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Unformatted text preview: 2t and e−3t of y +5y +6y = 0 are l.i. on (−∞, ∞). Let
k1 e−2t + k2 e−3t = 0 for all t ∈ (−∞, ∞). Then since it is true for all t, diﬀerentiate
k1 (−2e−2t ) + k2 (−3e−3t ) = 0, solve for k1 , k2 and ﬁnd k1 = k2 = 0.
Linear Independence and the Wronskian; W (f, g ) = f g − f g
We now discuss the relation between linear independence of f (t), g (t) and their
Wronskian without necessarily assuming that f (t) and g (t) are solutions of (6.15). Theorem 5 Let f (t) and g (t) be diﬀerentiable, on an open interval I .
(a) If W (f, g ) = 0 for some point t0 ∈ I , then f,g are l.i. on I.
(b) Alternatively, if f (t) and g (t) are l.d. on I , then W (f, g ) = 0 at every t on I . Proof:
(a) Consider k1 f (t)+k2 g (t) and suppose it is zero on I . Then k1 f (t)+k2 g (t) =
0 on I . Evaluate at t0 .
k 1 f ( t 0 ) + k 2 g ( t0 ) = 0 . k 1 f ( t 0 ) + k 2 g ( t0 ) = 0 , (6.21) But (6.21) is a pair of simultaneous algebraic equations with the coeﬃcient
matrix
f ( t0 ) g ( t0 )
f ( t0 ) g ( t 0 )
whose determinant is W (f, g ) at t0 . If W (f, g ) at t = t0 is nonzero, equations (6.21) have the unique solution
k1 = k2 = 0.
Hence f (t), g (t) are l.i.
(b) Let f (t) and g (t) be l.d. on I and suppose W (f, g ) is nonzero at some
point t0 , say, on I . Then from (6.21), k1 = k2 = 0 and thus f (t) and g (t)
are l.i. on I . Contradiction. Therefore the supposition that W (f, g ) is
nonzero at some point t0 on I is false and thus W (f, g ) = 0 at every point
on I .
Now if f (t) and g (t) are two solutions y1 (t) and y2 (t) of the second order ode
L[y ] = y + p(t)y + q (t)y = 0 (6.15) where p(t) and q (t) are continuous on I , we have the stronger result: Theorem 6 The solutions y1 (t), y2 (t) deﬁned above are l.i. on I if and only if
W (y1 , y2 ) at t is nonzero for all t in I .
19 The reason for the stronger result is (6.19): namely, if p(t) is continuous on I , then
t
W (t) = W (t0 ) exp −
p(t )dt for all t, t0 ∈ I
t0 This means that W (t) is either nonzero everywhere on I or zero everywhere on I .
Consider k1 y1 (t) + k2 y2 )(t) = 0 on I . Then, because we know that the ﬁrst
derivatives of solutions to (6.15) exist (when p, q are continuous on I ) we see that
k1 y1 (t) + k2 y2 (t) = 0 for all t ∈ I .
(1) If W (y1 , y2 ) = 0 on I then k1 = k2 = 0 and y1 (t), y2 (t) are l.i. on I .
(2) If y1 (t), y2 (t) are l.i. on I then W (y1 , y2 ) = 0 for all t on I . Why? Because
if W (y1 , y2 ) is zero at some point on I , it is zero for all points on I (because
of (6.19)). Then at every point t0 on I , the equations
k 1 y 1 ( t0 ) + k 2 y 2 ( t0 ) = 0 , k 1 y 1 ( t 0 ) + k 2 y 2 ( t0 ) = 0 have a nontrivial solution so that not both of k1 and k2 are zero. Therefore y1 and y2 are l.d. Contradiction. The...
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This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.
 Fall '08
 INDIK

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