Lecture Notes in Differential Equations - Bruce E. Shapiro

174 LESSON 20. THE WRONSKIAN
such that for some point t 0 the Wronskian
W (t 0 ) = y 1 (t 0 )y ′ 2(t 0 ) − y ′ 1(t 0 )y 2 (t 0 ) ≠ 0 (20.27)
then every solution of (20.26) has the form
y(t) = Ay 1 (t) + By 2 (t) (20.28)
for some numbers A and B. In this case y 1 and y 2 are said to form a
fundamental set of solutions to (20.26).
Theorem 20.4. Let f and g be functions. If their Wronskian is nonzero
at some point t 0 then they are linearly independent.
Proof. Suppose that the Wronskian is non-zero at some point t 0 . Then
W (f, g)(t 0 ) =
∣ f(t 0) g(t 0 )
f ′ (t 0 ) g ′ (t 0 ) ∣ ≠ 0 (20.29)
hence
f(t 0 )g ′ (t 0 ) − g(t 0 )f ′ (t 0 ) ≠ 0 (20.30)
We will prove the result by contradiction. Suppose that f and g are linearly
dependent. Then there exists some non-zero constants A and B such that
for all t,
Af(t) + Bg(t) = 0 (20.31)
Differentiating,
Af ′ (t) + Bg ′ (t) = 0 (20.32)
which holds for all t. Since (20.31) and (20.32) hold for all t, then they hold
for t = t 0 . Hence
Af(t 0 ) + Bg(t 0 ) = 0 (20.33)
Af ′ (t 0 ) + Bg ′ (t 0 ) = 0 (20.34)
We can write (20.33) as a matrix:
[ [ ]
f(t0 ) g(t 0 ) A
f ′ (t 0 ) g ′ = 0 (20.35)
(t 0 )]
B
From linear algebra, since A and B are not both zero, we know that the
only way that this can be true is the determinant equals zero. Hence
f(t 0 )g ′ (t 0 ) − g(t 0 )f ′ (t 0 ) = 0 (20.36)
This contradicts equation (20.30), so some assumption we made must be
incorrect. The only assumption we made was that f and g were linearly
dependent.
Hence f and g must be linearly independent.