Lecture Notes in Differential Equations - Bruce E. Shapiro

280 LESSON 29. REGULAR SINGULARITIES
because y 1 = t r = t (1−α)/2 .
Thus the second solution is y 2 = y 1 ln |t|, and the the general solution to
the Euler equation in case 2 becomes
y = C 1 t r + C 2 t r ln |t| (29.38)
Example 29.5. Solve the Euler equation t 2 y ′′ + 5ty ′ + 4y = 0.
The indicial equation is
0 = r(r − 1) + 5r + 4 = r 2 + 4r + 4 = (r + 2) 2 =⇒ r = −2 (29.39)
This is case 2 with r = −2, so y 1 = 1/t 2 and y 2 = (1/t 2 ) ln |t|, and the
general solution is
y = C 1
t 2 + C 2
ln |t| (29.40)
t2 Case 3: Complex Roots. If the roots of t 2 y ′′ + αty ′ + βy = 0 are a complex
conjugate pair then we may denote them as
r = λ ± iµ (29.41)
where λ and µ are real numbers. The two solutions are given by
Hence the general solution is
t r = t λ±iµ = t λ t ±iµ (29.42)
= t λ e ln(t±iµ )
(29.43)
= t λ e ±iµ ln t (29.44)
= t λ [cos(µ ln t) ± i sin(µ ln t)] (29.45)
y = C 1 t r1 + C 2 t r2 (29.46)
= C 1 t λ [cos(µ ln t) + i sin(µ ln t)]
+ C 2 t λ [cos(µ ln t) − i sin(µ ln t)] (29.47)
= t λ [(C 1 + C 2 ) cos(µ ln t) + (C 1 − iC 2 ) sin(µ ln t)] (29.48)
Thus for any C 1 and C 2 we can always find and A and B (and vice versa),
where
A = C 1 + C 2 (29.49)
B = C 1 − iC 2 (29.50)
so that
y = At λ cos(µ ln t) + Bt λ sin(µ ln t) (29.51)