惠州学院数学与统计学院，广东 惠州

收稿日期：2022年11月26日；录用日期：2022年12月22日；发布日期：2022年12月30日

摘要

本文讨论了几类三阶非线性微分方程的求解问题。通过综合运用变量变换和降阶的思想方法建立了方程的通解表达式。

关键词

三阶微分方程，非线性微分方程，解

Solutions of Several Third-Order Nonlinear Differential Equations

Huanxin Huang, Yubin Liu^*

School of Mathematics and Statistics, Huizhou University, Huizhou Guangdong

Received: Nov. 26^th, 2022; accepted: Dec. 22^nd, 2022; published: Dec. 30^th, 2022

ABSTRACT

This paper discusses several classes of third-order nonlinear differential equations. By using of variable transformation and order reduction, the general solution expression of the equations is established.

Keywords:Third-Order Differential Equations, Nonlinear Differential Equations, Solution

Copyright © 2022 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

高阶微分方程的求解是微分方程理论中一个重要的问题。一般情况下无法给出高阶微分方程的通解表达式，特别是对于非线性高阶微分方程。因此，寻找一些可积的非线性高阶微分方程类型是个有趣的课题。至今有不少学者在这方面做出了不少结果，提出了各种有效的方法。如不变量解法 [1] [2] [3]、算子解法 [4]、常数变易法 [5] [6] [7]、降阶法 [8] [9] [10] [11] [12] 等等。

变量变换是求解微分方程的重要思想方法。文献 [13] 通过多次运用线性代换，将四阶非微分方程转化成变量分离的微分方程或是一阶非齐次线性微分方程，进而求出方程的通解；文献 [14] 中，通过两次线性变换，将三阶变系数微分方程转化为变量分离的微分方程或是二阶常系数微分方程。更多运用变量变换的思想方法求解微分方程的工作见文献 [15] [16] [17] [18] [19] 及其中的参考文献。

本文运用变换思想结合伯努利方程、恰当方程等特殊方程的求解方法，给出了几类新的可积三阶非线性微分方程并给出其通解公式。本文的结构安排如下：第二节考虑4类三阶非线性微分方程的求解问题，建立方程的通解表达式；第三节通过例子说明第二节所得结果在具体方程求解中的运用。

2. 主要结果

本节给出几类新的可积的三阶非线性微分方程及其通解表达式。

定理1 若存在连续函数 $r\left(x\right),Q\left(x\right)$，及二阶连续可导函数 $p\left(x\right)$，使得

$A\left(x\right)=p\left(x\right)+Q\left(x\right)+1,$

$B\left(x\right)=2p^{\prime }\left(x\right)+p\left(x\right)+p\left(x\right)Q\left(x\right)+Q\left(x\right),$

$C\left(x\right)=p^{″}\left(x\right)+p^{\prime }\left(x\right)+p^{\prime }\left(x\right)Q\left(x\right)+p\left(x\right)Q\left(x\right),$

$f\left(x,y,y^{\prime },y^{″}\right)=r\left(x\right){\left[y^{″}+\left(p\left(x\right)+1\right)y^{\prime }+\left(p^{\prime }\left(x\right)+p\left(x\right)\right)y\right]}^n,$

则方程

$y^{‴}+A\left(x\right)y^{″}+B\left(x\right)y^{\prime }+C\left(x\right)y=f\left(x,y,y^{\prime },y^{″}\right).$ (1)

的通解为：

$y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}\left\{{\text{e}}^{-x}\left[{\displaystyle \int {\text{e}}^x}{\left[{\text{e}}^{{\displaystyle \int \left(n-1\right)Q\left(x\right)\text{d}x}}\left({\displaystyle \int \left(1-n\right)r\left(x\right){\text{e}}^{{\displaystyle \int \left(1-n\right)Q\left(x\right)\text{d}x}}\text{d}x+c_1}\right)\right]}^{\frac{1}{1-n}}\text{d}x+c_2\right]\right\}\text{d}x+c_3\right\},$

其中 $c_1,c_2,c_3$ 为任意常数。

证明：将微分方程(1)变形为：

$\begin{array}{l}\frac{\text{d}\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\prime }+\left(y^{\prime }+p\left(x\right)y\right)\right]}{\text{d}x}+Q\left(x\right)\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\prime }+\left(y^{\prime }+p\left(x\right)y\right)\right]\\ =r\left(x\right){\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\prime }+\left(y^{\prime }+p\left(x\right)y\right)\right]}^n.\end{array}$

令 $y^{\prime }+p\left(x\right)y=z$，则上述方程可转化为：

$\frac{\text{d}\left(z^{\prime }+z\right)}{\text{d}x}+Q\left(x\right)\left(z^{\prime }+z\right)=r\left(x\right){\left(z^{\prime }+z\right)}^n.$

再令 $z^{\prime }+z=u$，则有

$u^{\prime }+Q\left(x\right)u=r\left(x\right)u^n.$

该方程是伯努利方程，因此其解可表示为：

$u={\left[{\text{e}}^{{\displaystyle \int \left(n-1\right)Q\left(x\right)\text{d}x}}\left({\displaystyle \int \left(1-n\right)r\left(x\right){\text{e}}^{{\displaystyle \int \left(1-n\right)Q\left(x\right)\text{d}x}}\text{d}x}+c_1\right)\right]}^{\frac{1}{1-n}},$

其中 $c_1$ 是任意常数。

再由方程 $z^{\prime }+z=u$ 及 $y^{\prime }+p\left(x\right)y=z$ 易得方程(1)的通解：

$y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}\left\{{\text{e}}^{-x}\left[{\displaystyle \int {\text{e}}^x}{\left[{\text{e}}^{{\displaystyle \int \left(n-1\right)Q\left(x\right)\text{d}x}}\left({\displaystyle \int \left(1-n\right)r\left(x\right){\text{e}}^{{\displaystyle \int \left(1-n\right)Q\left(x\right)\text{d}x}}\text{d}x+c_1}\right)\right]}^{\frac{1}{1-n}}\text{d}x+c_2\right]\right\}\text{d}x+c_3\right\},$

其中 $c_2,c_3$ 为任意常数。

定理2 若 $p\left(x\right)$ 存在连续二阶导数，n是一正整数，则三阶微分方程

$\begin{array}{l}\frac{1}{n}{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}-2}\{\left(y^{\prime }+p\left(x\right)y\right)\left[y^{‴}+\left(1+p\left(x\right)\right)y^{″}+\left(2p^{\prime }\left(x\right)+p\left(x\right)\right)y^{\prime }+\left(p^{\prime }\left(x\right)+p^{″}\left(x\right)\right)y\right]\begin{array}{c}\\ \end{array}\\ +\left(\frac{1}{n}-1\right){\left[y^{″}+p\left(x\right)y^{\prime }+p^{\prime }\left(x\right)y\right]}^2\}=f\left\{\frac{1}{nx}{\left[y^{\prime }+p\left(x\right)y\right]}^{\frac{1}{n}-1}\left[y^{″}+p\left(x\right)y^{\prime }+p^{\prime }\left(x\right)y\right]+\frac{1}{x}{\left[y^{\prime }+p\left(x\right)y\right]}^{\frac{1}{n}}\right\}.\end{array}$

通解为：

$\{\begin{array}{l}y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}{\left[{\text{e}}^{-x}{\displaystyle \int {\text{e}}^{x}u\text{d}x}+c_2\right]}^n\text{d}x+c_3\right\},\\ {\displaystyle \int \frac{1}{f\left(\frac{u}{x}\right)-\frac{u}{x}}\text{d}\frac{u}{x}}=\ln \left|x\right|+c_1.\end{array}$

其中 $c_1,c_2,c_3$ 为任意常数。

证明：三阶微分方程可变为：

$\frac{\text{d}\left\{{\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right]}^{\prime }+{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right\}}{\text{d}x}=f\left\{\frac{1}{x}{\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right]}^{\prime }+\frac{1}{x}{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right\}.$ (2)

令 $y^{\prime }+p\left(x\right)y=z^n$，方程(2)可转化为：

$\frac{\text{d}\left(z^{\prime }+z\right)}{\text{d}x}=f\left(\frac{1}{x}\left(z^{\prime }+z\right)\right).$ (3)

再令 $z^{\prime }+z=u$，方程(3)变为：

$\frac{\text{d}u}{\text{d}x}=f\left(\frac{u}{x}\right).$ (4)

令 $u=wx$，方程(4)变为：

${\displaystyle \int \frac{1}{f\left(w\right)-w}\text{d}w}=\ln \left|x\right|+c_1,其中c_1为任意常数.$

所以：

${\displaystyle \int \frac{1}{f\left(\frac{u}{x}\right)-\frac{u}{x}}\text{d}\frac{u}{x}}=\ln \left|x\right|+c_1,其中c_1为任意常数.$

对于方程 $z^{\prime }+z=u$，求得通解：

$z={\text{e}}^{-x}\left[{\displaystyle \int \left(u{\text{e}}^x\right)\text{d}x}+c_2\right],其中c_2为任意常数.$

代入到 $y^{\prime }+p\left(x\right)y=z^n$ 中求得：

$y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}{\left[{\text{e}}^{-x}{\displaystyle \int u{\text{e}}^x\text{d}x}+c_2\right]}^n\text{d}x+c_3\right\},其中c_3为任意常数.$

则原方程的解是：

$\{\begin{array}{l}{\displaystyle \int \frac{1}{f\left(\frac{u}{x}\right)-\frac{u}{x}}\text{d}\frac{u}{x}}=\ln \left|x\right|+c_1,\\ y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}{\left[{\text{e}}^{-x}{\displaystyle \int u{\text{e}}^x\text{d}x}+c_2\right]}^n\text{d}x+c_3\right\}.\end{array}$

定理3 若 $p\left(x\right)$ 有连续二阶导数， $q\left(x\right),r\left(x\right)$ 是连续函数，n是一正整数，则三阶微分方程

$\begin{array}{l}\frac{1}{n}{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}-2}\{\left(y^{\prime }+p\left(x\right)y\right)\left[y^{‴}+\left(1+p\left(x\right)\right)y^{″}+\left(2p^{\prime }\left(x\right)+p\left(x\right)\right)y^{\prime }+\left(p^{\prime }\left(x\right)+p^{″}\left(x\right)\right)y\right]\begin{array}{c}\\ \end{array}\\ +\left(\frac{1}{n}-1\right){\left[y^{″}+p\left(x\right)y^{\prime }+p^{\prime }\left(x\right)y\right]}^2\}\\ =q\left(x\right)\left\{\frac{1}{n}{\left[y^{\prime }+p\left(x\right)y\right]}^{\frac{1}{n}-1}\left[y^{″}+p\left(x\right)y^{\prime }+p^{\prime }\left(x\right)y\right]+{\left[y^{\prime }+p\left(x\right)y\right]}^{\frac{1}{n}}\right\}+r\left(x\right).\end{array}$

的通解是：

$y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}{\left\{{\text{e}}^{-x}\left[{\displaystyle \int {\text{e}}^x}\left[{\text{e}}^{{\displaystyle \int q\left(x\right)\text{d}x}}\left({\displaystyle \int r\left(x\right){\text{e}}^{{\displaystyle \int -q\left(x\right)\text{d}x}}\text{d}x}+c_1\right)\right]+c_2\right]\right\}}^n+c_3\right\},其中c_1,c_2,c_3为任意常数.$

证明：微分方程可转化为：

$\frac{\text{d}\left\{{\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right]}^{\prime }+{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right\}}{\text{d}x}=q\left(x\right)\left\{{\left[{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right]}^{\prime }+{\left(y^{\prime }+p\left(x\right)y\right)}^{\frac{1}{n}}\right\}+r\left(x\right).$ (5)

令 $y^{\prime }+p\left(x\right)y=z^n$，方程(5)可转化为：

$\frac{\text{d}\left(z^{\prime }+z\right)}{\text{d}x}=q\left(x\right)\left(z^{\prime }+z\right)+r\left(x\right).$ (6)

再令 $z^{\prime }+z=u$，方程(6)变为：

$\frac{\text{d}u}{\text{d}x}=q\left(x\right)u+r\left(x\right).$ (7)

求得方程(7)的解：

$u={\text{e}}^{{\displaystyle \int q\left(x\right)\text{d}x}}\left({\displaystyle \int r\left(x\right){\text{e}}^{{\displaystyle \int -q\left(x\right)\text{d}x}}\text{d}x}+c_1\right),其中c_1为任意常数.$

代入方程 $z^{\prime }+z=u$，求得：

$z={\text{e}}^{-x}\left\{{\displaystyle \int {\text{e}}^x}\left[{\text{e}}^{{\displaystyle \int q\left(x\right)\text{d}x}}\left({\displaystyle \int r\left(x\right){\text{e}}^{{\displaystyle \int -q\left(x\right)\text{d}x}}\text{d}x}+c_1\right)\right]+c_2\right\},其中c_2为任意常数.$

最后代入到 $y^{\prime }+p\left(x\right)y=z^n$ 中求得：

$y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}{\left\{{\text{e}}^{-x}\left[{\displaystyle \int {\text{e}}^x}\left[{\text{e}}^{{\displaystyle \int q\left(x\right)\text{d}x}}\left({\displaystyle \int r\left(x\right){\text{e}}^{{\displaystyle \int -q\left(x\right)\text{d}x}}\text{d}x}+c_1\right)\right]+c_2\right]\right\}}^n\right\},其中c_3为任意常数.$

定理4 若 $p\left(x\right)$ 存在二阶连续导数， $\frac{\partial M\left(x,u\right)}{\partial u}=\frac{\partial N\left(x,u\right)}{\partial x}$ 成立，则三阶微分方程：

$\begin{array}{l}M\left[x,y^{″}+\left(p\left(x\right)+1\right)y^{\prime }+\left(p\left(x\right)+p^{\prime }\left(x\right)\right)y\right]+N\left[x,y^{″}+\left(p\left(x\right)+1\right)y^{\prime }+\left(p\left(x\right)+p^{\prime }\left(x\right)\right)y\right]\\ \left[y^{‴}+\left(p\left(x\right)+1\right)y^{″}+\left(p\left(x\right)+2p^{\prime }\left(x\right)\right)y^{\prime }+\left(p^{\prime }\left(x\right)+p^{″}\left(x\right)\right)y\right]=0.\end{array}$

的通解为：

$\{\begin{array}{l}{\displaystyle \int M\left(x,u\right)\text{d}x+{\displaystyle \int \left[N\left(x,u\right)-\frac{\partial }{\partial u}{\displaystyle \int M\left(x,u\right)\text{d}x}\right]\text{d}u=c_1,}}\\ y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}\left[{\text{e}}^{-x}{\displaystyle \int u{\text{e}}^x\text{d}x}+c_2\right]\text{d}x+c_3\right\}.\end{array}其中c_1,c_2,c_3为任意常数.$

证明：令 $y^{″}+\left(p\left(x\right)+1\right)y^{\prime }+\left(p\left(x\right)+p^{\prime }\left(x\right)\right)y=u$，方程可化为：

$M\left(x,u\right)\text{d}x+N\left(x,u\right)\text{d}u=0.$ (8)

由于 $\frac{\partial M\left(x,u\right)}{\partial u}=\frac{\partial N\left(x,u\right)}{\partial x}$，则方程(8)是恰当微分方程，解为：

${\displaystyle \int M\left(x,u\right)\text{d}x+{\displaystyle \int \left[N\left(x,u\right)-\frac{\partial }{\partial u}{\displaystyle \int M\left(x,u\right)\text{d}x}\right]\text{d}u}}=c_1,c_1为任意常数.$

又因为

$y^{″}+\left(p\left(x\right)+1\right)y^{\prime }+\left(p\left(x\right)+p^{\prime }\left(x\right)\right)y=u.$ (9)

令 $y^{\prime }+p\left(x\right)y=z$，方程(9)变为：

$z^{\prime }+z=u.$ (10)

方程(10)的解为：

$z={\text{e}}^{-x}\left[{\displaystyle \int \left(u{\text{e}}^x\right)\text{d}x}+c_2\right],c_2为任意常数.$

代入到 $y^{\prime }+p\left(x\right)y=z$ 中求得：

$y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}\left[{\text{e}}^{-x}{\displaystyle \int u{\text{e}}^x\text{d}x}+c_2\right]\text{d}x+c_3\right\},$

$c_3$ 为任意常数。于是原方程解为：

$\{\begin{array}{l}{\displaystyle \int M\left(x,u\right)\text{d}x+{\displaystyle \int \left[N\left(x,u\right)-\frac{\partial }{\partial u}{\displaystyle \int M\left(x,u\right)\text{d}x}\right]\text{d}u=c_1,}}\\ y={\text{e}}^{{\displaystyle \int -p\left(x\right)\text{d}x}}\left\{{\displaystyle \int {\text{e}}^{{\displaystyle \int p\left(x\right)\text{d}x}}}\left[{\text{e}}^{-x}{\displaystyle \int u{\text{e}}^x\text{d}x}+c_2\right]\text{d}x+c_3\right\}.\end{array}$

3. 应用例子

例1 求解方程

$y^{‴}+\left(\frac{2}{x}+1\right)y^{″}+\left(-\frac{2}{x^2}+\frac{2}{x}+\frac{1}{x^2}\right)y^{\prime }+\left(\frac{2}{x^3}-\frac{1}{x^3}\right)y=\frac{1}{x}{\left[y^{″}+\left(\frac{1}{x}+1\right)y^{\prime }+\left(-\frac{1}{x^2}+\frac{1}{x}\right)y\right]}^2.$

解：取 $p\left(x\right)=\frac{1}{x},Q\left(x\right)=\frac{1}{x},r\left(x\right)=\frac{1}{x},p^{\prime }\left(x\right)=-\frac{1}{x^2},p^{″}\left(x\right)=\frac{2}{x^3}$，显然 $p\left(x\right)$ 存在连续二阶导数， $r\left(x\right),Q\left(x\right)$ 是连续函数，则运用定理1求出其解：

$y=\frac{1}{x}\left[{\displaystyle \int x{\text{e}}^{-x}\left({\displaystyle \int \frac{{\text{e}}^x}{1-c_{1}x}\text{d}x}+c_2\right)\text{d}x}+c_3\right],c_1,c_2,c_3是任意常数.$

例2 求解方程

$\begin{array}{l}\frac{1}{2}{\left(y^{\prime }+xy\right)}^{-\frac{3}{2}}\left\{\left(y^{\prime }+xy\right)\left[y^{‴}+\left(1+x\right)y^{″}+\left(2+x\right)y^{\prime }+y\right]-\frac{3}{2}{\left[y^{″}+x{y}^{\prime }+y\right]}^2\right\}\\ =\frac{1}{x}{\left[y^{\prime }+xy\right]}^{-\frac{3}{2}}\left[y^{″}+x{y}^{\prime }+y\right]+\frac{2}{x}{\left[y^{\prime }+xy\right]}^{\frac{1}{2}}.\end{array}$

解：取 $p\left(x\right)=x,n=2$，由定理2得：

$\begin{array}{c}y=c_1^2{x}^3-3c_1^{2}x-4c_1^2{x}^2+8c_1^2+4c_1^{2}x+4c_{1}x-8c_1^2+c_3+{\displaystyle \int {\text{e}}^{\frac{1}{2}{x}^2}}(3c_1^2-4c_1+2c_1{c}_2{x}^2{\text{e}}^{-x}\\ -4c_1{c}_{2}x{\text{e}}^{-x}+42c_1{c}_2{\text{e}}^{-x}+c_2^2{\text{e}}^{-2x})\text{d}x,其中c_1,c_2,c_3是任意常数.\end{array}$

例3 求解方程

$\begin{array}{l}\frac{1}{2}{\left(y^{\prime }+y\right)}^{-\frac{3}{2}}\left\{\left(y^{\prime }+y\right)\left[y^{‴}+\left(1+x^2\right)y^{″}+y^{\prime }\right]-\frac{1}{2}{\left(y^{″}+y^{\prime }\right)}^2\right\}\\ =\left\{\frac{1}{2}{\left(y^{\prime }+y\right)}^{-\frac{1}{2}}\left(y^{″}+y^{\prime }\right)+\frac{1}{x}{\left(y^{\prime }+y\right)}^{\frac{1}{2}}\right\}+x.\end{array}$

解：取 $p\left(x\right)=1,q\left(x\right)=1,r\left(x\right)=x$，显然 $p\left(x\right)$ 有连续二阶导数， $q\left(x\right),r\left(x\right)$ 是连续函数，则由定理3可得方程通解：

$y=\frac{1}{12}{c}_1^2{\text{e}}^{2x}-c_2^2{\text{e}}^{-2x}+\frac{1}{4}{c}_1{\text{e}}^x-c_2{x}^2{\text{e}}^{-x}+c_3{\text{e}}^{-x}+x^2-2x-\frac{1}{2}{c}_{1}x+c_1{c}_2,其中c_1,c_2,c_3为任意常数.$

例4 求解方程

$\frac{y^{″}}{x^2}+\left(\frac{1}{x^3}+\frac{1}{x^2}\right)y^{\prime }+\left(\frac{1}{x^3}-\frac{1}{x^4}\right)y-1-\left[\frac{y^{‴}}{x^3}+\left(\frac{1}{x^3}-\frac{1}{x^4}\right)y^{″}-\left(\frac{4}{x^5}+\frac{1}{x^4}\right)y^{\prime }+\left(\frac{4}{x^6}-\frac{3}{x^5}\right)y\right]=0.$

解：取 $u=y^{″}+\left(\frac{1}{x}+1\right)y^{\prime }+\left(\frac{1}{x}-\frac{1}{x^2}\right)y$，$M\left(x,u\right)=\frac{u}{x^2}-1$，$N\left(x,u\right)=-\frac{1}{x}$，且 $\frac{\partial M}{\partial u}=\frac{\partial N}{\partial x}$ 成立，

则运用定理4求得三阶微分方程的解为：

$y=-\frac{1}{4}{x}^3-\frac{2}{3}{x}^2-x+\frac{1}{3}{c}_1{x}^2-\frac{1}{2}{c}_{1}x-c_2{\text{e}}^{-x}-\frac{c_2{\text{e}}^{-x}}{x}+\frac{c_3}{x},\left(c_1,c_2,c_3是任意常数\right).$

基金项目

国家自然科学基金项目(No. 11601180)，惠州学院自然科学基金项目(hzu201806)。

文章引用

黄焕鑫,刘玉彬. 几类三阶非线性微分方程的解
Solutions of Several Third-Order Nonlinear Differential Equations[J]. 理论数学, 2022, 12(12): 2189-2195. https://doi.org/10.12677/PM.2022.1212235

参考文献