一种求解Riccati方程的方法  

A Method for Solving the Riccati Equation

作　　者：宋华兵 SONG Huabing(School of Mathematics and Statistics,Zhaoqing University,Zhaoqing,Guangdong 526061,China)

机构地区：[1]肇庆学院数学与统计学院,肇庆526061

出　　处：《肇庆学院学报》2023年第5期74-79,共6页Journal of Zhaoqing University

摘　　要：由于Riccati方程为非线性方程,常用的初等积分方法难以获得其解析解,但如果知道Riccati方程一个特解,则可通过变换将其简化为一阶线性非齐次微分方程求解.文章以实例形式分析了一阶线性微分方程与Riccati方程之间存在相同特解的情况,在求解思路上,提出了将一阶线性微分方程作为Riccati方程求解的引导方程,分析了引导方程与Riccati方程之间存在共同特解的条件,给出了寻求可解Riccati方程的方法,并通过示例验证了此方法的可行性.Since the Riccati equation is a nonlinear equation,it is difficult to obtain its exact solution by the usual elementary integral method.However,if one special solution of Riccati equation is known,it can be simpli-fied to a first-order linear non-homogeneous differential equation by transformation.In this paper,the same spe-cial solutions between the first-order linear differential equation and the Riccati equation are analyzed with an ex-ample.In the solution of Riccati equation,the first order linear differential equation is considered as the leading equation of Riccati equation.The condition for the existence of the same special solution between the leading equation and the Riccati equation is analyzed,and a method for solving Riccati equation is given.The feasibili-ty is verified by examples.

关 键 词：RICCATI方程 可解条件 引导方程 

分 类 号：O193[理学—数学]