, but the solution of many equations involves advanced mathematical techniques, while some can only be solved by numerical methods. A systematic study of standard methods of solving differential equations is of great importance. Solutions are not obtained by guesswork or by trial and error but by a systematic procedure and the student will develop ability to recognize the type of equation involved and to choose an appropriate method of solution. This ability is most readily acquired by working through large numbers of examples.

In this opening chapter we give some definitions and consider such preliminary ideas as are necessary for a study of the subject. We also give some examples to show how a physical problem can be expressed as a differential equation with appropriate initial and/or boundary conditions. We further show how a differential equation can be formed by elimination of constants from a mathematical equation and consider the reverse process of finding a solution of a differential equation when this can be done by direct integration.

1.2 Definitions

The order of a differential equation is defined as the order of the highest differential coefficient present. The degree of a differential equation is the degree of the highest differential coefficient present when the equation has been made rational and integral as far as the differential coefficients and the dependent variable are concerned.

A linear differential equation is one which is linear in the dependent variable and all its derivatives. The general form of a linear differential equation of the nth order is

where the functions pr(x), r = 0, 1, 2,..., n, and q(x) are functions of x only. The coefficient of dny/dxn in this equation is often taken as unity without loss of generality. A linear equation is, of course, of the first degree, but the term non-linear is used to describe an equation in which y or any of its derivatives is of degree higher than the first. Considering the examples,

(2) is ordinary linear of the second order and first degree, (3) is ordinary non-linear of the second order and second degree and (4) is ordinary non-linear of the second order and first degree.

A solution (or integral) of a differential equation is a relation between the variables, not involving differential coefficients, which satisfies the differential equation. Thus the well-known differential equation of simple harmonic motion

has solution x = C sin (ωt + ϕ) where C and ϕ are any constants. This is easily verified, since we have

Equation (6) furnishes the general solution of the differential equation (5). In an actual problem we usually require a particular solution which satisfies certain initial conditions. Thus we may be given that y = 2 and dy/dt = 3ω when t = 0. If the constants C and ϕ are such that the general solution satisfies these conditions, we have from (6) and (7),

and tan ϕ = 2/3. Thus the initial conditions determine the amplitude and phase (but not the frequency) of the motion.

Notation, It is sometimes convenient to use primes to denote derivatives. Thus

When the time t is the independent variable it is customary to denote differentiation with respect to t by dots. Thus

1.3 Formation of differential equations

Some remarks on the translation of a physical problem into a mathematical one have been made in the introductory paragraphs of this chapter. Here we illustrate by examples some of the methods used in forming differential equations from the physical data.

Example 1. A particle moves in a straight line, being attracted to a fixed point by a force which is proportional to its distance from the point. Form the differential equation of the motion.

Let x be the distance of the particle from the fixed point at time t; then the force is μx directed towards the fixed point, μ being a constant. The velocity υ at any instant is the rate of change of x with respect to t and the acceleration f is the rate of change of υ with respect to t. That is

Equating the product of the mass m of the particle and its acceleration to the force, we have the differential equation

This differential equation has the same form as equation (5) of §1.2 and a solution is

where ω² = μ/m.

Example 2. In a chemical reaction A → B, the velocity of reaction is proportional to the amount remaining of A. Form the differential equation of the reaction and verify that if a is the initial amount of A and k the constant of proportionality, the amount of A at time t is ae−kt.

If x is the amount of A remaining at time t the rate of change of x with respect to t . We have therefore

It is easily verified that if x = Ce−kt, where C is any constant,

Thus, since x = a when t = 0, it follows that C = a and the solution is x = ae−kt.

Example 3. A uniform beam of length I and weight wl is supported at its ends in a horizontal position. Form the differential equation for the deflexion y of the centre line of the beam and find a solution which satisfies the end conditions.

The bending moment at a distance x due to the distributed load. A general expression for the bending moment of a deflected beam is Elyn, where E is Young’s modulus for the material and I is the second moment of area of the crosssection of the beam about its neutral axis. We have therefore

This equation may be integrated twice with respect to x, giving

where C and D are any constants. If now we consider the initial conditions that y = 0 when x = 0 and when x = l we find that D = 0 and C = wl³/24, and the solution becomes

1.4 Elimination of constants

The solution of an ordinary differential equation is a relation between two variables and will involve one or more arbitrary constants. When a relation between the variables is given, constants may be eliminated by differentiating and forming a differential equation.

For example, the equation y = x(a − x) is the equation of a parabola whose axis is parallel to the y. Differentiating we have

Eliminating the constant a between the two equations we have

that is,

The equation (1) is a differential equation whose solution is y = x(A − x), where A is an arbitrary constant, and thus the differential equation represents a family of parabolas whose axes are parallel to the y-axis. It is easily seen that if an equation involves n constants the constants can, in general, be eliminated between the original equation and the equations which give the first n derivatives of y with respect to x thus forming a differential equation of the nth order. We would expect, therefore, the solution of a differential equation of the nth order to contain n arbitrary constants.

We may, therefore, think of the solution of an ordinary differential equation as being the equation of a curve which passes through certain fixed points determined by the initial conditions. Such a curve is called an integral curve of the differential equation. The constants determined by the initial conditions are eliminated in the differential equation which thus represents a whole family of curves of similar shape but passing through different points.

For example, the equation y² = 3x + 7 is the equation of a parabola. It is also an integral curve of the differential equation

The general solution of this differential equation is y² = Ax + B, which can represent any one of a family of parabolas.

Example 4. Form a differential equation to eliminate the constants a, b and r in the equation of the circle (x − a)² + (y − b)² = r².

By repeated differentiation we have

Eliminating b between the last two equations we have

This differential equation which represents all circles in the x, y plane may be obtained more simply by stating that the curvature of a circle is constant, that is

Differentiation of this equation leads to the differential equation (2). If the constant is taken as 1/r, the equation (3) is the differential equation of all circles of radius r and could be obtained by eliminating a and b between the original equation and its first two derivatives.

1.5 Taylor series expansion of solutions

Suppose that the solution of a linear differential equation is such that the independent variable y can be expressed as a Taylor series in (x − a) near x = a, that is

where (y)a, (dy/dx)a, etc. denote the values of y, dy/dx, etc. when x = a.

Let the differential equation be of the nth order of the form

where the coefficients pi(x), i = 0, 1,..., n, and q(x) have finite differential coefficients of all orders for x = a and p0(a) ≠ 0. Putting x = a in (2) we have

Equation (3) gives the value of (dny/dxn)a in terms of (dn−1y/dxn−1)a and lower derivatives. Differentiating equation (2) and putting x = a we have

Equations (4) and (3) determine the value of (dn + ¹y/dxn + ¹)a in terms of (dn−1y/dxn−1)a and lower derivatives. By repeated differentiation all further coefficients of the Taylor series can be obtained in terms of (y)a and the first n − 1 derivatives of y at x = a. Thus a solution may be obtained in which these n quantities are chosen arbitrarily. In other words, the general solution of the nth order linear differential equation contains n arbitrary constants.

This theorem is of great importance. In particular, it enables us to recognize that we have obtained the most general solution by seeing that it contains the requisite number of arbitrary constants.

In advanced treatises on differential equations it is proved that the most general solution of a differential equation of the nth order, whether linear or not, contains n arbitrary constants. Theorems, known as existence theorems, are proved giving the precise analytical conditions under which the equation has a solution.

The solution of an ordinary differential equation of order n which contains n arbitrary constants is called the complete primitive or the general solution of the equation.

For most differential equations any solution that can be found to satisfy the equation can be derived from the general solution by giving suitable values to the arbitrary constants. In certain cases, however, it is possible to find a solution which cannot be derived in this way and such a solution is called a singular solution. Some singular solutions of differential equations are considered in §2.11.

Example 5. Find a Taylor series expansion valid near the origin for the solution of the differential equation .

Let the value of y when x = 0 be a. The value of (y′)0 is evidently zero. We can differentiate both sides of the differential equation n times, and we have

Therefore

and, in particular,

and so on.

Since

it is evident that all the odd derivatives are zero and hence we have

Exercises 1(a)

1. Write down the order and degree of the following differential equations:

(a,

(b,

(c,

(d.

2. Find the ordinary differential equations of which the following are the general solutions, A and B being arbitrary constants:

(a) y = A cosh nx + B sinh nx ,

(b) y = e − kt ( A cos nt + B sin nt ,

(c) y = ( A + Bx ) e mx ,

(d) y ² = Ax + B ,

(e) y ² = Ax + A ² .

is the general solution of the differential equation

4. Find the differential equation of all circles that pass through the origin.

5. In successive chemical reactions A → B → C ..., the velocity of each reaction is proportional to the amount remaining of each concentration. Prove that if x is the concentration of A and y the concentration of C at time t , y being initially zero,

where a and b are the initial concentrations of A and B respectively.

6. A particle moves on a smooth horizontal plane in a medium which causes a retardation kυ ² , where υ is the velocity. Show that if x is the displacement at time t

and verify that the general solution of this equation is

Show that if the initial velocity is V and the initial displacement is zero, A = 1 and B = kV.

7. Find a Maclaurin expansion in ascending powers of x for the solution of the differential equation

8. Show that if y = A sin −1 x + B ,

By finding a Maclaurin expansion for the solution in ascending powers of x

9. Prove Euler’s theorem on homogeneous functions, that is, if z = f ( x , y ), where f ( x , y ) is a homogeneous function of x and y of degree n , then z satisfies the differential equation

. By substituting y = a 0 + a 1 x + a 2 x ² + ... in the above differential equation, or by any other method, obtain the first three non-vanishing terms in the expansion of y in ascending powers of x . (L.U.)

1.6 The differential equation y ( n ) = f ( x )

(a) First order equations

The differential equation

in which y is explicitly absent, has clearly the solution

where A is an arbitrary constant. The solution may be written as a definite integral

Here a is an arbitrary constant and is the value of x corresponding to the value y = 0.

The differential equation

in which x is explicitly absent, is solved in the same way since

and the solution is

Example 6. A particle is projected vertically upwards with velocity V and the air resistance causes a retardation kυ, where υ is the velocity at time t. Find the height reached and the time to reach this height.

The retardation is g + kυ, where υ where x is the height above the ground. The acceleration is dv/dt and we have

The solution must satisfy the initial condition that υ = V when t = 0, therefore

giving

The velocity becomes equal to zero and the particle has reached its highest point when

From (1)

Therefore

and the initial condition that x = 0 when t = 0 gives the value of B and the solution

At the highest point kt = loge(1 + kV/g) and e−kt = g/(g + kV), therefore

(b) Non-linear equations

Non-linear differential equations of the first order sometimes arise in which one of the variables is explicitly absent. A method of solving such equations may be seen from the following example.

Example 7. Solve the differential equation .

We have

which can be written

This gives, on integration,

that is

and the general solution is y² = tan² (x−A).

(c) Equations of higher order

The differential equation

is solved in a similar manner by repeated integration. We have

and so on, an arbitrary constant being introduced at each integration. Differential equations of this type occur in the theory of deflexions of a beam where a general expression for the bending moment is EI d²y/dx. Here y is the deflexion, assumed small, at distance x along the beam, I is the second moment of area of the cross-section about its neutral axis and E is Young’s modulus for the material. If the bending moment or the loading at any point can be expressed as a function of x the resulting differential equation can be solved by direct integration, the constants of integration being determined by the slope, bending moment, etc. at the ends of the beam.

Example 8. A light beam of uniform cross-section and length I is supported at its ends in a horizontal position. The ends are clamped horizontally and a load is placed on the beam in such a way that the intensity of loading increases uniformly from zero at one end to w at the other. Find an expression for the deflexion of the beam at any point of its length.

The loading at distance x from one end is wx/l and we have the differential equation

Integrating we have

The initial conditions to be satisfied by the solution are

the last two conditions expressing the fact that the beam is horizontal at the ends. The conditions (i) and (iii) make C = D = 0, and (ii) and (iv) give

Hence, A = −3wl/20, B = wl²/30, and we have the solution

Exercises 1(b)

, with the initial condition that y = 0 when x = 0.

, with the initial condition that y = 0 when x = 1.

3. In a chemical reaction A → B + C , the reaction rate at time t is proportional to the concentration x of A and x = a when t and hence that x = ae − kt .

4. In a second order chemical reaction A + B → C + D , the concentrations of A and B are initially a and b respectively and x is the concentration change at time t . Prove that

Show that if b ≠ a, x = ab{e(b − a)kt − 1}/{be(b−a)kt−a}.

5. A particle is projected vertically upwards with velocity V , where υ is the velocity at time t . Show that

and hence that

, with the condition that y = 0 when x = 0.

, with the condition that y = 1 when x = 0.

.

, with the condition that y = 1 when x = 1.

when x = 0.

, with the condition that y = 1 when x = 0.

, with the condition that y = 0 when x = 0.

13. The bending moment in a uniform cantilever of length l at distance x from the wall. Show that the deflexion y of the centre line is given by the equation

14. A uniform cantilever of length l and weight wl projects horizontally from a wall and its free end is propped level with the wall end. Show that the deflexion y of the centre line at distance x from the wall is given by the equation

min. If it starts from rest find its greatest velocity in this half minute and the distance traversed. Find, also, the velocity of the particle when the force on it becomes zero. (L.U.)

1.7 Exact equations

Let the solution of a first order differential equation be ϕ(x,y) = A, a constant. Differentiating this equation with respect to x we have

that is

where P and Q are functions of x and y and

Conversely, if a function f(x, y) is expressed as a quotient of two functions P and Q, so that f(x, y) = P(x, y)/Q(x, y), a solution of the general equation of the first order and first degree

is ϕ(x, y) = A, provided that ϕ(x, y) satisfies the relations (2). Given P and Q, the necessary and sufficient condition for this is that

. The sufficiency of the condition is seen from the method of solving the equation (3) when the condition is satisfied.

Let

that is, a function ψ(x, y) can be obtained by integrating P with respect to x, treating y as a constant. Then from (4)

and integrating with respect to x

where f(y) is a function of y alone. Then the solution is ϕ(x, y) = A, where

since the function ϕ .

The differential equation (1) is usually written in the form

and is called an exact equation if P and Q satisfy the relation (4). The solution is ϕ(x, y) = A, where

and also

g(y) being a function of y only and h(x) a function of x only. The precise form of the solution is found by seeing that the two values of ϕ(x, y) are identical.

An explicit formula can be found for ϕ(x, y) as follows.

Let

Then

and

Thus

where

Similarly,

Example 9. Solve the differential equation .

Writing the equation in the form

, and hence the equation is exact. Then if ϕ(x, y) = A is the solution,

Since the two values of ϕ must be identical we have ϕ = x² + y² + xy and the solution is

Alternatively, using the formulae (6) and (7) we have

Then

and

A differential equation Pdx + Qdy = 0 which is not exact can be made exact by multiplying by a function F(x, y), called an integrating factor. Assuming that there is a solution ϕ(x, y) = A. Multiplying by an integrating factor is equivalent to writing the function f(x, y) in (3) as RP/RQ, so that any first order equation can be written as an exact equation and solved if the correct quotient is found.

If the equation RPdx + RQdy = 0 is exact we have

that is

This partial differential equation for R is not readily soluble, but a value of R which satisfies this equation may sometimes be obtained by inspection. An infinite number of integrating factors exist, since if R is an integrating factor so also are ϕ. R, ϕ². R, etc.

Example 10. Solve the differential equation 2xy dx + (y² − x²)dy = 0.

It is easily seen that this equation is not exact. If R(x, y) is an integrating factor, we have

that is

If we try a value of R independent of x, we find

and a solution found by inspection is R = 1/y². Then the exact equation is

and if ϕ(x, y) = A is the solution

Hence, h(x) = 0, g(y) = y, and the solution is x² + y² = Ay.

Exercises 1(c)

Show that the following equations are exact and find their solutions:

1. ydx + xdy = 0.

2. x − l dx − tan ydy = 0.

3. ( 3x ² + y ² ) dx + 2 xydy = 0.

4. cosh x cosh ydx + sinh x sinh ydy = 0.

5. log e ydx + xy −1 dy = 0.

6. (2 x ³ + xy ² ) dx + (2 y ³ + x ² y ) dy = 0.

7. x ( x ² + y ² ) dx + y ( x ² + y ² ) dy = 0.

8. (3 x ² − 2 y ² ) dx − (4 xy − 2 y ) dy = 0.

9. ( ax + by + c ) dx + ( bx + dy + e ) dy = 0.

10. Show that Jacobi’s equation

is an exact equation if 2a1 = b2 + c3 and b1 = c1 = 0, and find its solution in this case.

11. Show that the equation (2 x − y ) dx + (2 y + x ) dy = 0 can be made exact by the integrating factor 1/( x ² + y ² ) and find its solution.

12. Using the integrating factor cos x cos y , solve the equation

13. Using the integrating factor 1/( x + y ) ² , solve the equation

14. Find an integrating factor which is a function of x only and solve the equation y (1 − x ) dx − xdy = 0.

15. Show that possible integrating factors of the equation ydx − xdy = 0 are x −2 , y −2 , x −2 + y −2 , ( x − y ) −2 and find the solution.

16. Show that ( x − y ) −2 is an integrating factor of the equation

and find the solution.

17. Find an integrating factor which is