Have I found ALL the solutions to this diff eq & boundary conditions? If we find a solution to a differential equation and its boundary conditions, how can we know if we have found ALL the solutions?
For example, let g(x) be a smooth continuous function of x:
(Eq 1) $$g'^{2}=e^{-g}-1+(1-e^{-1})g
 $$ $$g(0)=0
 $$ $$g'(0)=0
 $$
The trivial solution to this is $g(x)=0
 $ for all x, and this is exactly what a numerical solver returns. However, if we differentiate both sides by x and divide out $g'
 $ we get:
(Eq 2) $$2g''=-e^{-g}+1-e^{-1}
 $$
Combinging Eqs 1&2 yields a valid diverging solution.  So I have a multi-part question because this illustrates some diff eq principles I've never firmly understood.


*

*If I hadn't known beforehand about the Eq2 solution, how could I have known that Eq1 had other solutions? 

*Is there something inherently different about the two solutions that makes the numerical solver find one and not the other? 

*Have we found ALL the solutions now, or do have test out some even higher derivatives of Eq2? 

*How can we know when we've found them all?


Thank you.
 A: To elaborate a little around @Winther's comments:
For a first order ODE with an initial condition, if the right-hand side is Lipschitz continuous then, by the Picard-Lindelöf theorem, a unique solution exists (i.e. if you find one solution, you have found all solutions).
Here you have specified two conditions on your solution. Generally speaking, if you don't impose the correct number of conditions there are three things that can happen: 
(1) You are lucky and a solution exists anyway. In your case this is so, but one of the conditions is redundant and can be ignored.
(2) If too many conditions are specified and are inconsistent, no solution exist. In your case, if we (for example) impose $g(0) = 0$ and $g'(0) = 1$ we would not find a solution.
(3) If too few conditions are specified we might find ourselves without uniqueness, and instead with a possibly infinite set of solutions. In your case, without any conditions imposed, we see (for example) that both $g=0$ and $g=1$ are solutions.
A lesson to be learned here is: If your problem is ill-posed (essentially if we use wrong number of conditions), do not expect a sensible result from any standard numerical solver. Always analyse the ODE first!
