## Firstorder Linear Differential Equations With Constant Coefficient And Constant Term

The first derivative dy/dt is the only one that can appear in a first-order differential equation, but it may enter in various powers: dy/dt, {dy/dt)2, or (dy/dt)3. The highest power attained by the derivative in the equation is referred to as the degree of the differential equation. In case the derivative dy/dt appears only in the first degree, and so does the dependent variable v, and furthermore, no product of the iorm y(dy / dt) occurs, then the equation is said to be linear. Thus a first-order linear differential equation will generally take the form*

where u and w are two functions of t. as is v. In contrast to dy/dt and y, however, no restriction whatsoever is placed on the independent variable Thus the function u and w may very well represent such expressions as t2 and e' or some more complicated functions of t\ on the other hand, u and w may also be constants.

This last point leads us to a further classification. When the function u (the coefficient of the dependent variable y) is a constant, and when the function w is a constant additive term, (14.1) reduces to the special case of a first-order linear differential equation with constant coefficient and constant term. In this section, we shall deal only with this simple variety of differential equations.

The Homogeneous Case

If u and w are constant functions and if w happens to be identically zero, (14.1) will become

where a is some constant. This differential equation is said to be homogeneous, on account of the zero constant term (compare with homogeneous-equation systems). More accurately, this equation is homogeneous because every term therein is uniformly in the first degree in terms of y and dy/df, in particular, the constant 0 — unlike any other constant—can be regarded as in the first degree in terms of y because 0 v = 0.

Equation (14.2) can be written alternatively as

But you will recognize that the differential equation (13.16) we met in the Domar model is precisely of this form. Therefore, by analogy, we should be able to write the solution of (14.2) or (14.2') immediately as follows:

In (14.3), there appears an arbitrary constant A; therefore it is a general solution.

* Note that the derivative term dy/dt in (14.1) has a unit coefficient. This is not to imply that it can never actually have a coefficient other than one, but when such a coefficient appears, we can always "normalize" the equation bv dividing each term bv the said coefficient. For this reason, the form given in (14.1) may nonetheless be regarded as a general representation.

When any particular value is substituted for A, the solution becomes a particular solution of (14.2). There is an infinite number of particular solutions, one for each possible value of A, including the value y(0). This latter value, however, has a special significance: v(0) is the only value that can make the solution satisfy the initial condition. Since this represents the result of definitizing the arbitrary constant, we shall refer to (14.3') as the definite solution of the differential equation (14.2) or (14.2').

You should observe two things about the solution of a differential equation: (1) the solution is not a numerical value, but rather a function y(t)—a time path if t symbolizes time; and (2) the solution y(t) is free of any derivative or differential expressions, so that as soon as a specific value of t is substituted into it, a corresponding value of y can be calculated directly.

The Nonhomogeneous Case

When a nonzero constant takes the place of the zero in (14.2), we have a nonhomogeneous linear differential equation

The solution of this equation will consist of the sum of two terms, one of which is called the complementary function (which we shall denote by yc), and the other known as the particular integral (to be denoted by y ). As will be shown, each of these has a significant economic interpretation. Here, we shall present only the method of solution; its rationale will become clear later.

Even though our objective is to solve the non homogeneous equation (14.4), frequently we shall have to refer to its homogeneous version, as shown in (14.2). For convenient reference, we call the latter the reduced equation of (14.4). The nonhomogeneous equation (14.4) itself can accordingly be referred to as the complete equation. It turns out that the complementary function^ is nothing but the general solution of the reduced equation, whereas the particular integral y is simply any particular solution of the complete equation.

Our discussion of the homogeneous case has already given us the general solution of the reduced equation, and we may therefore write yc = Ae-"' [by (14.3)]

What about the particular integral? Since the particular integral is any particular solution of the complete equation, we can first try the simplest possible type of solution, namely, y being some constant (y = k). If y is a constant, then it follows that dy/dt = 0, and (14.4) will become ay = b, with the solution y = b/a. Therefore, the constant solution will work as long as a j= 0. In that case, we have yp = ~ (a*0)

The sum of the complementary function and the particular integral then con stitutes the general solution of the complete equation (14.4):

(14.5) y(t) = yc + yp = Ae~a' + ^ [general solution, case of a + 0]

What makes this a general solution is the presence of the arbitrary constant A. We may, of course, definitize this constant by means of an initial condition. Let us say that 7 takes the value >(0) when t = 0. Then, by setting t = 0 in (14.5), we find that h h v(0) = A+ - and A = y(0) -a a

Thus we can rewrite (14.5) into

It should be noted that the use of the initial condition to definitize the arbitrary constant is—and should be—undertaken as the final step, after we have found the general solution to the complete equation. Since the values of both yc and yp are related to the value of _y(0), both of these must be taken into account in definitizing the constant A.

Example 1 Solve the equation dy/dt + 2y = 6, with the initial condition y(0) = 10. Here, we have a = 2 and b = 6; thus, by (14.5'), the solution is y(t) = [10 - 3}e'2' + 3 = le'1' + 3

Example 2 Solve the equation dy/dt + 4y = 0, with the initial condition y(0) = 1. Since a = 4 and b = 0, we have y(t) = [1 - 0]e~4' + 0 = e-4'

The same answer could have been obtained from (14.3'), the formula for the homogeneous case. The homogeneous equation (14.2) is merely a special case of the nonhomogeneous equation (14.4) when b = 0. Consequently, the formula (14.3') is also a special case of formula (14.5') under the circumstance that b = 0.

What if a = 0, so that the solution in (14.5') is undefined? In that case, the differential equation is of the extremely simple form

By straight integration, its general solution can be readily found to be (14.7) y(t) = bt + c where c is an arbitrary constant. The two component terms in (14.7) can, in fact, again be identified as the complementary function and the particular integral of the given differential equation, respectively. Since a = 0, the complementary function can be expressed simply as yc = Ae~al = Ae° = A (A = an arbitrary constant)

As to the particular integral, the fact that the constant solution y = k fails to work in the present case of a = 0 suggests that we should try instead a noncon-stant solution. Let us consider the simplest possible type of the latter, namely, y = kt.liy = kt, then dy/dt = k, and the complete equation (14.6) will reduce to k = b, so that we may write yp = bt (a = 0)

Our new trial solution indeed works! The general solution of (14.6) is therefore (14.7') y(t) = y\ + y = A + bt [general solution, case of a = 0]

which is identical with the result in (14.7), because c and A are but alternative notations for an arbitrary constant. Note, however, that in the present case, is a constant whereas yp is a function of time—the exact opposite of the situation in (14.5).

By definitizing the arbitrary constant, we find the definite solution to be (14.7") y(t) = y(0) + bt [definite solution, case of a = 0]

Example 3 Solve the equation dy/dt = 2, with the initial condition y(0) = 5. The solution is, by (14.7"), y(t) = 5 + 2/

Verification of the Solution

It is true of all solutions of differential equations that their validity can always be checked by differentiation.

If we try that on the solution (14.5'), we can obtain the derivative dy dt m--

When this expression for dy/dt and the expression for>'(?) as shown in (14.5') are substituted into the left side of the differential equation (14.4), that side should reduce exactly to the value of the constant term b on the right side of (14.4) if the solution is correct. Performing this substitution, we indeed find that m-l

Thus our solution is correct, provided it also satisfies the initial condition. To check the latter, let us set t = 0 in the solution (14.5'). Since the result

is an identity, the initial condition is indeed satisfied.

It is recommended that, as a final step in the process of solving a differential equation, you make it a habit to check the validity of your answer by making sure (1) that the derivative of the time path y(t) is consistent with the given differentia] equation and (2) that the definite solution satisfies the initial condition.

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