定理1.2 若存在一函数 \(\varphi(x,y)\in C^1\) 使得 \(\operatorname{div} X=f(\varphi(x,y))X(\varphi(x,y))\),则
\[\mu(\varphi(x, y))=\operatorname{e}^{-\int f(\varphi(x, y)) \mathrm{d} \varphi(x, y)}
\]
是方程(1)的积分因子.
证明：
\[\begin{align*}
X\mu(\varphi(x,y))=&Q \frac{\partial(\mu(\varphi))}{\partial x}-P \frac{\partial(\mu(\varphi))}{\partial y}\\
=&\mu'(\varphi)Q\frac{\partial \varphi}{\partial x}-P\frac{\partial \varphi}{\partial y}\\
=&-f(\varphi)e^{-\int f(\varphi(x,y)){\rm d}\varphi(x,y)}X(\varphi)\\
=&-f(\varphi)e^{-\int f(\varphi(x,y)){\rm d}\varphi(x,y)}\frac{\operatorname{div} X}{f(\varphi)}\\
=&-\mu(\varphi(x,y))\operatorname{div} X
\end{align*}
\]
这个定理很重要, 由它可以导出许多有用的结论.
推论 1.1
对于函数 \(\varphi ( x , y ) = x ^ { n } + y ^ { m }\) , 若
\[\frac {\operatorname {div} X}{n Q x ^ {n - 1} - m P y ^ {m - 1}} = f \left(x ^ {n} + y ^ {m}\right)
\]
则有积分因子
\[\mu (x, y) = \mathrm{e} ^ {- \int f (x ^ {n} + y ^ {m}) \mathrm {d} (x ^ {n} + y ^ {m})}
\]
推论 1.2
对于函数 \(\varphi ( x , y ) = x y\) , 若
\[\frac {\operatorname {div} X}{y Q - x P} = f (x y)
\]
则有积分因子
\[\mu (x, y) = \mathrm {e} ^ {- \int f (x y) \mathrm {d} (x y)}
\]
推论 1.3
对于函数 \(\varphi ( x , y ) = \mathbf { e } ^ { \int f ( x ) \mathrm { d } x + g ( y ) \mathrm { d } y }\) , 若
\[\frac {\operatorname {div} X}{X (\varphi (x , y))} = \frac {- 1}{\varphi (x , y)}
\]
则有积分因子
\[\mu (x, y) = \mathrm {e} ^ {\int f (x) \mathrm {d} x + g (y) \mathrm {d} y}
\]
例 1.5 求微分方程 \(( x ^ { 2 } y + y ^ { 2 } ) \mathrm { d } x - x ^ { 3 } \mathrm { d } y = 0\) 的积分因子.
解：已知
\[X=-x^{3} \frac{\partial}{\partial x}-\left(x^{2} y+y^{2}\right) \frac{\partial}{\partial y}
\]
\[\operatorname{div} X=-4x^2-2y
\]
令 \(\varphi(x,y)=xy\) 可以得到
\[\frac{\operatorname{div} X}{X\varphi}=\frac{2}{\varphi}
\]
于是
\[\mu(x, y)=\mathrm{e}^{-\int \frac{2}{\varphi} {\rm d} \varphi}=\mathrm{e}^{-2 \ln |\varphi|}=\frac{1}{x^{2} y^{2}}
\]
例 1.6 求 \(( x + y ) \mathrm { d } x - ( x - y ) \mathrm { d } y = 0\) 的积分因子.
解:已知
\[X = (y - x) \frac {\partial}{\partial x} - (x + y) \frac {\partial}{\partial y} \quad \operatorname {div} X = - 2
\]
令 \(\varphi ( x , y ) = x ^ { 2 } + y ^ { 2 }\) , 可以得到
\[\frac {\operatorname {div} X}{X \varphi} = \frac {1}{\varphi}
\]
于是
\[\mu (x, y) = \mathrm {e} ^ {- \int \frac {1}{\varphi} \mathrm {d} \varphi} = \frac {1}{x ^ {2} + y ^ {2}}
\]
例1.7 \(x ( 4 y \mathrm { d } x + 2 x \mathrm { d } y ) + y ^ { 3 } ( 3 y \mathrm { d } x + 5 x \mathrm { d } y ) = 0\)
解:原式化为
\[(4 x y + 3 y ^ {4}) \mathrm {d} x + (2 x ^ {2} + 5 x y ^ {3}) \mathrm {d} y = 0
\]
已知
\[X = \left(2 x ^ {2} + 5 x y ^ {3}\right) \frac {\partial}{\partial x} - \left(4 x y + 3 y ^ {4}\right) \frac {\partial}{\partial y} \quad \operatorname {d i v} X = - 7 y ^ {3}
\]
令 \(\varphi ( x , y ) = x ^ { a } y ^ { b }\) , 可以得到
\[\frac {\operatorname {div} X}{X \varphi} = \frac {- 7 y ^ {3}}{- x ^ {a} y ^ {b} \left[ (4 b - 2 a) x + (3 b - 5 a) y ^ {3} \right]}
\]
令 \(4 b - 2 a = 0 , 3 b - 5 a = - 7 ,\) , 则 \(a = 2 , b = 1\) .
\[\frac {\operatorname {div} X}{X \varphi} = - \frac {1}{\varphi}
\]
于是
\[\mu (x, y) = \mathrm {e} ^ {- \int - \frac {1}{\varphi} \mathrm {d} \varphi} = x ^ {2} y
\]
乘上积分因子, 原方程变为
\[(4 x ^ {3} y ^ {2} + 3 x ^ {2} y ^ {5}) \mathrm {d} x + (2 x ^ {4} y + 5 x ^ {3} y ^ {4}) \mathrm {d} y = 0
\]
\[\mathrm {d} \left(x ^ {3} y ^ {5}\right) + \mathrm {d} \left(x ^ {4} y ^ {2}\right) = 0
\]
\[x ^ {3} y ^ {5} + x ^ {4} y ^ {2} = C
\]
例 1.8 \(\left( x + y ^ { 3 } \right) \mathrm { d } x + \left( x ^ { 3 } + y \right) \mathrm { d } y = 0\)
解:注意到
\[\frac {\operatorname {div} X}{X \varphi} = \frac {3 (y ^ {2} - x ^ {2})}{\frac {\partial \varphi}{\partial x} (x ^ {3} + y) - \frac {\partial \varphi}{\partial y} (x + y ^ {3})}
\]
令 \(\varphi ( x , y ) = x ^ { a } + y ^ { a }\) , 可以得到
\[\frac {\operatorname {div} X}{X \varphi} = \frac {3 \left(y ^ {2} - x ^ {2}\right)}{a \left(x y ^ {a - 1} + y ^ {a + 2} - x ^ {a + 2} - x ^ {a - 1} y\right)}
\]
显然取 \(a = 2\).积分因子为 \(( x ^ { 2 } + y ^ { 2 } ) ^ { - \frac { 3 } { 2 } }\) . 原方程变为
\[\frac{x+y^{3}}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}} {\rm d} x+\frac{x^{3}+y}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}} {\rm d} y=0
\]
\[-\frac{1}{\sqrt{x^{2}+y^{2}}}+y \sin \left(\arctan \frac{x}{y}\right)=C
\]
定理 1.3 若存在一函数 \(\Omega ( x , y ) \in C ^ { 1 }\) , 使得 \({ \frac { \operatorname {div} ( \Omega ( x , y ) X ) } { \Omega ( x , y ) Q } } = H ( x ) .\) , 则
\[\mu (x, y) = \Omega (x, y) e ^ {- \int H (x) \mathrm{d}x}
\]
是方程 (1) 的积分因子.
若存在一函数 \(\Omega ( x , y ) \in C ^ { 1 }\) , 使得 \(\frac { \mathrm {div} ( \Omega ( x , y ) X ) } { - \Omega ( x , y ) P } = R ( y ) ,\) , 则
\[\mu (x, y) = \Omega (x, y) e ^ {- \int R (y) \mathrm{d}y}
\]
是方程 (1) 的积分因子.
证明: 见[胡彦霞. “一阶常微分方程积分因子解法 (原文有一些计算错误)”. In: 井冈山大学学报：自然科学版 40.6(2019), p. 5.]
例1.9 求解 Bernoulli 方程 \(\frac { \mathrm { d } y } { \mathrm{d} x } = p ( x ) y + q ( x ) y ^ { n } , n \neq 0 , 1\) 的积分因子.
解:已知
\[\begin{align*}
X&=-\frac{\partial}{\partial x}-\left(p(x) y+q(x) y^n\right) \frac{\partial}{\partial y}\\
\operatorname{div} X&=-p(x)-n q(x) y^{n-1}
\end{align*}
\]
令 \(\Omega ( x , y ) = - y ^ { n }\) 可得
\[\frac {\operatorname {div} (\Omega x)}{\Omega Q} = (1 - n) p (x)
\]
于是
\[\mu (x, y) = y ^ {- n} \mathrm {e} ^ {- \int (1 - n) p (x) \mathrm {d} x}
\]
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