L a CCLDO with characteristic polynomial f(r); so L = f(D).
To solve Ly = e^{sx} recall that Le^{sx} = f(s)e^{sx}, so if f(s) is nonzero a particular solution is given by
xp = e^{sx}/f(s).
First example: y" + 4y' + 4y = e^{3x}: f(r) = (r+2)^2, f(3) = 25, so xp = e^{3x}/25 and the general solution is
x = e^{3x}/25 + ae^{-2x} + bxe^{-2x}.
Second example: Ly = sin(3x), say with the same L = D^2 + 4D + 4I. Recommended procedure: replace this with a different problem: Lz = e^{3ix}. Then y = Im z. s = 3i; f(3i) = (3i+2)^2 = -5 + 12i, so
zp = e^{3ix}/f(3i) = -(5 + 12i)(cos(3x) + i sin(3x))/169
yp = Im zp = -(12 cos(3x) + 5 sin(3x))/169.
y = -(12 cos(3x) + 5 sin(3x))/169 + ae^{-2x} + bxe^{-2x}.
In Ly = e^{sx} for a general CCLDO L = f(D), try y = e^{sx}u, where if s happens to be a root of f(r) the function u may not be constatn. Then by the ESL
e^{sx} = f(D)(e^{sx}u) = e^{sx}f(D+sI)u
so we wish to solve 1 = f(D+sI)u: we've eliminated the exponential at the expense of changing the CCLDO. This generalizes instantly to: To solve Ly = e^{sx}q(x), the expression y = e^{sx}u leads to
e^{sx}q(x) = f(D)(e^{sx}u) = e^{sx}f(D+sI)u
so we wish to solve f(D+sI)u = q(x). If q(x) happens to be polynomial, then Monday's work does this for us.
Example: y" + 4y' + 4y = e^{-2x}(x^2 + 1). Write y = e^{-2x}u, so that
e^{-2x}(x^2 + 1) = f(D)(e^{-2x}u) = e^{-2x}f(D-2I)u.
Since f(D-2I) = ((D-2I)+2I)^2 = D^2, we are faced with D^2 u = x^2 + 1. This (we know from Monday's work) has a unique solution up = ax^4 + bx^3 + cx^2, which is easy to find in this case by integrating twice: up = (x^4)/12 + (x^2)/2. Thus
yp = e^{-2x}((x^4)/12 + (x^2)/2).
The general expression is given in the handout, but I recommend using this algorithm employing the ESL.
Application: Mass-dashpot-spring driven by a periodic force, with period omega0/(2 pi). We can get the general such driving term by using linear combinations of the real and imaginary parts of e^{i omega0 t), so let's focus on
y" + py' + qy = e^{i omega0 t}.
f(r) = r^2 + pr + q, and as long as f(i \omega0) is not zero a solution is given by
xp = e^{i omega0 t}/f(i omega0).
f(i omega0) = (q - (omaga0)^2) + i p omega0 so our assumption is (p,q) is not (0,(omega0)^2). Express f(i omega0) in polar terms:
There is a triangle with adjacent side q - (omega0)^2, opposite side p omega0, angle phi, and hypotenuse r. Then
f(i omega0) = r e^{i phi}
xp = e^{i(omega0 t - phi)}/r
That is, we have a periodic solution, with amplitude 1/r times the driving amplitude, and with phase lag phi.
Plotted against omega0, the amplitude multiplier starts at 1/q when omega0 = 0, grows to a maximum when omega0 = sqrt(q - (p^2)/2), and then falls back to 0 as omega0 --> infinity.
The phase lag starts at 0 with slope zero, grows through a point of inflection and approaches pi as omega0 --> infinity.