1.2 THE METHOD OF VARIATION OF PARAMETERS 9
1.2.1 The Wronskian
Before we carry on, let’s pause to discuss some further properties of the Wronskian.
Recall that if V is a vector space over R, then two elements v
1
, v
2
∈ V are linearly
dependent if ∃ α
1
,α
2
∈ R, with α
1
and α
2
not both zero, such that α
1
v
1
+α
2
v
2
= 0.
Now let V = C
1
(a, b) be the set of once-differentiable functions over the interval
a<x<b.Ifu
1
,u
2
∈ C
1
(a, b) are linearly dependent, ∃ α
1
,α
2
∈ R such that
α
1
u
1
(x)+α
2
u
2
(x)=0∀x ∈ (a, b). Notice that, by direct differentiation, this also
gives α
1
u
1
(x)+α
2
u
2
(x) = 0 or, in matrix form,
u
1
(x) u
2
(x)
u
1
(x) u
2
(x)
α
1
α
2
=
0
0
.
These are homogeneous equations of the form
Ax = 0,
which only have nontrivial solutions if det(A) = 0, that is
W =
u
1
(x) u
2
(x)
u
1
(x) u
2
(x)
= u
1
u
2
− u
1
u
2
=0.
In other words, the Wronskian of two linearly dependent functions is identically
zero on (a, b). The contrapositive of this result is that if W ≡ 0on(a, b), then u
1
and u
2
are linearly independent on (a, b).
Example
The functions u
1
(x)=x
2
and u
2
(x)=x
3
are linearly independent on the interval
(−1, 1). To see this, note that, since u
1
(x)=x
2
, u
2
(x)=x
3
, u
1
(x)=2x, and
u
2
(x)=3x
2
, the Wronskian of these
two functions is
W =
x
2
x
3
2x 3x
2
=3x
4
− 2x
4
= x
4
.
This quantity is not identically zero, and hence x
2
and x
3
are linearly independent
on (−1, 1).
Example
The functions u
1
(x)=f(x) and u
2
(x)=kf(x), with k a constant, are linearly
dependent on any interval, since their Wronskian is
W =
fkf
f
kf
=0.
If the functions u
1
and u
2
are solutions of (1.2), we can show by differentiating
W = u
1
u
2
− u
1
u
2
directly that
dW
dx
+ a
1
(x)W =0.
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