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mathwonk
#2
Mar15-06, 09:18 PM
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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.