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( t0 ) = y 0 .
dt y ( t0 ) = y 0 , (6.16) Theorem 3 Given p(t), q (t) continuous on an open interval I . Then there is exactly
one solution y = φ(t) of (6.15) with initial conditions (6.16) and t0 ∈ I and the
solution exists throughout the interval I . Theorem 4 This result is also true for the inhomogeneous equations (6.1) with
initial conditions (6.16) providing g (t) is also continuous on I . For constant coeﬃcient equations, solutions of (6.15) can be found explicitly, are
continuous and have continuous ﬁrst and second derivatives for all t, −∞ < t < ∞.
Principle of Superposition and Wronskians
We have already proved that if y1 (t) and y2 (t) solve (6.15), then so does y (t) =
Ay1 (t) + By2 (t).
Question: In what circumstances can we use a linear combination of solutions of
(6.15) to ﬁnd a unique solution to the initial value problem (6.15) and (6.16)?
Answer: When a certain functional of y1 (t) and y2 (t), the Wronskian
W (y1 , y2 ) = y1 (t)y2 (t) − y1 (t)y2 (t) = det y 1 ( t)
y 1 ( t)
y 2 ( t)
y 2 ( t) (6.17) estimated at t0 is nonzero.
Proof: Let y1 (t), y2 (t) solve (6.15). Then so does y (t) = Ay1 (t) + By2 (t). Apply
initial conditions
y0 = Ay1 (t0 ) + By2 (t0 )
y0 =
Ay1 (t0 ) +
17
By2 (t0 ). (6.18) Solve for A and B ,
y 0 y 2 ( t0 )
y 1 ( t0 ) y 0
det
det
y 0 y 2 ( t0 )
y 1 ( t0 ) y 0
A=
,B =
,
(W (y1 , y2 ))t=t0
(W (y1 , y2 ))t=t0 The solution (A, B ) of (6.18) is unique if
W (y1 , y2 )t=t0 = 0.
Note: If W (y1 , y2 )t=t0 is zero, the solution (A, B ) can still exist but it is not unique.
Why? Recall:
x−y =1 2x − 2y = b
1 1
is zero, either no solutions exist or, if b = 2, then there are inﬁnite
2 2
solutions (α, α − 1) for all α.
Since det The Wronskian W (y1 , y2 ) satisﬁes a diﬀerential equation.
Proof: Let W = W (y1 (t), y2 (t)) = y1 (t)y2 (t) − y1 (t)y2 (t) dW
dt =
=
= Thus,
y 1 ( t) y 2 ( t) − y 1 ( t) y 2 ( t)
y1 (t)(−p(t)y2 (t) − q (t)y2 (t)) − y2 (t)(−p(t)y1 (t) − q (t)y1 (t)) − p( t ) W t
W (t) = W (t0 ) exp −
p(t )dt . (6.19) t0 So we have the interesting result that if p(t) is continuous on an interval I , then if
W (t0 ) is nonzero, W (t) is nonzero throughout I .
This means that no matter what point in I we choose to give initial conditions, the
solution of (6.15) and (6.16) is always unique if at any point t0 ∈ I, W (t0 ) = 0.
The notion of linear dependence, linear independence and the Wronskian
We say two functions f (t) and g (t) are linearly independent (l.d.) on an interval I
if there exist two constants k1 and k2 , not both zero, such that
k1 f (t) + k2 g (t) = 0 for all t ∈ I. (6.20) We say two functions f (t) and g (t) are linearly independent (l.i.) on an interval I
if they are not l.d. on I , i.e. an equation such as (6.20) holds for all t ∈ I and only
if k1 = k2 = 0.
18 Example: The solutions e...
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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 University of Arizona Tucson.
 Fall '08
 INDIK

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