We consider the language $L=\{+, -, ' , T, 0, 1\}$
Let $\text{Exp}(\mathbb{C})$ (the exponential sums) be the structure in that we interpret $L$.
We define $\text{Exp}(\mathbb{C})$ as the set of expresions $$\alpha=\alpha_1 e^{\mu_i z}+\dots \alpha_N e^{\mu_N z}$$ where $\alpha_i \in \mathbb{C}$ and $\mu_i \in \mathbb{C}$. (The $\mu_i$'s are pairwise distinct.)
Let $T(x)$ be the predicate that $x$ is non-constant.
Consider the expression $$\exists x \left (ax'(z)+bx(z)=y(z) \land T(x) \right ) $$
Since we are looking at the structure $\text{Exp}(\mathbb{C})$, $x$ and $y$ will have the folowing form $$x(z)=\sum_i \beta_i e^{k_i z}, \ \ y(z)=\sum_i \alpha_i e^{\mu_i z}$$
Substituting this at the differential equation we get $$\sum_i \{(a\mu_i +b)\beta_i -\alpha_i \}e^{\mu_i z} =0$$ So that we can solve for $\beta_i$, which means that we can find the solution $x(z)$, it must hold that $$a\mu_i +b \neq 0 \text{ for all } i \text{ for which } \alpha_i \neq 0$$
Is this correct?
So we have to be sure $T(x)$ holds for tthee solutions that we find, right?
How can we do that?
Or do we maybe have to take cases for $a$ and $b$ and find at each case the solution?
Also how can we write the above condition as a quantifier-free formula since there is the index $i$ ?