12Differential equations and difference equations

We start our considerations with the following introductory example.

Example 12.1 We wish to determine all functions y = y(x) having the rate of change

√

3 2x + 1. Using the definition of the rate of change (see Chapter 4.4), this problem can be formulated as follows. Determine all functions y = y(x) which satisfy the equation

y(x) = y (x) = 3√2x + 1. y(x)

The latter equation can be rewritten as

y (x) = 3√2x + 1 · y(x).

This is a relationship between the independent variable x, the function y(x) and its derivative y (x). Such an equation is called a differential equation.

In this chapter, we discuss some methods for solving special types of differential equations (and later related difference equations). In what follows, as well as y(x), y (x), . . . we also use the short forms y, y , . . . First, we formally define a so-called ordinary differential equation.

between the independent variable x, a function y(x) and its derivatives is called an ordinary differential equation. The order of the differential equation is determined by the highest order of the derivatives appearing in the differential equation.

If representation (12.1) can be solved for y(n), we get the explicit representation

y(n) = f (x, y, y , y , . . . , y(n−1))

Differential and difference equations 445

of a differential equation of order n. The notion of an ordinary differential equation indicates that only functions depending on one variable can be involved, in contrast to partial differential equations which include also functions of several variables and their partial derivatives. Hereafter, we consider only ordinary differential equations and for brevity, we skip ‘ordinary’.

Definition 12.2 A function y(x) for which relationship F(x, y, y , y , . . . , y(n)) = 0 holds for all values x Dy is called a solution of the differential equation. The set

S = {y(x) | F(x, y, y , y , . . . , y(n)) = 0 for all x Dy}

is called the set of solutions or general solution of the differential equation.

12.1 DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

A first-order differential equation can be written in implicit form:

F(x, y, y ) = 0

or in explicit form:

y = f (x, y).

In the rest of this section, we first discuss the graphical solution of a differental equation of the first order and then a special case of such an equation which can be solved using integration.

12.1.1 Graphical solution

Consider the differential equation y = f (x, y). At any point (x0, y0), the value y = f (x0, y0) of the derivative is given, which corresponds to the slope of the tangent at point (x0, y0). This means that we can roughly approximate the function in a rather small neighbourhood by the tangent line. To get an overview on the solution, we can determine the curves, along which the derivative y has the same value (i.e. we can consider the isoquants with y being constant). In this way, we can ‘approximately’ graph the (infinitely many) solutions of the given differential equation by drawing the so-called direction field. This procedure is illustrated in Figure 12.1 for the differential equation y = x/2.

For any x R (and arbitrary y), the slope of the tangent line is equal to x/2, i.e. for x = 1, we have y = 1/2 (that is, along the line x = 1 we have the slope 1/2 of the tangent line), for x = 2, we have y = 1 (i.e. along the line x = 2 the slope of the tangent line is equal to 1) and so on. Similarly, for x = −1, we get y = −1/2, for x = −2 we get y = −1 and so on. Looking at many x values, we get a rough impression of the solution. In the given case, we get parabolas (in Figure 12.1 the parabolas going through the points (0, −2), (0, 0) and (0, 2) are drawn), and the general solution is given by

y = x2 + C

4

with C being an arbitrary real constant.

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For each point on the line y = x, we have y = −1, and for each point on the line y = −x, we have y = 1. Analogously, for each point on the line y = x/2, we get y = −2 and for each point on the line y = −x/2, we get y = 2. Continuing in this way, we get the direction field given in Figure 12.2. Thus, the solutions are circles with the origin of the coordinate system as the centre, i.e.

x2 + y2 = C (C > 0).

(In Figure 12.2 the circles with radius 2, i.e. C = 4, and radius 4, i.e. C = 16, are given.) Implicit differentiation of y2 = C − x2 confirms the result:

2y · y = −2x

y = − x . y

12.1.2 Separable differential equations

We continue with such a type of differential equation, where a solution can be found by means of integration. Let us consider the following special case of a first-order differential equation:

y = f (x, y) = g(x) · h( y),

i.e. function f (x, y) can be written as the product of a function g depending only on variable x and a function h depending only on variable y. In this case, function f = f (x, y) has a special structure, and we say that this is a differential equation with separable variables. Solving this type of differential equation requires only integration techniques. First, we rewrite

y = dy = g(x) · h( y) dx

in the form

dy = g(x) dx. h( y)

That is, all terms containing function y are now on the left-hand side and all terms containing variable x are on the right-hand side (i.e. we have separated the variables). Then we integrate both sides

and obtain

H ( y) = G(x) + C.

448 Differential and difference equations

If we eliminate y, we get the general solution y(x) including constant C. A particular (or definite) solution yP (x) is obtained when C is assigned a particular value by considering an initial condition y(x0) = y0. The latter problem of finding a particular solution is also referred to as an initial value problem. We illustrate this approach by the following two examples.

Now we get on the left-hand side the natural logarithm function as an antiderivative. Considering the integral on the right-hand side, we apply the substitution

u = 2x + 1 which yields

du = 2. dx

Now we can find the integral:

, √2x + 1 dx = 1 , √u du = 1 · u3/2 + C .

23

Therefore, equation (12.2) leads to

ln |y| = (2x + 1)3/2 + C .

Thus, we get the general solution

|y| = e(2x+1)3/2+C = e(2x+1)3/2 · eC

Here we have written the term eC as a new constant C > 0. By dropping the absolute values (they are now not necessary since the logarithmic terms have been removed), we can rewrite the solution obtained as

y(x) = C · e(2x+1)3/2

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with C being an arbitrary constant. If now the initial condition y(0) = e is also given, we obtain from the general solution

e = C · e1,

i.e. C = 1. This yields the particular solution

yP (x) = e(2x+1)3/2

of the initial value problem.

Example 12.3 We wish to determine all functions y = y(x) whose elasticity is given by the function

y(x) = 2x + 1.

Applying the definition of elasticity (see Chapter 4.4), we obtain

y (x) = 2x + 1. y(x)

We can rewrite this equation as

yx

or, equivalently,

dy = 2 + 1 dx.

yx

This is a differential equation with separable variables and we can integrate the functions on both sides. This yields:

, dy = , 2 + 1 dx

yx

ln |y| = 2x + ln |x| + C

|y| = e2x+ln |x|+C = e2x · eln |x| · eC

|y| = e2x · |x| · C, C > 0.

By dropping the absolute values, we can rewrite the solution obtained as y(x) = C · x · e2x

with C being an arbitrary real constant.

450 Differential and difference equations

Finally, we mention that we can reduce certain types of differential equations by an appropriate substitution to a differential equation with separable variables. This can be done e.g. for a differential equation of the type:

y = g(ax + by + c).

We illustrate this procedure by an example.

Example 12.4 Let the differential equation

y = (9x + y + 2)2

be given. (In this form it is not a differential equation with separable variables.) Applying the substitution

u = 9x + y + 2,

we get by differentiation

du = 9 + dy dx dx

which can be rewritten as

y = dy = du − 9. dx dx

After substituting the corresponding terms, we get

du − 9 = u2. dx

This is a differential equation with separable variables, and we get

du

= dx.

u2 + 9

Integrating both sides, we obtain

1 · arctan u = x + C

33

or, correspondingly,

arctan u = 3(x + C) 3

u = tan 3(x + C) 3

u = 3 tan 3(x + C).

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After substituting back, we get

Equation (12.3) is called linear since function y and all derivatives occur only in the first power and there are no products of the n + 1 functions y, y , . . . , y(n). The functions ai(x), i = 1, 2, . . . , n − 1, depend on variable x and can be arbitrary.

Definition 12.4 If q(x) ≡ 0 in equation (12.3), i.e. function q = q(x) is identically equal to zero for all values x, the differential equation (12.3) is called homogeneous. Otherwise, i.e. if q(x) ≡ 0, the differential equation is called non-homogeneous.

Function q = q(x) is also known as the forcing term. Next, we present some properties of solutions of a linear differential equation of order n and then we discuss an important special case.

12.2.1 Properties of solutions

Homogeneous differential equations

First, we present some properties of solutions of linear homogeneous differential equations of order n.

THEOREM 12.1 Let y1(x), y2(x), . . . , ym(x) be solutions of the linear homogeneous differential equation

Then the linear combination

y(x) = C1 y1(x) + C2 y2(x) + · · · + Cm ym(x) is a solution as well with C1, C2, . . . , Cm R.

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Non-homogeneous differential equations

We present the following theorem that describes the structure of the general solution of a linear non-homogeneous differential equation.

THEOREM 12.4 Let

SH = {yH (x) | yH (x) = C1 y1(x) + C2 y2(x) + · · · + Cn yn(x), C1, C2, . . . , Cn R}

be the general solution of the homogeneous equation (12.4) of order n and yN (x) a particular solution of the non-homogeneous equation (12.3). Then the general solution of a linear non-homogeneous differential equation (12.3) can be written as

S = {y(x) | y(x) = yH (x) + yN (x), yH (x) SH }

= {y(x) | y(x) = C1 y1(x) + C2 y2(x) + . . . + Cn yn(x) + yN (x), C1, C2, . . . , Cn R}.

Initial value problems

If an initial value problem of the linear differential equation of order n is considered, n initial conditions

are given from which the n constants C1, C2, . . . , Cn of the general solution can be determined. This yields the particular solution yP (x). The following theorem gives a sufficient condition for the existence and uniqueness of a particular solution.

with constant coefficients ai, i {0, 1, . . . , n − 1}, be given. Then this linear differential equation has exactly one particular solution yP (x).

We only note that for arbitrary functions ai(x), i {1, 2, . . . , n − 1}, an initial value problem does not necessarily have a solution.

12.2.2 Differential equations with constant coefficients

Now, we consider the special case of a linear differential equation of order n with constant coefficients, i.e. ai(x) = ai for i = 0, 1, 2, . . . , n − 1.

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Table 12.2 Settings for special forcing terms

Non-homogeneous differential equations

Now we discuss how a particular solution yN (x) of a non-homogeneous linear differential equation with constant coefficients can be found. We describe the method of undetermined coefficients and consider here only three specific types of forcing terms q. In these cases, a solution can be found by a special setting of the solution function yN (x) in dependence on function q in equation (12.7).

It can be seen that the setting in Table 12.2 follows the structure of the forcing term given on the left-hand side. We emphasize that the small letters in Table 12.2 are fixed in dependence on the given function q = q(x) whereas capital letters are parameters which have to be determined such that yN (x) is a solution of the given differential equation. Using the corresponding setting above, we now insert yN (x) into the non-homogeneous differential equation and determine the coefficients A, B and Ai, respectively, by a comparison of the coefficients of the left-hand and right-hand sides. It should be noted that the modifications described in cases (b) of Table 12.2 are necessary to guarantee that the suggested approach works (otherwise we would not be able to determine all coefficients by this approach). Moreover, we mention that this approach can also be used for sums or products of the above forcing terms, where the setting has to follow the structure of the given forcing term. We illustrate the determination of the general solution by the following examples.

Differential and difference equations 457

By setting y(x) = eλx and substituting y(x), y (x) and y (x) into the homogeneous equation, we get the characteristic equation

Since we have two real roots of multiplicity one, case (a) in Table 12.1 applies for each root and we get the complementary function

yH (x) = C1ex + C2e−2x.

Since function q is a polynomial of degree two, we use the setting

yN (x) = Ax2 + Bx + C

in order to find a particular solution of the non-homogeneous differential equation. We obtain

( yN ) (x) = 2Ax + B

( yN ) (x) = 2A.

Substituting the above terms into the differential equation (12.10), we get

2A + (2Ax + B) − 2(Ax2 + Bx + C) = 2x2 − 4x + 1.

Comparing now the coefficients of all powers of x, i.e. the coefficients of x2, x1 and x0, we obtain:

From the first equation we obtain A = −1, and then from the second equation B = 1 and, finally, from the third equation C = −1/2. The general solution of the non-homogeneous differential equation (12.10) is given by

y(x) = yH (x) + yN (x) = C1ex + C2e−2x − x2 + 2x − 1 . 2

Example 12.7 We solve the non-homogeneous differential equation

y + 2y + 5y = cos 2x.

458 Differential and difference equations

Considering the homogeneous equation and again setting y = eλx, we obtain the characteristic equation

λ3 + 2λ2 + 5λ = λ · (λ2 + 2λ + 5) = 0.

From the latter equation we get

λ1 = 0, λ2 = −1 + √1 − 5 = −1 + 2i and λ3 = −1 − √1 − 5 = −1 − 2i,

i.e. there is a real root of multiplicity one and two (conjugate) complex roots of multiplicity one. Therefore, we get the complementary function

yH (x) = C1e0x + C2e−x cos 2x + C3e−x sin 2x

= C1 + e−x(C2 cos 2x + C3 sin 2x).

To find a particular solution yN of the non-homogeneous differential equation, we use the setting

yN (x) = A cos 2x + B sin 2x

and obtain

( yN ) (x) = −2A sin 2x + 2B cos 2x (yN ) (x) = −4A cos 2x − 4B sin 2x (yN ) (x) = 8A sin 2x − 8B cos 2x.

Substituting yN (x) and its derivatives into the non-homogeneous differential equation, we obtain

(8A sin 2x − 8B cos 2x) + 2(−4A cos 2x − 4B sin 2x)

+ 5(−2A sin 2x + 2B cos 2x) = cos 2x.

Comparing now the coefficients of sin 2x and cos 2x on both sides of the above equation, we obtain:

sin 2x : 8A − 8B − 10A = −2A − 8B = 0 cos 2x : −8B − 8A + 10B = −8A + 2B = 1.

This is a system of two linear equations with the two variables A and B which can be easily solved:

Differential and difference equations 459

Example 12.8 Let the non-homogeneous differential equation

y − 4y = e2x

be given. In order to solve the homogeneous equation, we consider the characteristic equation

λ3 − 4λ = λ(λ2 − 4) = 0

which has the three roots

λ1 = 0, λ2 = 2 and λ3 = −2.

This yields the complementary function

yH (x) = C1e0x + C2e2x + C3e−2x = C1 + C2e2x + C3e−2x.

To get a particular solution of the non-homogeneous equation, we have to use case (b) with k = 1 for an exponential forcing term e2x in Table 12.2 since λ2 = 2 is a root of multiplicity one of the characteristic equation, and we set

yN (x) = Axe2x.

(Indeed, if we used case (a) and set yN (x) = Ae2x, we would be unable to determine coefficient A, as the reader can check.) We obtain

= Ae2x(1 + 2x)

( yN ) (x) = 2Ae2x(1 + 2x) + 2Ae2x = 2Ae2x(2 + 2x) = 4Ae2x(1 + x) (yN ) (x) = 8Ae2x(1 + x) + 4Ae2x = 4Ae2x(3 + 2x).

Substituting the terms for ( yN ) and (yN ) into the non-homogeneous equation, we get

4Ae2x(3 + 2x) − 4Ae2x(1 + 2x) = e2x.

Comparing the coefficients of e2x on both sides, we get

12A − 4A = 1

which gives A = 1/8 and thus the particular solution

yN (x) = 1 xe2x. 8

Therefore, the general solution of the given differential equation is given by

y(x) = yH (x) + yN (x) = C1 + C2e2x + C3e−2x + 1 xe2x. 8

462 Differential and difference equations and obtain a system of differential equations:

y1(x) = y2(x) y2(x) = y3(x)

yn−1(x) = yn(x)

yn(x) = −a0y1(x) − a1y2(x) − · · · − an−1yn−1(x) + q(x).

Under additional assumptions (which we do not discuss here), there is even an equivalence between a linear differential equation of order n with constant coefficients and a system of n linear differential equations of the first order with constant coefficients, i.e. a linear differential equation of order n can be transformed into a system of n linear differential equations of the first order with constant coefficients, and vice versa. By the following example, we illustrate this process of transforming a given system into a differential equation of order n by eliminating systematically all but one of the functions yi(x).

Example 12.10 Let the system of first-order differential equations

y1 = y2 + y3 + ex y2 = y1 − y3

y3 = y1 + y2 + ex

be given with the initial conditions

y1(0) = 0, y2(0) = 0 and y3(0) = 1.

We transform this system containing three functions y1(x), y2(x), y3(x) and its derivatives into one linear differential equation of order three. To this end, we differentiate the first equation which yields

y1 = y2 + y3 + ex

=(y1 − y3) + (y1 + y2 + ex) + ex

=2y1 + y2 − y3 + 2ex.

Differentiating the latter equation again, we get

y1 = 2y1 + y2 − y3 + 2ex

=2(y2 + y3 + ex) + (y1 − y3) − (y1 + y2 + ex) + 2ex

=y2 + y3 + 3ex.

Now we can replace the sum y2 + y3 by means of the first equation of the given system, and we obtain the third-order differential equation

y1 − y1 = 2ex.

Differential and difference equations 463

Using the setting

y1(x) = eλx,

we obtain the characteristic equation

λ3 − λ = 0

with the roots

λ1 = 0, λ2 = 1 and λ3 = −1.

This yields the following general solution of the homogeneous equation:

y1H = C1 + C2 · ex + C3 · e−x.

In order to find a particular solution of the non-homogeneous equation, we set

y1N = Axex

(notice that number 1 is a root of the characteristic equation, therefore we use case (b) in the setting for an exponential forcing term according to Table 12.2), which yields

and therefore

Aex(x + 3) − Aex(x + 1) = 2ex,

i.e. we obtain

ex(Ax + 3A − Ax − A) = 2ex from which we obtain

A = 1.

Therefore, the general solution is given by

y1(x) = C1 + C2ex + C3e−x + xex.

We still have to determine the solutions y2(x) and y3(x). This can be done by using the first equation of the given system and the equation obtained for y1 . We get

y2 + y3 = y1 − ex

y2 − y3 = y1 − 2y1 − 2ex.

Adding both equations and dividing the resulting equation by 2 gives