Since
and
are linearly independent it follows that the Wronskian
must satisfy
for all
where the differential equation is defined, say
. Without loss of generality, suppose that
. Then
So at
and either
and
are both positive or both negative. Without loss of generality, suppose that they are both positive. Now, at
and since
and
are successive zeros of
it causes
. Thus, to keep
we must have
. We see this by observing that if
then
would be increasing (away from the
-axis), which would never lead to a zero at
. So for a zero to occur at
at most
(i.e.,
and it turns out, by our result from the Wronskian that
). So somewhere in the interval
the sign of
changed. By the Intermediate Value Theorem there exists
such that
.
On the other hand, there can be only one zero in
, because otherwise
would have two zeros and there would be no zeros of
in between, and it was just proved that this is impossible.