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.

中文翻译：

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一类时滞微分方程的Malmquist型定理

我们证明，如果以下时滞微分方程的有理系数

$$ \ begin {aligned} w ^ k（z）\ sum _ {\ mu = 1} ^ se_ \ mu（z）w（z + c_ \ mu）+ a（z）\ frac {w ^ {（n }}（z）} {w（z）} = \ frac {\ sum _ {i = 0} ^ pa_i（z）w ^ i} {\ sum _ {j = 0} ^ qb_j（z）w ^ j } \ end {aligned} $$

接受一个超阶的超验整体解w小于1，然后将其简化为有理系数的时滞微分方程

$$ \ begin {aligned} w ^ k（z）\ sum _ {\ mu = 1} ^ se_ \ mu（z）w（z + c_ \ mu）+ a（z）\ frac {w ^ {（n ）}（z）} {w（z）} = \ frac {1} {w（z）} \ sum _ {i = 0} ^ {k + 2} A_ {i}（z）w ^ i（z ），\ end {aligned} $$

这改进了张和黄最近获得的一些已知定理。构造了一些例子来表明我们的结果是准确的。