Lecture Notes in Differential Equations - Bruce E. Shapiro

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the differential equation. Over what domain do these two functions form a
fundamental set of solutions?
Calculating the Wronskian, we find that
W (t) =
t 1/2
∣1/(2t )
∣
1/t ∣∣∣
−1/t 2 (27.160)
= − t1/2
t 2 − 1
t(2t 1/2 )
(27.161)
= −3
2t 3/2 (27.162)
This Wronskian is never equal to zero. Thus these two solutions form a
fundamental set on any open interval over which they are defined, namely
t > 0 or t < 0.
Theorem 27.10 (Abel’s Formula). The Wronskian of
is
a n (t)y (n) + a n−1 (t)y (n−1) + · · · + a 1 (t)y + a 0 (t) = 0 (27.163)
W (t) = Ce − ∫ p(t)dt
(27.164)
where p(t) = a n−1 (t)/a n (t). In particular, for n = 2, the Wronskian of
y ′′ + p(t)y ′ + q(t)y = 0 (27.165)
is also given by the same formula, W (t) = Ce − ∫ p(t)dt .
Lemma 27.11. Let M be an n × n square matrix with row vectors m i ,
determinant M, and let d(M, i) be the same matrix with the ith row vector
replaced by dm i /dt. Then
Proof. For n=2,
dM
dt
d
dt det M =
n ∑
i=1
det d(M, i) (27.166)
= d dt (m 11m 22 − m 12 m 21 ) (27.167)
= m 11 m ′ 22 + m ′ 11m 22 − m 12 m ′ 21 − m ′ 12m 21 (27.168)
Now assume that (27.166) is true for any n × nmatrix, and let M be any
n + 1 × n + 1 matrix. Then if we expand its determinant by the first row,
n+1
∑
M = (−1) 1+i m 1i min(m 1i ) (27.169)
i=1