A NOTE ON LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

Richard E. Bayne, James E. Joseph, Myung H. Kwack, and Thomas H. Lawson

As is well–known, when a ≠ b,

(1)  y = c_{1}e^{ax} + c_{2}e^{bx}

is a general solution of the differential equation

(2)  y′′ − (a + b)y′ + aby = 0.

In [1], Euler points out the fact that

(3)  y = c_{1}e^{ax} + c_{2}xe^{ax}

is a general solution of the differential equation

(4)  y′′ − 2ay′ + a^{2}y = 0,

although easily checked, is not so easily motivated. He provided a technique to show how this fact arises. In this paper, we offer a technique which shows how solutions (1) and (3) arise as general solutions of (2) and (4), respectively. We then use this technique, along with induction, to give the form of a general solution of the nth order linear differential equation with complex coefficients.

Motivation for Technique. If a and b are complex constants, a general solution of the differential equation

(5)  y′′ − (a + b)y′ + aby = q(x)

is

(6)  y = e^{ax} ∫ e^{(b−a)x}(∫ e^{−bx}q(x)dx)dx.

\underline{Proof}. Writing the equivalent form

(7)  (y′ − ay)′ − b(y′ − ay) = q(x)

for the differential equation in (5) and multiplying through by the integrating factor e^{−bx}, we find that

(8)  y′ − ay = e^{bx} ∫ e^{−bx}q(x)dx

is a general solution of (7). Multiplying through by the integrating factor e^{−ax} in (8), we obtain

(9)  y = e^{ax} ∫ e^{(b−a)x}(∫ e^{−bx}q(x)dx)dx

as a general solution of (5).

Utilization of the same technique and induction yields the following theorem.

\underline{Theorem}. If n is a positive integer, a_{1}, …, a_{n} are complex constants, s_{0} = 1, and for k = 1, …, n, s_{k} = ∑_{ϕ} ∏ a_{ϕ(j)}, where ϕ is a strictly increasing function from {1, …, k} to {1, …, n}, then a general solution to

(10)  ∑_{k=0}^{n}(−1)^{k}s_{k}y^{(n−k)} = q(x)

is

y = e^{a_{n}x} ∫ (∫ e^{(a_{n−1}−a_{n})x}(∫ ⋯ (∫ e^{(a_{1}−a_{2})x} ∫ e^{−a_{1}x}q(x)dx)dx ⋯ dx)dx)dx.

\underline{Proof}. (By Induction). For n = 1, the hypothesis of the statement to be proved is that

(11)  y′ − ay = q(x).

The equation in (11) has

(12)  y = e^{ax} ∫ e^{−ax}q(x)dx

as a general solution. Suppose the statement holds for the integer n and that

(13)  ∑_{k=0}^{n+1}(−1)^{k}s_{k}y^{(n+1−k)} = q(x),

where a_{k} for k = 1, …, n + 1 and s_{k} are as described above. Then by the linearity of differentiation,

(14)  (∑_{k=0}^{n}(−1)^{k}s_{k}y^{(n−k)})′ − a_{1}(∑_{k=0}^{n}(−1)^{k}s_{k}y^{(n−k)}) = q(x),

where s_{0} = 1 and for each k = 1, …, n, s_{k} = ∑_{ϕ} ∏ a_{ϕ(j)}, where each ϕ is a strictly increasing function from {1, …, k} to {2, …, n+1}. From the case for 1, a general solution of the equation in (14) is

∑_{k=0}^{n}(−1)^{k}s_{k}y^{(n−k)} = e^{a_{1}x} ∫ e^{−a_{1}x}q(x)dx

and from the induction hypothesis, we obtain

y = e^{a_{n+1}x} ∫ [∫ e^{(a_{n}−a_{n+1})x}[∫ ⋯ [∫ e^{(a_{1}−a_{2})x} ∫ e^{−a_{1}x}q(x)dx]dx ⋯ dx]dx]dx

as a general solution of (13) and the proof is complete.

The following corollaries arise naturally from the Theorem. The proofs are omitted.

\underline{Corollary}. If n is a positive integer and a is a complex constant, a general solution to

∑_{k=0}^{n} \binom{n}{k}(−1)^{k}a^{k}y^{(n−k)} = q(x)

is

y = e^{ax} \underbrace{∫ (∫ ⋯ (∫}_{n integrals} e^{−ax}q(x)dx) ⋯ dx)dx.

\underline{Corollary}. If n is a positive integer and a is a complex constant, then

∑_{k=0}^{n} \binom{n}{k}(−1)^{k}a^{k}y^{(n−k)} = 0

has

y = e^{ax} ∑_{k=0}^{n−1} c_{k}x^{n−1−k}

as a general solution.

When viewed in the terminology of differential operators, the proof of the theorem is perhaps a bit more transparent. In this terminology, (10) may be written as

(∏_{k=1}^{n}(D − a_{k}))y = q(x)

and

(∏_{k=1}^{n+1}(D − a_{k}))y = D([ ∏_{k=2}^{n+1}(D − a_{k})]y) − a_{1}(∏_{k=2}^{n+1}(D − a_{k}))y,

from which the inductive step follows.

\underline{Reference}

1. R. Euler, “A Note on a Differential Equation,” Missouri Journal of Mathematical Sciences, 1 (1989), 26–27.

Richard E. Bayne James E. Joseph Myung H. Kwack Department of Mathematics Howard University Washington, D.C. 20059 email: bayne@scs.howard.edu

Thomas H. Lawson Center for Academic Reinforcement Howard University Washington, D.C. 20059