2.3 Linear Equations

INTRODUCTION

We continue our search for solutions of first-order DEs by next examining linear equations. Linear differential equations are an especially “friendly” family of differential equations in that, given a linear equation, whether first-order or a higher-order kin, there is always a good possibility that we can find some sort of solution of the equation that we can look at.

A Definition

The form of a linear first-order DE was given in (7) of Section 1.1. This form, the case when n = 1 in (6) of that section, is reproduced here for convenience.

DEFINITION 2.3.1 Linear Equation

A first-order differential equation of the form

(1)

is said to be a linear equation in the dependent variable y.

When g(x) = 0, the linear equation (1) is said to be homogeneous; otherwise, it is nonhomogeneous.

Standard Form

By dividing both sides of (1) by the lead coefficient a1(x) we obtain a more useful form, the standard form, of a linear first-order equation

. (2)

We seek a solution of (2) on an interval I for which both functions P and f are continuous.

In the discussion that follows, we illustrate a property and a procedure and end up with a formula representing the form that every solution of (2) must have. But more than the formula, the property and the procedure are important, because these two concepts carry over to linear equations of higher order.

The Property

The differential equation (2) has the property that its solution is the sum of the two solutions, y = yc + yp, where yc is a solution of the associated homogeneous equation

(3)

and yp is a particular solution of the nonhomogeneous equation (2). To see this, observe

.

The Homogeneous DE

The homogeneous equation (3) is also separable. This fact enables us to find yc by writing (3) as

and integrating. Solving for y gives yc = ce–∫P(x)dx. For convenience let us write yc = cy1(x), where y1 = e–∫P(x) dx. The fact that dy1/dx + P(x)y1 = 0 will be used next to determine yp.

The Nonhomogeneous DE

We can now find a particular solution of equation (2) by a procedure known as variation of parameters. The basic idea here is to find a function u so that is a solution of (2). In other words, our assumption for yp is the same as yc = cy1(x) except that c is replaced by the “variable parameter” u. Substituting yp = uy1 into (2) gives

so that                                                                                                     

Separating variables and integrating then gives

    and     

From the definition of y1(x), we see 1/y1(x) = e∫P(x)dx. Therefore

and (4)

Hence if (2) has a solution, it must be of form (4). Conversely, it is a straightforward exercise in differentiation to verify that (4) constitutes a one-parameter family of solutions of equation (2).

You should not memorize the formula given in (4). There is an equivalent but easier way of solving (2). If (4) is multiplied by

(5)

and then (6)

is differentiated, (7)

we get (8)

Dividing the last result by e∫P(x)dx gives (2).

Method of Solution

The recommended method of solving (2) actually consists of (6)–(8) worked in reverse order. In other words, if (2) is multiplied by (5), we get (8). The left side of (8) is recognized as the derivative of the product of e∫P(x)dx and y. This gets us to (7). We then integrate both sides of (7) to get the solution (6). Because we can solve (2) by integration after multiplication by , we call this function an integrating factor for the differential equation. For convenience we summarize these results. We again emphasize that you should not memorize formula (4) but work through the following two-step procedure each time.

  Guidelines for Solving a Linear First-Order Equation

  1. Put a linear first-order equation of form (1) into standard form (2) and then determine P(x) and the integrating factor e∫P(x)dx.
  2. Multiply (2) by the integrating factor. The left side of the resulting equation is automatically the derivative of the product of the integrating factor and y. Write

    and then integrate both sides of this equation.

EXAMPLE 1 Solving a Linear DE

Solve

SOLUTION

This linear equation can be solved by separation of variables. Alternatively, since the equation is already in the standard form (2), we see that the integrating factor is e∫(–3)dx = e–3x. We multiply the equation by this factor and recognize that

    is the same as    

Integrating both sides of the last equation,

gives e–3x y = −2e–3x + c. Thus a solution of the differential equation is , . ≡

When a1, a0, and g in (1) are constants, the differential equation is autonomous. In Example 1, you can verify from the normal form dy/dx = 3(y + 2) that −2 is a critical point and that it is unstable and a repeller. Thus a solution curve with an initial point either above or below the graph of the equilibrium solution y = −2 pushes away from this horizontal line as x increases.

Constant of Integration

Notice in the general discussion and in Example 1 we disregarded a constant of integration in the evaluation of the indefinite integral in the exponent of e∫P(x)dx. If you think about the laws of exponents and the fact that the integrating factor multiplies both sides of the differential equation, you should be able to answer why writing ∫P(x) dx + c is unnecessary. See Problem 55 in Exercises 2.3.

General Solution

Suppose again that the functions P and f in (2) are continuous on a common interval I. In the steps leading to (4) we showed that if (2) has a solution on I, then it must be of the form given in (4). Conversely, it is a straightforward exercise in differentiation to verify that any function of the form given in (4) is a solution of the differential equation (2) on I. In other words, (4) is a one-parameter family of solutions of equation (2), and every solution of (2) defined on I is a member of this family. Consequently, we are justified in calling (4) the general solution of the differential equation on the interval I. Now by writing (2) in the normal form y′ = F(x, y) we can identify F(x, y) = −P(x)y + f(x) and ∂F/∂y = −P(x). From the continuity of P and f on the interval I, we see that F and ∂F/∂y are also continuous on I. With Theorem 1.2.1 as our justification, we conclude that there exists one and only one solution of the first-order initial-value problem

(9)

defined on some interval I0 containing x0. But when x0 is in I, finding a solution of (9) is just a matter of finding an appropriate value of c in (4); that is, for each x0 in I there corresponds a distinct c. In other words, the interval I0 of existence and uniqueness in Theorem 1.2.1 for the initial-value problem (9) is the entire interval I.

EXAMPLE 2 General Solution

Solve

SOLUTION

By dividing by x we get the standard form

(10)

From this form we identify P(x) = −4/x and f(x) = x5ex and observe that P and f are continuous on the interval (0, ). Hence the integrating factor is

Here we have used the basic identity = N, N > 0. Now we multiply (10) by x–4,

,     and obtain    

It follows from integration by parts that and so

gives x–4y = xex − ex + c. Thus the general solution defined on is . ≡

Singular Points

Except in the case when the lead coefficient is 1, the recasting of equation (1) into the standard form (2) requires division by a1(x). Values of x for which a1(x) = 0 are called singular points of the equation. Singular points are potentially troublesome. Specifically in (2), if P(x) (formed by dividing a0(x) by a1(x)) is discontinuous at a point, the discontinuity may carry over to functions in the general solution of the differential equation.

EXAMPLE 3 General Solution

Find the general solution of

SOLUTION

We write the differential equation in standard form

(11)

and identify P(x) = x/(x2 − 9). Although P is continuous on (−, −3), on (−3, 3), and on (3, ), we shall solve the equation on the first and third intervals. On these intervals the integrating factor is

After multiplying the standard form (11) by this factor, we get

and integrating gives

Thus on either (−, −3) or (3, ), the general solution of the equation is . ≡

Notice in the preceding example that x = 3 and x = −3 are singular points of the equation and that every function in the general solution is discontinuous at these points. On the other hand, x = 0 is a singular point of the differential equation in Example 2, but the general solution y = x5ex − x4ex + cx4 is noteworthy in that every function in this one-parameter family is continuous at x = 0 and is defined on the interval (−, ) and not just on (0, ) as stated in the solution. However, the family y = x5ex − x4ex + cx4 defined on (−, ) cannot be considered the general solution of the DE, since the singular point x = 0 still causes a problem. See Problems 50 and 51 in Exercises 2.3. We will study singular points for linear differential equations in greater depth in Section 5.2.

EXAMPLE 4 An Initial-Value Problem

Solve the initial-value problem

SOLUTION

The equation is in standard form, and P(x) = 1 and f(x) = x are continuous on the interval (−, ). The integrating factor is e∫dx = ex, and so integrating

gives exy = xex − ex + c. Solving this last equation for y yields the general solution . But from the initial condition we know that y = 4 when x = 0. Substituting these values in the general solution implies c = 5. Hence the solution of the problem on the interval (−, ) is

y = x − 1 + 5e–x. (12) ≡

Recall that the general solution of every linear first-order differential equation is a sum of two special solutions: yc, the general solution of the associated homogeneous equation (3), and yp, a particular solution of the nonhomogeneous equation (2). In Example 4 we identify yc = ce–x and yp = x − 1. FIGURE 2.3.1, obtained with the aid of a graphing utility, shows (12) in blue along with other representative solutions in the family y = x − 1 + ce–x. It is interesting to observe that as x gets large, the graphs of all members of the family are close to the graph of yp = x − 1, which is shown in green in Figure 2.3.1. This is because the contribution of yc = ce–x to the values of a solution becomes negligible for increasing values of x. We say that yc = ce–x is a transient term since yc → 0 as x → . While this behavior is not a characteristic of all general solutions of linear equations (see Example 2), the notion of a transient is often important in applied problems.

Eight solution curves are graphed. The curve labeled c equals 0 is a line that goes up and to the right through approximately (negative 4, negative 5) and (0, negative 1). The family of curves c greater than 0 is above the curve c equals 0. The curves go down and to the right with decreasing steepness to a low point, then go up and to the right, approaching the curve c equals 0. The family of curves c less than 0 is below the curve c equals 0. The curves go up and to the right with decreasing steepness and approach the curve c equals 0.

FIGURE 2.3.1 Some solutions of the DE in Example 4

Piecewise-Linear Differential Equation

In the construction of mathematical models (especially in the biological sciences and engineering) it can happen that one or more coefficients in a differential equation is a piecewise-defined function. In particular, when either P(x) or f(x) in (2) is a piecewise-defined function the equation is then referred to as a piecewise-linear differential equation. In the next example, f(x) is piecewise continuous on the interval [0, ) with a single jump discontinuity at x = 1. The basic idea is to solve the initial-value problem in two parts corresponding to the two intervals over which f(x) is defined; each part consists of a linear equation solvable by the method of this section. As we will see, it is then possible to piece the two solutions together at x = 1 so that y(x) is continuous on [0, ). See Problems 33–38 in Exercises 2.3.

EXAMPLE 5 An Initial-Value Problem

Solve     where    

SOLUTION

The graph of the discontinuous function f is shown in FIGURE 2.3.2. We solve the DE for y(x) first on the interval [0, 1] and then on the interval (1, ). For 0 ≤ x ≤ 1 we have

    or, equivalently,    

A graph representing a discontinuous function consists of a line. The line begins at (0, 1) and goes right to a point marked at (1, 1).

FIGURE 2.3.2 Discontinuous f(x) in Example 5

Integrating this last equation and solving for y gives y = 1 + c1e–x. Since y(0) = 0, we must have c1 = −1, and therefore y = 1 − e–x, 0 ≤ x ≤ 1. Then for x > 1, the equation

leads to y = c2e–x. Hence we can write

By appealing to the definition of continuity at a point it is possible to determine c2 so that the foregoing function is continuous at x = 1. The requirement that y(x) = y(1) implies that c2e–1 = 1 − e–1 or c2 = e − 1. As seen in FIGURE 2.3.3, the piecewise defined function

(13)

is continuous on the interval [0, ). ≡

A curve begins at the origin and goes up and to the right with decreasing steepness to a high point. It then goes down and to the right with decreasing steepness and approaches the horizontal axis.

FIGURE 2.3.3 Graph of function in (13) of Example 5

It is worthwhile to think about (13) and Figure 2.3.3 a little bit; you are urged to read and answer Problem 53 in Exercises 2.3.

Error Function

In mathematics, science, and engineering, some important functions are defined in terms of nonelementary integrals. Two such special functions are the error function and complementary error function:

See Appendix A.

(14)

From the known result , we can write . Using the additive interval property of definite integrals we can rewrite the last result in the alternative form

(15)

It is seen from (15) that the error function erf(x) and complementary error function erfc(x) are related by the identity

. (16)

Because of its importance in probability, statistics, and applied partial differential equations, the error function has been extensively tabulated. Note that erf(0) = 0 is one obvious function value. Numerical values of erf(x) can also be found using a CAS such as Mathematica.

If we are solving an initial-value problem (9) and recognize that indefinite integration of the right-hand side of (7) would lead to a nonelementary integral, then as we saw in Example 6 of Section 2.2 it is convenient to use instead definite integration over the interval [x0, x]. The last example illustrates that this procedure automatically incorporates the initial condition at x0 into the solution of the DE, in other words, we do not have to solve for the constant c in its general solution.

EXAMPLE 6 The Error Function

Solve the initial-value problem .

SOLUTION

The differential equation is already in standard form, and so we see that the integrating factor is e∫(−2x dx) = e−x2. Multiplying both sides of the equation by this factor then gives , which is the same as

. (17)

Because indefinite integration of both sides of equation (17) leads to the nonelementary integral , we identify x0 = 0 and use definite integration over the interval [0, x]:

Using the initial condition y(0) = 1 the last expression yields the solution

. (18)

Then by inserting the factor into this solution in the following manner:

we see from (14) that (18) can be rewritten in terms of the error function as

. (19)

The graph of solution (19), shown in FIGURE 2.3.4, was obtained with the aid of a CAS. ≡

A curve enters the third quadrant and goes up and to the right with decreasing steepness through approximately (negative 1, negative 2) to (0, 1). It then goes up and to the right with increasing steepness through (0.5, 2).

FIGURE 2.3.4 Graph of (19) in Example 6

See Problems 41–48 in Exercises 2.3.

Use of Computers

Some computer algebra systems are capable of producing explicit solutions for some kinds of differential equations. For example, to solve the equation y′ + 2y = x, we use the input commands

Translated into standard symbols, the output of each program is

REMARKS

(i) Occasionally a first-order differential equation is not linear in one variable but is linear in the other variable. For example, the differential equation

is not linear in the variable y. But its reciprocal

    or    

is recognized as linear in the variable x. You should verify that the integrating factor e∫(–1)dy = e–y and integration by parts yield an implicit solution of the first equation: x = −y2 − 2y − 2 + cey.

(ii) Because mathematicians thought they were appropriately descriptive, certain words were “adopted” from engineering and made their own. The word transient, used earlier, is one of these terms. In future discussions the words input and output will occasionally pop up. The function f in (2) is called the input or driving function; a solution of the differential equation for a given input is called the output or response.

(iii) The term special functions mentioned in conjunction with the error function on page 58 also applies to the sine integral function and the Fresnel sine integral function introduced in Problems 47 and 48 in Exercises 2.3. “Special Functions” is actually a well-defined field of study in mathematics. The Fresnel sine integral function is one of two such special integral functions that the French engineer/physicist Augustin-Jean Fresnel (1788–1827) encountered in his study of optics. More special functions are studied in Section 5.3.

2.3 Exercises Answers to selected odd-numbered problems begin on page ANS-2.

In Problems 1–24, find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.

In Problems 25–32, solve the given initial-value problem. Give the largest interval I over which the solution is defined.

  1. ,
  2. ,
  3. ; L, R, E, and i0 constants
  4. ; K, Tm, and T0 constants
  5. ,
  6. ,
  7. ,
  8. ,
    [Hint: In your solution let .]

In Problems 33–36, proceed as in Example 5 to solve the given initial-value problem. Use a graphing utility to graph the continuous function y(x).

  1. where

  2. where

  3. where

  4. where

In Problems 37 and 38, proceed as in Example 5 to solve the given initial-value problem. Use a graphing utility to graph the continuous function y(x).

  1. where

  2. where

In Problems 39 and 40, the given differential equation is not linear in y. Nevertheless, find a general solution of the equation.

In Problems 41 and 42, proceed as in Example 6 and express the solution of the given initial-value problem in terms of erf(x) (Problem 41) and erfc(x) (Problem 42).

In Problems 43–46, proceed as in Example 6 and express the solution of the given initial-value problem in terms of an integral-defined function.

  5. The sine integral function is defined as

    where the integrand is defined to be 1 at x = 0. Express the solution of the initial-value problem

    in terms of Si(x).

  6. The Fresnel sine integral function is defined as

    Express the solution of the initial-value problem

    in terms of S(x).

Discussion Problems

  1. Reread the discussion following Example 1. Construct a linear first-order differential equation for which all nonconstant solutions approach the horizontal asymptote y = 4 as x → .
  2. Reread Example 2 and then discuss, with reference to Theorem 1.2.1, the existence and uniqueness of a solution of the initial-value problem consisting of xy′ − 4y = x6ex and the given initial condition.

    (a) y(0) = 0

    (b) y(0) = y0,     y0 > 0

    (c) y(x0) = y0,     x0 > 0,     y0 > 0

  3. Reread Example 3 and then find the general solution of the differential equation on the interval (−3, 3).
  4. Reread the discussion following Example 4. Construct a linear first-order differential equation for which all solutions are asymptotic to the line y = 3x − 5 as x → .
  5. Reread Example 5 and then discuss why it is technically incorrect to say that the function in (13) is a solution of the IVP on the interval [0, ).

  6. (a) Construct a linear first-order differential equation of the form xy′ + a0(x)y = g(x) for which yc = c/x3 and yp = x3. Give an interval on which y = x3 + c/x3 is the general solution of the DE.

    (b) Give an initial condition y(x0) = y0 for the DE found in part (a) so that the solution of the IVP is y = x3 − 1/x3. Repeat if the solution is y = x3 + 2/x3. Give an interval I of definition of each of these solutions. Graph the solution curves. Is there an initial-value problem whose solution is defined on the interval (−, )?

    (c) Is each IVP found in part (b) unique? That is, can there be more than one IVP for which, say, y = x3 − 1/x3, x in some interval I is the solution?

  7. In determining the integrating factor (5), there is no need to use a constant of integration in the evaluation of ∫P(x) dx. Explain why using ∫P(x) dx + c has no effect on the solution of (2).

Mathematical Models

  1. Radioactive Decay Series     The following system of differential equations is encountered in the study of the decay of a special type of radioactive series of elements:

    where λ1 and λ2 are constants. Discuss how to solve this system subject to x(0) = x0, y(0) = y0. Carry out your ideas.

  2. Heart Pacemaker     A heart pacemaker consists of a switch, a battery of constant voltage E0, a capacitor with constant capacitance C, and the heart as a resistor with constant resistance R. When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart. During the time the heart is being stimulated, the voltage E across the heart satisfies the linear differential equation

    Solve the DE subject to E(4) = E0.

Computer Lab Assignments

  1. (a) Use a CAS to graph the solution curve of the initial-value problem in Problem 42 on the interval

    (b) Use tables or a CAS to find the value y(2).

  2. (a) Use a CAS to graph the solution curve of the initial-value problem in Problem 47 on the interval [

    (b) Use a CAS to find the value of the absolute maximum of the solution y(x) on the interval.

  3. (a) Use a CAS to graph the solution curve of the initial-value problem in Problem 48 on the interval

    (b) It is known that Fresnel sine integral as and What does the solution approach as As

    (c) Use a CAS to find the values of the absolute maximum and the absolute minimum of the solution y(x) on the interval.