Lecture Notes in Differential Equations - Bruce E. Shapiro

285
This is a homogeneous linear equation with constant coefficients; the solution
is
y = C 1 cos x + C 2 sin x (30.18)
= C 1 cos ln t + C 2 sin ln t. (30.19)
The solution we found to (30.5) in example 30.2 is not even defined, much
less analytic at t = 0. So there is no power series solution. However, it
turns out that we can still make use of the result we found in example 30.1,
which said that the power series solution only “exists” when k 2 + 1 = 0,
if we drop the requirement that k be an integer, since then we would have
k = ±i, and our “series” solution would be
y = ∑ k∈S
c k t k = c −i t −i + c i t i (30.20)
where the sum is taken over the set S = {−i, i}.
Example 30.3. Show that the “series” solution given by (30.20) is equivalent
to the solution we found in example 30.2.
Rewriting (30.20),
y = c −i t −i + c i t i (30.21)
where
= c −i exp ln t −i + c i exp ln t i (30.22)
= c −i exp(−i ln t) + c i exp(i ln t) (30.23)
= c −1 [cos(−i ln t) + i sin(−i ln t)] + c i [cos(i ln t) + i sin(i ln t)] (30.24)
= c −1 cos(i ln t) − ic −1 sin(i ln t) + c i cos(i ln t) + ic i sin(i ln t) (30.25)
= (c i + c −1 ) cos(i ln t) + i(c i − c −1 ) sin(i ln t) (30.26)
= C 1 cos(i ln t) + C 2 sin(i ln t) (30.27)
C 1 = c −i + c i (30.28)
C 2 = i(c i − c −i ) (30.29)
Since this is a linear system and we can solve for c i and c −i , namely
c i = 1 2 (C 1 − iC 2 ) (30.30)
c −i = 1 2 (C 1 + iC 2 ) (30.31)
Given either solution, we can find constants that make it the same as the
other solution, hence the two solutions are identical.