This is a homogeneous linear equation, so by the uniqueness theorem the only solution that can vanish at any point is y=0.
Say we wish to solve on *.
1) is non-vanishing i.e. for any .
Then we can write and thus .
This leads to with
2) is the zero function but that is definitely a solution.
3) vanishes at some point i.e. for some .
This is were the difficultly is. If it vanishes at and in the open interval containing then I think this leads to being the zero function**. However, if does not vanish on any open interval containing it then it means there is a point such that . But by continuity it means there is an interval containing such that is non-vanishing on . By applying #1 to this restricted function, , we find that must be of the form . And this perhaps shows that is everywhere.*** Therefore is non-vanishing everywhere. A contradiction if is a non-zero function. Thus, this case does not exist for non-zero functions.
How do we complete the proof?
And is there a general theorem that tells us we can divide the functions we are solving for without worrying it can vanish.
*)This can be solved without seperation of variables i.e. with integrating fractor.
But with this something interesting happens.
**)Though I did not find the complete proof. I guess the proof comes down to a analogue of the identity theorem from complex analysis applied to real functions. This is because the solution to the differencial equation is and we can possibly even proof its analyticity. With analyticity we can have an analogue of the identity theorem.
***)Again by identity theorem for real function - by using power series.