# 积分因子存在性的充要条件及其应用The Necessary and Sufficient Condition for the Existence of Integrating Factors with Its Application

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In this paper, we give the necessary and sufficient condition for the existence of integrating factors and their proof. Meanwhile, some examples are given for its application.

1. 引言

$M\left(x,y\right)\text{d}x+N\left(x,y\right)\text{d}y=0$ (1)

2. 定理的证明

$\frac{\partial u}{\partial x}\text{d}x+\text{d}y\frac{\partial u}{\partial y}=0$ (2)

$\frac{\text{d}y}{\text{d}x}=-\frac{\partial u}{\partial x}/\frac{\partial u}{\partial y}.$ (3)

$\frac{\text{d}y}{\text{d}x}=-\frac{M}{N}$ (4)

$\frac{\partial u}{\partial x}/\frac{\partial u}{\partial y}=\frac{M}{N},$

$\frac{\partial u}{\partial x}/M=\frac{\partial u}{\partial y}/N.$

$\frac{\partial u}{\partial x}=\mu M$ , $\frac{\partial u}{\partial y}=\mu N.$

$\mu \left(x,y\right)$ 乘以(1)两端，得

$\mu M\text{d}x+\mu N\text{d}y=0,$

$\frac{\partial u}{\partial x}\text{d}x+\frac{\partial u}{\partial y}\text{d}x=0,$

$\begin{array}{c}\frac{\partial }{\partial y}\left[\mu \phi \left(u\right)M\right]=\phi \left(u\right)\frac{\partial \left(uM\right)}{\partial y}+\mu M{\phi }^{\prime }\left(u\right)\frac{\partial u}{\partial y}\\ =\phi \left(u\right)\frac{\partial \left(uM\right)}{\partial y}+\mu M{\phi }^{\prime }\left(u\right)\mu N\\ =\phi \left(u\right)\frac{\partial \left(uM\right)}{\partial y}+{\mu }^{2}MN{\phi }^{\prime }\left(u\right)\end{array}$ (5)

$\begin{array}{c}\frac{\partial }{\partial y}\left[\mu \phi \left(u\right)N\right]=\phi \left(u\right)\frac{\partial \left(uN\right)}{\partial x}+\mu N{\phi }^{\prime }\left(u\right)\frac{\partial u}{\partial x}\\ =\phi \left(u\right)\frac{\partial \left(uN\right)}{\partial x}+\mu N{\phi }^{\prime }\left(u\right)\mu M\\ =\phi \left(u\right)\frac{\partial \left(uN\right)}{\partial x}+{\mu }^{2}MN{\phi }^{\prime }\left(u\right)\end{array}$ (6)

$\frac{\partial \left(uM\right)}{\partial y}=\frac{\partial \left(uN\right)}{\partial x},$

$\frac{\partial }{\partial y}\left[\mu \phi \left(u\right)M\right]=\frac{\partial }{\partial x}\left[\mu \phi \left(u\right)N\right].$

$\mu M\text{d}x+\mu N\text{d}y=\text{d}u$$\stackrel{¯}{\mu }M\text{d}x+\stackrel{¯}{\mu }N\text{d}y=\text{d}v,$

$\mu M=\frac{\partial u}{\partial x},\text{\hspace{0.17em}}\mu N=\frac{\partial u}{\partial y},\text{\hspace{0.17em}}\stackrel{¯}{\mu }M=\frac{\partial v}{\partial x},\text{\hspace{0.17em}}\stackrel{¯}{\mu }N=\frac{\partial v}{\partial y}$

$\frac{\partial \left(u,y\right)}{\partial \left(x,y\right)}=|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial \nu }{\partial x}& \frac{\partial \nu }{\partial y}\end{array}|=|\begin{array}{cc}\mu M& \mu N\\ \stackrel{¯}{\mu }M& \stackrel{¯}{\mu }N\end{array}|\equiv 0$

$\frac{\partial u}{\partial x}\cdot \frac{\partial u}{\partial y}=\mu M\cdot \mu N\overline{)\equiv }0,$

$v\left(x,y\right)=\psi \left(u\left(x,y\right)\right).$

$\stackrel{¯}{\mu }\left(M\text{d}x+N\text{d}y\right)=\text{d}v=\text{d}\psi \left(u\right)={\psi }^{\prime }\left(u\right)\text{d}u={\psi }^{\prime }\left(u\right)\mu \left(M\text{d}x+N\text{d}y\right),$

3. 应用举例

${\mu }_{2}\left(M\text{d}x+N\text{d}y\right)=\text{d}u,$

$u\left(x,y\right)=c$ (任意常数)为方程(1)的通解。根据上述定理知 ${\mu }_{1}={\mu }_{2}\phi \left(u\right)$ ，其中 $\phi \left(t\right)$ 是t的可微函数，于是

$\frac{{\mu }_{1}}{{\mu }_{2}}=\phi \left(u\right).$ (7)

$0\equiv \text{d}\phi \left(u\right)\equiv {\phi }^{\prime }\left(u\right)\text{d}u\equiv {\phi }^{\prime }\left(u\right){\mu }_{2}\left(M\text{d}x+N\text{d}y\right).$

${\mu }_{2}\overline{)\equiv }0$$\phi \left(u\right)$ 不恒为常数，即 ${\phi }^{\prime }\left(u\right)\overline{)\equiv }0$ ，故 $M\text{d}x+N\text{d}y\equiv 0$ ，即对 $\phi \left(u\right)=c$ 微分后推导出 $M\text{d}x+N\text{d}y\equiv 0$ 。这说明由 $\phi \left(u\right)=c$ (任意常数)所确定的y与x的关系式是方程(1)的解，故 $\frac{{\mu }_{1}}{{\mu }_{2}}=c$ (任意常数)是(1)的通解。

$xM\left(x,y\right)+yN\left(x,y\right)=c$

$x\left(4y\text{d}x+2x\text{d}y\right)+{y}^{3}\left(3y\text{d}x+5x\text{d}y\right)=0.$

$x\left(4y\text{d}x+2x\text{d}y\right)=x\left[\left(2y\text{d}x+2x\text{d}y\right)+2x\text{d}y\right]=x\left[2\text{d}\left(xy\right)+2y\text{d}x\right]=2x\text{d}\left(xy\right)+2xy\text{d}x=\text{d}\left(2{x}^{2}y\right)$

${\mu }_{1}=1,\text{\hspace{0.17em}}{u}_{1}=2{x}^{2}y.$

${y}^{3}\left(3y\text{d}x+5x\text{d}y\right)=3{y}^{4}\text{d}x+5x{y}^{3}\text{d}y.$

$\frac{3}{x}\text{d}x+\frac{5}{y}\text{d}y=3\text{d}\left(\mathrm{ln}|x|\right)+5\text{d}\left(\mathrm{ln}|y|\right)=\text{d}\left(\mathrm{ln}|{x}^{3}{y}^{5}|\right),$

$1\cdot \phi \left(2{x}^{2}y\right)=\frac{1}{x{y}^{4}}\stackrel{˜}{\phi }\left(\mathrm{ln}|{x}^{3}{y}^{5}|\right).$

$\phi \left(2{x}^{2}y\right)=2{x}^{2}y$$\stackrel{˜}{\phi }\left(\mathrm{ln}|{x}^{3}{y}^{5}|\right)=2{x}^{3}{y}^{5}$ ，则得到原方程的一个积分因子 $\mu =2{x}^{2}y$ 。用它乘原方程两边，得

$2{x}^{2}y\text{d}\left(2{x}^{2}y\right)+6{x}^{2}{y}^{5}\text{d}x+10{x}^{3}{y}^{4}\text{d}y=0.$

$\frac{1}{2}\text{d}{\left(2{x}^{2}y\right)}^{2}+2{y}^{5}\text{d}{x}^{3}+2{x}^{3}\text{d}{y}^{5}=\text{d}\left(2{x}^{4}{y}^{2}\right)+\text{d}\left(2{x}^{2}{y}^{5}\right)=\text{d}\left(2{x}^{4}{y}^{2}+2{x}^{3}{y}^{5}\right)=0,$

${x}^{4}{y}^{2}+{x}^{3}{y}^{5}=c,$

$\left({x}^{3}y-2{y}^{2}\right)\text{d}x+{x}^{4}\text{d}y=0.$

$\left({x}^{3}y\text{d}x+{x}^{4}\text{d}y\right)+\left(-2{y}^{2}\text{d}x\right)=0,$

${x}^{3}\left(y\text{d}x+x\text{d}y\right)+\left(-2{y}^{2}\text{d}x\right)=0.$ (8)

$\frac{1}{{x}^{3}}\phi \left(xy\right)=\frac{1}{-2{y}^{2}}\stackrel{˜}{\phi }\left(x\right).$

$\phi \left(xy\right)=\frac{1}{{\left(xy\right)}^{2}}$$\stackrel{˜}{\phi }\left(x\right)=-\frac{2}{{x}^{5}}$ ，则得到原方程的一个积分因子

$\mu =\mu \left(x,y\right)=\frac{1}{{x}^{5}{y}^{2}}.$

$\frac{1}{{x}^{2}{y}^{2}}\text{d}\left(xy\right)-\frac{2}{{x}^{5}}\text{d}y=0,$

$\text{d}\left(-\frac{1}{xy}\right)+\text{d}\left(\frac{1}{2{x}^{4}}\right)=0,$

$\text{d}\left(-\frac{1}{xy}+\frac{1}{2{x}^{4}}\right)=0.$

$-\frac{1}{xy}+\frac{1}{2{x}^{4}}=c,$

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