Let the function be the unique solution of the initial-value problem
with where
is a constant,
and
are continuous functions on an interval
, and
with
any real number. If
and
prove that the function
for which
for all
is a solution of
if and only if for all
.
Proof. Assume
is a solution of
. Let
. Then,
(This division is allowed since on
implies
on
.) Therefore,
Thus, if is a solution of
then
is the unique solution of
Conversely, if
on
then
Therefore,
Then, since we know by hypothesis that is the unique solution of
we have
Substituting this into the above equation (and noting that our assumption that implies
) we have
Therefore is a solution of
with
In the question $k$ should be equal to $1-n$ not $n-1$ as presented