operator factorization

You know that factoring a number, like 12 = 3 times 4, breaks a hard object into simple pieces. The same trick works on a constant-coefficient differential operator. Instead of confronting a whole nth-order operator at once, you split it into a product of first-order pieces, each of which you can handle by the easy first-order linear method. Factoring trades one hard step for several easy ones.

Here is the mechanism. The operator L = D^2 - 3D + 2 factors exactly as its characteristic polynomial does: (D - 1)(D - 2). To read the factored form, work from the inside out — (D - 1)(D - 2)[y] = 0 means first form u = (D - 2)[y] = y' - 2y, then require (D - 1)[u] = u' - u = 0. The inner factor D - 2 already reveals the solution e^(2x) (it solves y' - 2y = 0), the outer factor D - 1 reveals e^(x), and the two combine into the full solution. Each linear factor (D - r) hands you one exponential e^(rx), so factoring the operator IS reading off the solutions — it is the characteristic-root method dressed in operator clothing.

Beyond solving, factoring lets you peel a high-order problem apart one order at a time, and it is the conceptual root of reduction of order and of why repeated factors (D - r)^m produce the x e^(rx) family. The crucial honesty: this clean factoring is a CONSTANT-coefficient privilege. For variable coefficients the factors do not commute, so (D - a(x))(D - b(x)) is generally not (D - b(x))(D - a(x)), and you cannot factor freely or treat the operator as a tidy polynomial. Reserve operator factorization for the constant-coefficient world.