A refinement of the Kovalevskaya theorem on analytic solvability of a Cauchy problem

Let
\begin{equation}\label{eq0} P=z_n^m+\sum_{\alpha\in A} a_{\alpha}z^{\alpha} \end{equation}
be a polynomial where $ A \subset \mathbb{Z}^{n-1}_{\geqslant 0}\times \{0,1,\ldots,m-1\} $ is a finite set of exponents. Consider a differential equation
\begin{equation}\label{eq1} P( \mathcal{D})y=f \end{equation}
with $f=\sum_{k\in \mathbb{Z}^n_{\geqslant 0} }{b_k x^k}$ given as a power series. Note that $\frac{\partial^m}{\partial x^m_n}$ is the highest derivative in $x_n$ but not necessarily the highest derivative in the equation. Consider a Cauchy problem for \eqref{eq1} with initial conditions
\begin{equation}\label{eeq4} \frac{\partial^{k}y}{ \partial x^k}(x',0)=y_k(x'), \;\; k=0,\ldots,m-1, \end{equation}

where $x'=(x_1,\ldots,x_{n-1})$.
Theorem. \textit{If the right hand side $f$ of \eqref{eq1} is an entire function of exponential type then the Cauchy problem \eqref{eq1}, \eqref{eeq4} has a unique analytic solution.}
A strict condition on $f$ is dictated by the fact that in the proof we apply the Borel transform to $f$. The condition that $f$ is an entire function of exponential type is crucial for the domain of convergence of this transform to be non-empty.
Note that relaxation of the condition on $A$ in \eqref{eq0} reflects in stricter conditions on the right hand side as demonstrated by the well-known example by Kovalevskaya about the heat transfer equation.