# Higherorder Linear Differential Equations

The methods of solution introduced above are readily extended to an n th-order linear differential equation. With constant coefficients and a constant term, such an equation can be written generally as

(15.50) /">(/) + ay-»(t) + • • ■ + an_xy'(t) + any = b

### Finding the Solution

In this case of constant coefficients and constant term, the presence of the higher derivatives does not materially affect the method of finding the particular integral discussed earlier.

If we try the simplest possible type of solution, y = k, we can see that all the derivatives from_y'(0 to_y(n)(i) will be zero; hence (15.50) will reduce to ank = b, and we can write yP = k=^ {an + 0) [cf. (15.3)]

In case an = 0, however, we must try a solution of the form y = kt. Then, since y'{t) = k, but all the higher derivatives will vanish, (15.50) can be reduced to an^xk = b, thereby yielding the particular integral y =kt=-^-t (a„ = 0;an_ , * 0) [cf. (15.3')]

If it happens that an = an_] = 0, then this last solution will fail, too; instead, a solution of the form^ = kt2 must be tried. Further adaptations of this procedure should be obvious.

As for the complementary function, inclusion of the higher-order derivatives in the differential equation has the effect of raising the degree of the characteristic equation. The complementary function is defined as the general solution of the reduced equation

(15.51) y»'(0 + "(0 + ■ • • + an_xy'{t) + any = 0

Trying y = Ae" (+ 0) as a solution and utilizing the knowledge that this implies y'(t) = rAer\ y"(t) = r2Aer',..., y{"\t) = r"Aerl, we can rewrite (15.51) as

Aen(r" + a{r"-] + ■ • ■ + a„ . ,r + a J = 0

This equation is satisfied by any value of r which satisfies the following (nth-degree polynomial) characteristic equation

(15.51') r" + + ■■■ + a„_xr + a„ = 0

There will, of course, be n roots to this polynomial, and each of these should be included in the general solution of (15.51). Thus our complementary function should in general be in the form

As before, however, some modifications must be made in case the n roots are not all real and distinct. First, suppose that there are repeated roots, say, r, = r2 = r3. Then, to avoid "collapsing," we must write the first three terms of the solution as A,er'' + A2ter'' + A3l2er'' [cf. (15.9)]. In case we have r4 = rx as well, the fourth term must be altered to A4t3er>', etc.

Second, suppose that two of the roots are complex, say, r5, r6 = h ± vi then the fifth and sixth terms in the above solution should be combined into the following expression:

By the same token, if two distinct pairs of complex roots are found, there must be two such trigonometric expressions (with a different set of values of h, v, and two arbitrary constants for each).* As a further possibility, if there happen to be two pairs of repeated complex roots, then we should use eht as the multiplicative term for one but use teh' for the other. Also, even though h and v have identical values in the repeated complex roots, a different pair of arbitrary constants must now be assigned to each.

Once y and yc are found, the general solution of the complete equation (15.50) follows easily. As before, it is simply the sum of the complementary function and the particular integral: y(t) = yc + y In this general solution, we can count a total of n arbitrary constants. Thus, to definitize the solution, as many as n initial conditions will be required.

### Example 1 Find the general solution of

* It is of interest to note that, inasmuch as complex roots always come in conjugate pairs, wc can be sure of having at 1east one real root when the differential equation is of an odd order, i.e., when n is an odd number.

v(4)(0 + 6 y"'(t) + 14 y"(t) + 16 y'(t) + 8>> = 24

The particular integral of this fourth-order equation is simply 24 *

Its characteristic equation is, by (15.51'), r4 + 6 r3 + 14 r2 + \6r + 8 = 0 which can be factored into the form

From the first two parenthetical expressions, we can obtain the double roots r, = r2 = — 2, but the last (quadratic) expression yields the pair of complex roots r3, r4 = — 1 ± i, with h = — 1 and v = 1. Consequently, the complementary function is yc = Ate~2' + A2te~2' + e~'(A3cost + ^4sin t) and the general solution is y(t) = Axe~2' + A2te2' + e '(A}cosl + ^4sin t) + 3

The four constants^,, A2, A3, and A4 can be definitized, of course, if we are given four initial conditions.

Note that all the characteristic roots in this example either are real and negative or are complex and with a negative real part. The time path must therefore be convergent, and the intertemporal equilibrium is dynamically stable.

### Convergence and the Routh Theorem

The solution of a high-degree characteristic equation is not always an easy task. For this reason, it should be of tremendous help if we can find a way of ascertaining the convergence or divergence of a time path without having to solve for the characteristic roots. Fortunately, there does exist such a method, which can provide a qualitative (though nongraphic) analysis of a differential equation. This method is to be found in the Routh theorem* which states that:

The real parts of all of the roots of the n th-degree polynomial equation a0r" + atrn~] + • • ■ + an__ + a„ = 0

are negative if and only if the first n of the following sequence of determinants a h