Lecture Notes in Differential Equations - Bruce E. Shapiro

175
Example 20.3. Show that y = sin t and y = cos t are linearly independent.
Their Wronskian is
W (t) = (sin t)(cos t) ′ − (cos t)(sin t) ′ (20.37)
= − sin 2 t − cos 2 t (20.38)
= −1 (20.39)
Since W (t) ≠ 0 for all t then if we pick any particular t, e.g., t = 0, we have
W (0) ≠ 0. Hence sin t and cos t are linearly independent.
Example 20.4. Show that y = sin t and y = t 2 are linearly independent.
Their Wronskian is
At t = π, we have
W (t) = (sin t)(t 2 ) ′ − (t 2 )(sin t) ′ (20.40)
= 2t sin t − t 2 cos t (20.41)
W (π) = 2π sin π − π 2 cos π = π 2 ≠ 0 (20.42)
Since W (π) ≠ 0, the two functions are linearly independent.
Corollary 20.5. If f and g are linearly dependent functions, then their
Wronskian must be zero at every point t.
Proof. If W (f, g)(t 0 ) ≠ 0 at some point t 0 then theorem (20.4) tells us that
f and g must be linearly independent. But f and g are linearly dependent,
so this cannot happen. Hence their Wronskian can never be nonzero.
Suppose that y 1 and y 2 are solutions of y ′′ + p(t)y ′ + q(t)y = 0. Then
Differentiating,
W (y 1 , y 2 )(t) = y 1 y ′ 2 − y 2 y ′ 1 (20.43)
d
dt W (y 1, y 2 )(t) = y 1 y 2 ′′ + y 1y ′ 2 ′ − y2y 1 ′′ − y 2y ′ 1 ′ (20.44)
= y 1 y 2 ′′ − y 2 y 1 ′′
(20.45)
Since y 1 and y 2 are both solutions, they each satisfy the differential equation:
y 1 ′′ + p(t)y 1 ′ + q(t)y 1 = 0 (20.46)
y 2 ′′ + p(t)y 2 ′ + q(t)y 2 = 0 (20.47)