(A handout in dvi, postscript, or pdf formats.)
Once we have
y" + p(x)y' + q(x)y = r(x) (*)
we become interested in what the LHS does to a general function y. This is an example of an "operator": y gets sent to Ly = y" + p(x)y' + q(x)y.
Other examples: Dy = y'; Ay = y+1; M_f y = f.y , where f is a given functional; and many others.
Warning: Dx^3 = 3x^2 isn't intended to imply that D8 = 12 ("take x=2"). D operates on the whole function; you can't compute the derivative at 2 knowing only the value at 2.
L is a "Differential Operator": it's made out if differentiation.
An operator F is "linear" if F(u+v) = Fu + Fv and F(cu) = cFu. For example D and M_f is linear, but A is not.
We'll study LDO's (linear differential operators), especially CCLDO's, those with constant coefficients. Such an operator L has a characteristic polynomial f , and L = f(D). Thus if we factor f(r) (as (r-r1)(r-r2) in the second order case) then we can factor L itself accordingly: f(D) = (D-r1I)(D-r2I).
Operators don't generally commute with each other. For example D M_f - M_f D = M_{f'}. But any two CCLDOs do.
Exponentials play an important role. De^{rx} = re^{rx}, and so
f(D)e^{rx} = f(r)e^{rx}.
In particular, f(D)e^{r1 x} = f(r1) e^{r1 x} = 0, something we have exploited repeatedly.
We'll want to generalize this to study f(D)(e^{rx} u) for a function u. First,
D(e^{rx} u) = e^{rx}(D + rI) u
and then using this repeatedly
D^n (e^{rx} u) = e^{rx}(D + rI)^n u.
Adding up,
f(D)(e^{rx} u) = e^{rx} f(D + rI) u (ESL)
the Exponential Shift Law. I used this to compute
D^2(e^{2x} cos(x))