Generally, initial conditions to an $n^{th}$ order ODE involve initial conditions only involving derivatives up to the degree $ n-1 \ (like \ y^{(n-1)}(0) \ = \ A).$
Even a basis of the space of solutions to an ODE consists of solutions satisfying the following initial conditions, none of which contain any condition on $y_i^{(n)}(0)$.
$$y_1(0)=1,y_1'(0)=0,y_1''(0)=0,…,y_1^{(n−1)}(0)=0 \\ y_2(0)=0,y_2'(0)=1,y_2''(0)=0,…,y_2^{(n−1)}(0)=0 \\ y_3(0)=0,y_3'(0)=0,y_3''(0)=1,…,y_3^{(n−1)}(0)=0 \\ … \\ y_n(0)=0,y_n'(0)=0,y_n''(0)=0,…,y_n^{(n−1)}(0)=1$$
Now I understand that we can have only $n$ total initial conditions on the ODE, but is there any technical reason for skipping $y_i^{(n)}(0)$ instead of $y_i(0)$ when writing the initial conditions?
Can we have a basis to the solution space with conditions like
$$y_1'(0)=1,y_1''(0)=0,y_1'''(0)=0,…,y_1^{(n)}(0)=0 \\ y_2'(0)=0,y_2''(0)=1,y_2'''(0)=0,…,y_2^{(n)}(0)=0 \\ y_3'(0)=0,y_3''(0)=0,y_3'''(0)=1,…,y_3^{(n)}(0)=0 \\ … \\ y_n'(0)=0,y_n''(0)=0,y_n'''(0)=0,…,y_n^{(n)}(0)=1$$
Your equation might be incompatible with that initial conditions. Consider for example $$ \begin{cases} y''+y'=0 , & t>0\\ y''(0)=1 \\ y'(0)=0 \end{cases} $$ If this problem had a solution, continuity would give at time $t=0$ the contradiction $$1=0.$$