DSolve does not return a solution when initial condition is added

Question

Consider the following equation $$ y''(x) = -2e^{-y}. $$

The following code

DSolve[y''[x] == -2 Exp[-y[x]], y[x], x] //FullSimplify

returns

{{y[x] -> Log[(2 (-1 + Cosh[Sqrt[C[1]] (x + C[2])]))/C[1]]}, 
 {y[x] -> Log[(2 (-1 + Cosh[Sqrt[C[1]] (x + C[2])]))/C[1]]}}

(They the same solution. Let's ignore that first.) If I impose an initial condition $y(0) = 0$ then Mathematica fails to return a solution

DSolve[{y''[x] == -2 Exp[-y[x]], y[0] == 0}, y[x], x]

with the error message

DSolve::bvfail: For some branches of the general solution, unable to solve the conditions.

But a solution does exist. One can choose $C[1] = 1$ and $C[2] = \cosh^{-1}(3/2)$ in the solution and verify that $y(0) = 0$.

Any idea why is this the case?

Answer 1

If you differentiate the ode you get y'''[x]==-y'[x] Exp[-y[x]]==-y'[x] y''[x].

Mathematica can handle this "equivalent" ode and evaluates a unique solution

Y = DSolveValue[{y'''[x] == -y'[x] y''[x],y[0]==0}, y, x]
(*Function[{x}, 2 Log[Cosh[(x Sqrt[C[1]])/Sqrt[2]]]]*)

The solution depends on one parameter C[1]. C[1] is choosen to fullfill the boundary condition

 Y''[0] == 1 (*Y''[0] ==Exp[-0]*)
 (* C[1]==1*) 
 Y/. %
 (*Function[{x}, 2 Log[Cosh[(x Sqrt[1])/Sqrt[2]]]]*)

As you can see no additional boundary condition can be required! This solution has slope Y'[0]==0

Answer 2

I assume because you are under-specifying the conditions, MMa cannot solve for two unknowns with only one equation. Are you sure that your choices of C[1] and C[2] give a unique solution to the problem. There are some instances where MMa can satisfy one condition by solving for one of the constants and leave the other unevaluated, but evidently not in this case. MMa finds a solution if you specify two conditions such as

DSolve[{y''[x] == -2 Exp[-y[x]], y[0] == 0, y'[0] == 0}, y[x], 
  x] // FullSimplify

{{y(x)->log(cos^2(x))},{y(x)->log(cos^2(x))},{y(x)->log(cos^2(x))}}*)

There are inverse function warnings, but we get a solution.