I have noticed the following trivial method for solving special constant coeff. ode's, but cannot find it in the usual ode books, (i had better look in courant though, as everything else is in there.) given a cc ode of form (D-a)(D-b)y = f, where f is a solution of some cc ode, i.e. where g is annihilated by some polynomial P(D) in D, if the polynomial P does not have D-a or D-b as a factor (which case is easily done separately), then when expressed as a polynomial in D-a, P has a non zero constant term, hence can be solved as follows: (D-a)Q(D) = c where c is not zero. Hence we get that (D-a)^-1 = (1/c)Q(D). Thus y = (1/c)Q(D)f. Repeat for D-b. example: This is not exactly the same type but is an easy example: To solve (D^2 +D + 1)y = f, where f is a polynomial of degree 2, just take y = (1-D)f. I.e. mod D^3, (D^2+D+1)^-1 = 1-D. This is just the usual principle that the inverse of an invertible matrix with given minimal polynomial, can be computed by solving the minimal polynomial for the non zero constant term, then dividing out the variable, and dividing by the constant term. e.g. if T^2 + T - 2 = 0, then T^2+T = 2, so T[T+1] = 2, so T^(-1) = (1/2)(T+1). Surely a solution method as obvious as this was standard hundreds of years ago, but i have not found it in any of over 15 books I have consulted. In the example above, if the RHS of the equation is a polynomial of degree < n, then D^n annihilates it, so the inverse of (1-D) is 1+D+D^2+...+D^n. This can be modified to give the inverse of any (D-a), hence any polynomial in D. On this forum there is surely someone who has seen this method. If so please give me a reference. This obviously related to the annihilator method, or undetermined coefficients method, but obviates the need to solve for any constants as they are provided automatically.