Abstract
We show that if the following delay differential equation of rational coefficients $$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z)w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}= \frac{\sum _{i=0}^pa_i(z)w^i}{\sum _{j=0}^qb_j(z)w^j} \end{aligned}$$ admits a transcendental entire solution w of hyper-order less than one, then it reduces into a delay differential equation of rational coefficients $$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{1}{w(z)}\sum _{i=0}^{k+2}A_{i}(z)w^i(z), \end{aligned}$$ which improves some known theorems obtained most recently by Zhang and Huang. Some examples are constructed to show that our results are accurate.
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