(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″− 4y = 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.
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