(Solution by factorization of the operator) We motivated the idea of seeking solutions to (1) in the form eλx by observing that the general solution of the first-order equation y′+ a1y = 0 is an exponential, , and wondering if higher-order equations might admit exponential solutions too. A more compelling approach is as follows. Having already seen that the first-order equation admits an exponential solution, consider the second-order equation (18).
Show that (18) can be written, equivalently, as
(D − λ1)(D − λ2)y = 0, (11.1)
where D denotes d/dx, and λ1 and λ2 are the two roots of λ2 + a1 λ + a2 = 0. NOTE: In (11.1) we accomplish a factorization of the original differential operator L = D2+a1 D+a2 as (D − λ1)(D − λ2). By the left-hand side of (11.1), we mean (D − λ1)((D − λ2)y). That is, first let the operator to the left of y (namely, D − λ2) act on y, then let the operator to the left of (D − λ2)y (namely, D − λ1) act on that.
NOTE: Similarly for higher-order equations. For instance, y′″− 2y″ −y′ + 2y = (D− 2)(D + 1 )(D − l)y = 0 can be solved by setting (D + 1)(D − 1 )y = u and solving (D − 2)u = 0 for u(x); then set (D − 1)y = v and solve (D + l)v = u for v(x). finally, solve (D − l)y = v for y(x). The upshot is that the solution of an nth-order linear homogeneous differential equation with constant coefficients can be reduced to the solution of a sequence of n first-order linear equations.
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