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