(Solution by factorization of the operator) We motivated the idea of seeking solutions to...

(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).

Solve y″+ 2y′ + y = 0 by factoring the operator as (D − 1 )(D − 2)y = 0. Solve the latter by the method outlined in (b): Setting (D − 2)y = u, solve (D − l)u = u′ − u = 0 for u(x). Then, knowing u(x), solve (D − 2)y − u, namely, y′ − 2y − u(x), for y(x).

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.