(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).
To solve (11.1), let (D − λ2)y = u, so that (18) reduces to the first-order equation
Solve (11.2) for u, put that u on the right−hand side of , which is again of first order, and solve the latter for y. Show that if λ1, λ2 are distinct, then the result is given by (23), whereas if they are repeated, then the result is.
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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