How did Maple obtain this solution that satisfies these initial conditions?

This is really a math question. But I can't figure how Maple did it.

Maple solves the ODE with 2 initial conditions correctly.

But if I use the general solution, and setup the set 2 equations, and tell Maple to solve _C1 and _C2, it says there is no solutions. Which is correct.

I would really like to know how Maple then managed to solve for these initial conditions.  This is problem from text book

I know how to solve it by hand to obtain general solution. But do not know how find solution that satisfies the IC's. when pluggin the initial conditions, the resulting 2 equation have no solution.

restart;
ode:=2*diff(y(x),x$2)=exp(y(x));
maple_general_sol:=dsolve(ode);
odetest(maple_general_sol,ode);
IC:=y(0)=0,D(y)(0)=1;
maple_sol_with_IC:=dsolve([ode,IC]);
odetest(maple_sol_with_IC,[ode,IC]);

The goal is to now take the general solution found by Maple above, and manually solve for _C1 and _C2:

eq1:=0=subs(x=0,rhs(maple_general_sol));
the_derivative:= diff(rhs(maple_general_sol),x):
eq2:=1=subs(x=0,the_derivative);

Two equations, two unknown. But using solve or PDEtools:-Solve produce no solution.

sol1:=solve([eq1,eq2],[_C1,_C2]);
sol2:=PDEtools:-Solve([eq1,eq2],[_C1,_C2]);

Looking at equation (1) above, we see that it is the same as 

eq1:=1=(tan(_C2/(2*_C1))^2 + 1)/_C1^2;

Because 0=ln(Z) means Z=1. Using the above simpler equation now gives

eq1:=1=(tan(_C2/(2*_C1))^2 + 1)/_C1^2;
the_derivative:= diff(rhs(maple_general_sol),x):
eq2:=1=subs(x=0,the_derivative);
sol1:=solve([eq1,eq2],[_C1,_C2]);

We see that there is indeed no solution. Second equation above says _C1= tan(.). Let tan(.)=Z. Plugging this in first equation gives Z^2= (Z^1+1) which has no solution for Z since this says 0=1 

So direct appliction of the initial conditions produces no solution. So how did Maple find the solution it obtained? I also tried using limits. But no luck. Book does not show how to obtain solution either. 

Any ideas?