线性方程及常数变易法求解的原理

方程

\[a(x)\frac{{\text{d}y}}{{\text{d}x}} + b(x)y = c(x)\]

叫做一阶线性微分方程.改写为

\[\frac{{\text{d}y}}{{\text{d}x}} + \frac{{b(x)}}{{a(x)}}y = \frac{{c(x)}}{{a(x)}}\]

\begin{equation} \frac{{\text{d}y}}{{\text{d}x}} + P(x)y = Q(x) \end{equation}

\[P(x) = \frac{{b(x)}}{{a(x)}},Q(x) = \frac{{c(x)}}{{a(x)}}\] \[\frac{{\text{d}y}}{{\text{d}x}} = {\rm{ - }}P(x)y + Q(x)\] \[y' = f(x,y)\]

$(1)$式叫做一阶线性微分方程的标准形式，原因在于：

$(1)$式是微分方程，其中$y$的最高阶导数的次数是1次，因此是一阶微分方程.

$y,$\(y'\)的最高次幂是1次，因此是线性的，即一阶线性微分方程.

特殊情况下，$x,y$等变量可以不出现，但是$y'$是必须出现的.

当$c(x) = 0$时，方程

$$a(x)\frac{{\text{d}y}}{{\text{d}x}} + b(x)y = 0$$

称为一阶齐次线性微分方程，其中$a(x)$和$b(x)$是自变量$x$的函数，这个方程的解可以通过积分求出.

等号两端同时乘以${\text{d}}x$，再除以$a(x)y$，整理得

$$\frac{1}{y}\text{d}y = - \frac{{b(x)}}{{a(x)}}\text{d}x$$

两端积分，得

$$\ln |y| = - \int {\frac{{b(x)}}{{a(x)}}\text{d}x} + K$$

其中$K$是任意常数.则

$$y = \pm {e^K}{e^{ - \int {[b(x)/a(x)]\text{d}x} }}$$

因为$K$是任意常数，从而$ \pm {e^K}$是非零的任意常数，可知$y=0$也是原方程的解，因此方程的解为

$$y = C{e^{ - \int {[b(x)/a(x)]\text{d}x} }}\\ = C{e^{ - \int {[P(x)\text{d}x} }} $$

其中$C$为任意常数.

上面的解法只适用于一阶齐次线性微分方程，当$c(x) \ne 0$时，方程

$$a(x)\frac{{\text{d}y}}{{\text{d}x}} + b(x)y = c(x)$$

的解会稍微复杂.我们来解这个方程，对于方程的左端，如果$b(x)$恰好是$a(x)$的导数，即$a' (x)=b(x)$，那么左端就是$a(x)y$的导数，即

$$\frac{{{\text{d}}[a\left( x \right)y]}}{{{\text{d}}x}} = a\left( x \right)\frac{{\text{d}y}}{{\text{d}x}} + a'\left( x \right)y$$

则

$$\frac{{\text{d}[a(x)y]}}{{\text{d}x}} = c(x)$$

两端积分，得

$$a(x)y = \int {c(x)} \text{d}x + K$$

其中$K$是任意常数，得

$$y = \frac{1}{{a(x)}}[\int {c(x)} \text{d}x + K]$$

上面的解法只适用于$y$的系数$b(x)$恰好是$\frac{{\text{d}y}}{{\text{d}x}}$的系数$a(x)$的导数的情况，这不是普遍情况，但是我们也许可以在一般的非齐次线性微分方程两端乘以某个函数$u(x)$，而将该方程转化为上面的情况，即

$$u(x)a(x)\frac{{\text{d}y}}{{\text{d}x}} + u(x)b(x)y = u(x)c(x)$$

这里选择的函数$u(x)$能够使$u(x)b(x)$为$u(x)a(x)$的导数，即

$$\frac{{\text{d}[u(x)a(x)]}}{{\text{d}x}}$$ $$ = u(x)\frac{{\text{d}a(x)}}{{\text{d}x}} + \frac{{\text{d}u(x)}}{{\text{d}x}}a(x) = u(x)b(x)$$

重新整理上式，得

$$a(x)\frac{{\text{d}u(x)}}{{\text{d}x}} + u(x)[\frac{{\text{d}a(x)}}{{\text{d}x}} - b(x)] = 0$$

这是一个关于函数$u(x)$的一阶齐次线性微分方程，如果我们把

$$\frac{{\text{d}a(x)}}{{\text{d}x}} - b(x)$$

看作函数$u(x)$的系数，则

$$\frac{{\text{d}u(x)}}{{u(x)}} = - \frac{1}{{a(x)}}[\frac{{\text{d}a(x)}}{{\text{d}x}} - b(x)]\text{d}x$$

两端积分，得

$$\int {\frac{{\text{d}u(x)}}{{u(x)}}} = $$ $$- \int {\frac{1}{{a(x)}}[\frac{{\text{d}a(x)}}{{\text{d}x}} - b(x)]\text{d}x} $$ $$\ln |u(x)| = $$ $$- \int {\{ \frac{1}{{a(x)}}[\frac{{\text{d}a(x)}}{{\text{d}x}} - b(x)]\} \text{d}x} + K$$

其中$K$是任意常数.

$$\left| {u(x)} \right| = {{\text{e}}^{ - \int {\left\{ {\frac{1}{{a\left( x \right)}}\left[ {\frac{{{\text{d}}a\left( x \right)}}{{{\text{d}}x}} - b\left( x \right)} \right]} \right\}{\text{d}}x} + K}}$$ $$u(x) = \pm {{\text{e}}^K} \cdot {{\text{e}}^{ - \mathop {\int {\left\{ {\frac{1}{{a(x)}}\left[ {\frac{{{\text{d}}a(x)}}{{{\text{d}}x}} - b(x)} \right]} \right\}{\text{d}}x} }\nolimits }}$$

其中$±e^K$是不为零的任意常数，因为$u(x)=0$也使方程

$$a(x)\frac{{\text{d}u(x)}}{{\text{d}x}} + u(x)[\frac{{\text{d}a(x)}}{{\text{d}x}} - b(x)] = 0$$

成立，因此用任意常数$k$代替$±e^K$，解得

$$u(x) = k{{\text{e}}^{ - \int {\left\{ {\frac{1}{{a(x)}}\left[ {\frac{{{\text{d}}a(x)}}{{{\text{d}}x}} - b(x)} \right]} \right\}{\text{d}}x} }}$$

从而，一阶非齐次线性微分方程

$$a(x)\frac{{\text{d}y}}{{\text{d}x}} + b(x)y = c(x)$$

在乘以函数$u(x)$后所得方程

$$u(x)a(x)\frac{{\text{d}y}}{{\text{d}x}} + u(x)b(x)y = u(x)c(x)$$

的解可以以如下形式表示

$$\text{d}\{ [u(x)a(x)]y\} = u(x)c(x)\text{d}x$$

两端积分，得

\[\begin{gathered} u(x)a(x)y = \int {u(x)c(x)\text{d}x} + C \hfill \\ y = \frac{1}{{u(x)a(x)}}[\int {u(x)c(x)\text{d}x} + C] \hfill \\ \end{gathered} \] \[y= \\ \frac{\int{kc\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}\text{d}x}+C}{ka\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}}\] \[y= \\ \frac{\int{c\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}\text{d}x}+\frac{C}{k}}{a\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}}\] \[y= \\ \frac{\int{c\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}\text{d}x}+C}{a\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}}\] \[y= \\ \frac{\int{c\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}\text{d}x}+C}{a\left( x \right){{\text{e}}^{-\mathop{\int{\left\{ \frac{1}{a\left( x \right)}\left[ \frac{\text{d}a\left( x \right)}{\text{d}x}-b\left( x \right) \right] \right\}\text{d}x}}^{}}}}\]

对于非齐次一阶线性微分方程

$$\frac{{\text{d}y}}{{\text{d}x}} + P(x)y = Q(x)$$

两端乘以$u(x)$，得

\[\begin{gathered} u(x)\frac{{\text{d}y}}{{\text{d}x}} + u(x)P(x)y = u(x)Q(x) \hfill \\ \frac{{\text{d}u(x)}}{{\text{d}x}} = u(x)P(x) \hfill \\ \frac{{\text{d}u(x)}}{{u(x)}} = P(x)\text{d}x \hfill \\ \ln |u(x)| = \int {P(x)\text{d}x} + C \hfill \\ u(x) = \pm {e^C}{e^{\int {P(x)\text{d}x} }} \hfill \\ u(x) = {C_1}{e^{\int {P(x)\text{d}x} }} \hfill \\ \frac{{\text{d}[u(x)y]}}{{\text{d}x}} = u(x)Q(x) \hfill \\ \text{d}[u(x)y] = u(x)Q(x)\text{d}x \hfill \\ u(x)y = \int {u(x)Q(x)\text{d}x} + {C_2} \hfill \\ y = \frac{1}{{u(x)}}[\int {u(x)Q(x)\text{d}x} + {C_2}] \hfill \\ y = \frac{1}{{{C_1}{e^{\int {P(x)\text{d}x} }}}} \cdot \hfill \\ [\int {{C_1}{e^{\int {P(x)\text{d}x} }}Q(x)\text{d}x} + {C_2}] \hfill \\ y = \frac{1}{{{C_1}{e^{\int {P(x)\text{d}x} }}}}\int {{C_1}{e^{\int {P(x)\text{d}x} }}Q(x)\text{d}x} \hfill \\ + \frac{{{C_2}}}{{{C_1}{e^{\int {P(x)\text{d}x} }}}} \hfill \\ y = \frac{1}{{{e^{\int {P(x)\text{d}x} }}}}\int {{e^{\int {P(x)\text{d}x} }}Q(x)\text{d}x} \hfill \\ + C{e^{ - \int {P(x)\text{d}x} }} \hfill \\ y = {e^{ - \int {P(x)\text{d}x} }}\int {{e^{\int {P(x)\text{d}x} }}Q(x)\text{d}x} \hfill \\ + C{e^{ - \int {P(x)\text{d}x} }} \hfill \\ y = C{e^{ - \int {P(x)\text{d}x} }} + \hfill \\ {e^{ - \int {P(x)\text{d}x} }}\int {Q(x){e^{\int {P(x)\text{d}x} }}\text{d}x} \hfill \\ \hfill \\ \end{gathered} \]