We have to solve on some open interval,

.

We want to divide by .

But we have to consider the case where in that case .

By substitution it does solve the differencial equation.

Second case consider *

And divide through,

Thus,

These are the necessary solutions, check them to show they all work.

*)Note there is a case where for some point in the open interval and non-zero in some point. If that where the case then since the function is continous there is an open interval where it is non-zero and hence the solution we have above. But then there is no way that the function will connect will the zero point for that will lead to non-continuity and hence non-differenciability. But that cannot be the case because we assume is differenciable on the open interval. Thus that cannot be the case.