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 Recognitions: Homework Help Science Advisor more details: method 1: suppose the RHS of our ode is a polynomial of degree < n. Then D^n =0 on that function, so if the LHS factors with factors such as D-a, then since (1-D)(1+D+D^2+D^3+...+D^[n-1]) = 1-D^n, then also (1-D/a)(1+D/a+[D/a]^2+[D/a]^3+...+[D/a]^[n-1]) = 1-[D/a]^n, hence a(1-D/a)(1+D/a+[D/a]^2+[D/a]^3+...+[D/a]^[n-1]) = (a-D)(1+D/a+[D/a]^2+[D/a]^3+...+[D/a]^[n-1]) = a(1-[D/a]^n). Hence (D-a)(-1/a)(1+D/a+[D/a]^2+[D/a]^3+...+[D/a]^[n-1]) = (1-[D/a]^n). Thus if we want to solve (D-a)y = f where f is a polynomial of degree < n,m then taking y = (-1/a)(1+D/a+[D/a]^2+[D/a]^3+...+[D/a]^[n-1])f, gives us (D-a)y = (1-[D/a]^n)f = f, since (D/a)^nf = 0. Repeat for each factor (D-a) of the differential operator. I.e. this method inverts any operator which is a product of operators of form D-a, i.e. any linear diff op with constant coefficients.