假设有 $n$ 阶非齐次常微分方程, $$ \begin{equation}\label{eq:n阶非齐次常微分方程} \begin{aligned} \left[\dfrac{\mathrm{d}^{n}}{\mathrm{d} t^{n}} + p_{n – 1}\left(t\right) \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} + \cdots + p_{1}\left(t\right) \dfrac{\mathrm{d}}{\mathrm{d} t} + p_{0}\left(t\right)\right] x\left(t\right) = q\left(t\right). \end{aligned} \end{equation} $$ 其中 $p_{0}\left(t\right)$, $\cdots$, $p_{n – 1}\left(t\right)$, $q\left(t\right)$ 在我们所研究之 $t$ 之取值区间 $\mathcal{O}$ 上连续. 构造如下矩阵, $$ \begin{equation}\label{eq:构造矩阵} \begin{aligned} & X\left(t\right) \equiv \begin{pmatrix} x\left(t\right) \\ \dfrac{\mathrm{d}}{\mathrm{d} t} x\left(t\right) \\ \vdots \\ \dfrac{\mathrm{d}^{n – 2}}{\mathrm{d} t^{n – 2}} x\left(t\right) \\ \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} x\left(t\right) \end{pmatrix}, \\ & \\ & P\left(t\right) \equiv \begin{pmatrix} 0 & -1 & 0 & \cdots & 0 \\ 0 & 0 & -1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1 \\ p_{0}\left(t\right) & p_{1}\left(t\right) & p_{2}\left(t\right) & \cdots & p_{n – 1}\left(t\right) \end{pmatrix}, \\ & \\ & Q\left(t\right) \equiv \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ q\left(t\right) \end{pmatrix}. \end{aligned} \end{equation} $$ 则方程 $\eqref{eq:n阶非齐次常微分方程}$ 可写为矩阵形式, $$ \begin{equation}\label{eq:n阶非齐次常微分方程写为矩阵形式} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} X\left(t\right) + P\left(t\right) X\left(t\right) = Q\left(t\right). \end{aligned} \end{equation} $$ 其所对应之齐次方程为 $$ \begin{equation}\label{eq:n阶非齐次常微分方程之矩阵形式所对应之齐次方程} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \tilde{X}\left(t\right) + P\left(t\right) \tilde{X}\left(t\right) = 0. \end{aligned} \end{equation} $$ 假设齐次方程 $\eqref{eq:n阶非齐次常微分方程之矩阵形式所对应之齐次方程}$ 已得通解, 在其解空间中选出一组基, 即选出 $n$ 个线性无关之解, $$ \begin{equation}\label{eq:n阶非齐次常微分方程所对应之齐次方程之n个线性无关解} \begin{aligned} \Phi_{1}\left(t\right), \cdots, \Phi_{n}\left(t\right). \end{aligned} \end{equation} $$ 它们均满足式 $\eqref{eq:n阶非齐次常微分方程之矩阵形式所对应之齐次方程}$, 即 $$ \begin{equation}\label{eq:n个齐次线性无关解均满足齐次方程} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi_{\alpha}\left(t\right) + P\left(t\right) \Phi_{\alpha}\left(t\right) = 0, \alpha = 1, \cdots, n. \end{aligned} \end{equation} $$ 将它们组成一个解矩阵, $$ \begin{equation}\label{eq:n个齐次线性无关解组成大矩阵} \begin{aligned} \Phi\left(t\right) \equiv \left[\Phi_{1}\left(t\right), \cdots, \Phi_{n}\left(t\right)\right]. \end{aligned} \end{equation} $$ 则 $\Phi\left(t\right)$ 亦满足齐次方程 $\eqref{eq:n阶非齐次常微分方程之矩阵形式所对应之齐次方程}$, $$ \begin{equation}\label{eq:n个齐次线性无关解组成之大矩阵满足齐次方程} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi\left(t\right) + P\left(t\right) \Phi\left(t\right) = 0. \end{aligned} \end{equation} $$ 由于各个 $\Phi_{\alpha}\left(t\right)$ 线性无关, 因此至少存在一个 $t_{0} \in \mathcal{O}$ 使得 $\Phi\left(t_{0}\right)$ 之秩为 $n$, 从而 $\mathrm{det}\Phi\left(t_{0}\right) \neq 0$, 即 $\Phi\left(t_0\right)$ 可逆. 为了证明 $\forall t \in \mathcal{O}$ 之 $\Phi\left(t\right)$ 均可逆, 我们利用行列式微分公式, $$ \begin{equation}\label{eq:行列式微分公式} \begin{aligned} \mathrm{d} \left[\mathrm{det} \Phi\left(t\right)\right] = \mathrm{tr}\left\{\mathrm{adj}\left[\Phi\left(t\right)\right] \mathrm{d} \Phi\left(t\right)\right\}. \end{aligned} \end{equation} $$ 由式 $\eqref{eq:行列式微分公式}$ 可知, $$ \begin{equation}\label{eq:行列式微商公式} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \left[\mathrm{det} \Phi\left(t\right)\right] = \mathrm{tr}\left\{\mathrm{adj}\left[\Phi\left(t\right)\right] \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi\left(t\right)\right\}. \end{aligned} \end{equation} $$ 将式 $\eqref{eq:n个齐次线性无关解组成之大矩阵满足齐次方程}$ 代入式 $\eqref{eq:行列式微商公式}$, 可知, $$ \begin{equation}\label{eq:解矩阵所满足之方程代入行列式微商公式} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \left[\mathrm{det} \Phi\left(t\right)\right] = – \mathrm{tr}\left\{\mathrm{adj}\left[\Phi\left(t\right)\right] P\left(t\right) \Phi\left(t\right)\right\}. \end{aligned} \end{equation} $$ 其中 $\mathrm{adj}\left[\Phi\left(t\right)\right]$ 为 $\Phi\left(t\right)$ 之伴随矩阵, 利用矩阵乘积求迹运算之轮换不变性可知, $$ \begin{equation}\label{eq:利用矩阵乘积求迹运算之轮换不变性化简行列式微商公式} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \left[\mathrm{det} \Phi\left(t\right)\right] = {} & – \mathrm{tr}\left\{\mathrm{adj}\left[\Phi\left(t\right)\right] P\left(t\right) \Phi\left(t\right)\right\} \\ = {} & – \mathrm{tr}\left\{P\left(t\right) \Phi\left(t\right) \mathrm{adj}\left[\Phi\left(t\right)\right]\right\} \end{aligned} \end{equation} $$ 由于任意方阵 $A$ (不要求可逆) 均满足 $$ \begin{equation}\label{eq:任意矩阵与其伴随矩阵之乘积均满足之公式} \begin{aligned} A \mathrm{adj} \left(A\right) = {} & \mathrm{adj} \left(A\right) A \\ = {} & I \mathrm{det} A, \end{aligned} \end{equation} $$ 其中 $I$ 为单位矩阵, 将式 $\eqref{eq:任意矩阵与其伴随矩阵之乘积均满足之公式}$ 代入式 $\eqref{eq:利用矩阵乘积求迹运算之轮换不变性化简行列式微商公式}$ 即可知, $$ \begin{equation}\label{eq:行列式微商公式化简} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \left[\mathrm{det} \Phi\left(t\right)\right] = – \mathrm{det} \Phi\left(t\right) \mathrm{tr}\left[P\left(t\right)\right]. \end{aligned} \end{equation} $$ 由此可解得 $$ \begin{equation}\label{eq:行列式微商公式之解} \begin{aligned} \mathrm{det} \Phi\left(t\right) = \mathrm{det} \Phi\left(t_{0}\right) \mathrm{exp} \left\{- \int_{t_{0}}^{t} \mathrm{d} t^{\prime} \mathrm{tr}\left[P\left(t^{\prime}\right)\right]\right\}. \end{aligned} \end{equation} $$ 由于指数因子 $\mathrm{exp} \left\{- \int_{t_{0}}^{t} \mathrm{d} t^{\prime} \mathrm{tr}\left[P\left(t^{\prime}\right)\right]\right\}$ 一定不为零, 且 $\mathrm{det} \Phi\left(t_{0}\right) \neq 0$, 因此 $$ \begin{equation}\label{eq:任意时刻之解矩阵行列式均不为零} \begin{aligned} \mathrm{det} \Phi\left(t\right) \neq 0, \forall t \in \mathcal{O}. \end{aligned} \end{equation} $$ 从而 $\forall t \in \mathcal{O}$ 之 $\Phi\left(t\right)$ 均可逆. 令 $$ \begin{equation}\label{eq:对因变量作变换} \begin{aligned} Y\left(t\right) \equiv \Phi^{-1}\left(t\right) X\left(t\right). \end{aligned} \end{equation} $$ 计算 $\dfrac{\mathrm{d}}{\mathrm{d} t} Y\left(t\right)$, $$ \begin{equation}\label{eq:对变换后之因变量求微商} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} Y\left(t\right) = \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi^{-1}\left(t\right) X\left(t\right) + \Phi^{-1}\left(t\right) \dfrac{\mathrm{d}}{\mathrm{d} t} X\left(t\right). \end{aligned} \end{equation} $$ 由于 $\Phi\left(t\right) \Phi^{-1}\left(t\right) = I$, 其中 $I$ 为单位矩阵, $I = \mathrm{diag}\left(1, \cdots, 1\right)$, 因此有 $$ \begin{equation}\label{eq:单位矩阵求微商} \begin{aligned} 0 = {} & \dfrac{\mathrm{d}}{\mathrm{d} t} I \\ = {} & \dfrac{\mathrm{d}}{\mathrm{d} t} \left[\Phi\left(t\right) \Phi^{-1}\left(t\right)\right] \\ = {} & \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi\left(t\right) \Phi^{-1}\left(t\right) + \Phi\left(t\right) \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi^{-1}\left(t\right). \end{aligned} \end{equation} $$ 由此可知, $$ \begin{equation}\label{eq:解矩阵逆之微商} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi^{-1}\left(t\right) = – \Phi^{-1}\left(t\right) \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi\left(t\right) \Phi^{-1}\left(t\right). \end{aligned} \end{equation} $$ 将式 $\eqref{eq:解矩阵逆之微商}$ 代入式 $\eqref{eq:对变换后之因变量求微商}$ 即可得, $$ \begin{equation}\label{eq:将解矩阵逆之微商代入变换后因变量之微商} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} Y\left(t\right) = – \Phi^{-1}\left(t\right) \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi\left(t\right) \Phi^{-1}\left(t\right) X\left(t\right) + \Phi^{-1} \dfrac{\mathrm{d}}{\mathrm{d} t} X\left(t\right). \end{aligned} \end{equation} $$ 由式 $\eqref{eq:n个齐次线性无关解组成之大矩阵满足齐次方程}$ 可知, $$ \begin{equation}\label{eq:n个齐次线性无关解组成之大矩阵所满足之齐次方程进行变换} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} \Phi\left(t\right) \Phi^{-1}\left(t\right) = – P\left(t\right). \end{aligned} \end{equation} $$ 将式 $\eqref{eq:n个齐次线性无关解组成之大矩阵所满足之齐次方程进行变换}$ 代入式 $\eqref{eq:将解矩阵逆之微商代入变换后因变量之微商}$ 即可得, $$ \begin{equation}\label{eq:变换后因变量微商继续化简} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} Y\left(t\right) = {} & \Phi^{-1}\left(t\right) P\left(t\right) X\left(t\right) + \Phi^{-1} \dfrac{\mathrm{d}}{\mathrm{d} t} X\left(t\right) \\ = {} & \Phi^{-1}\left(t\right) \left[\dfrac{\mathrm{d}}{\mathrm{d} t} X\left(t\right) + P\left(t\right) X\left(t\right)\right]. \end{aligned} \end{equation} $$ 将式 $\eqref{eq:n阶非齐次常微分方程写为矩阵形式}$ 代入式 $\eqref{eq:变换后因变量微商继续化简}$ 即可得, $$ \begin{equation}\label{eq:变换后因变量微商进行最终化简} \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d} t} Y\left(t\right) = {} & \Phi^{-1}\left(t\right) \left[\dfrac{\mathrm{d}}{\mathrm{d} t} X\left(t\right) + P\left(t\right) X\left(t\right)\right] \\ = {} & \Phi^{-1}\left(t\right) Q\left(t\right). \end{aligned} \end{equation} $$ 对其积分即可得, $$ \begin{equation}\label{eq:对变换后因变量微商积分} \begin{aligned} Y\left(t\right) = {} & C + \int \mathrm{d} t \Phi^{-1}\left(t\right) Q\left(t\right), \end{aligned} \end{equation} $$ 其中 $C$ 为一个常数列矩阵. 由式 $\eqref{eq:对因变量作变换}$ 与式 $\eqref{eq:对变换后因变量微商积分}$ 可得方程 $\eqref{eq:n阶非齐次常微分方程写为矩阵形式}$ 之通解, $$ \begin{equation}\label{eq:方程矩阵形式之通解} \begin{aligned} X\left(t\right) = {} & \Phi\left(t\right) \left[C + \int \mathrm{d} t \Phi^{-1}\left(t\right) Q\left(t\right)\right], \end{aligned} \end{equation} $$ 其中常数列矩阵 $C$ 需由初始条件确定. $X\left(t\right)$ 通解之第一个矩阵元即为方程 $\eqref{eq:n阶非齐次常微分方程}$ 之通解 $x\left(t\right)$. 具体写出来即为 $$ \begin{equation}\label{eq:方程之通解} \begin{aligned} x\left(t\right) = \sum_{\alpha = 1}^{n} \phi_{\alpha}\left(t\right) \left(C_{\alpha} + \int \mathrm{d} t \dfrac{\mathcal{W}_{\alpha}\left(t\right)}{\mathcal{W}\left(t\right)}\right), \end{aligned} \end{equation} $$ 其中各 $\phi_{\alpha}$ 为方程 $\eqref{eq:n阶非齐次常微分方程}$ 之各齐次通解 (即 $\Phi_{\alpha}\left(t\right)$ 之第一个矩阵元), $$ \begin{equation}\label{eq:方程之齐次通解} \begin{aligned} \left[\dfrac{\mathrm{d}^{n}}{\mathrm{d} t^{n}} + p_{n – 1}\left(t\right) \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} + \cdots + p_{1}\left(t\right) \dfrac{\mathrm{d}}{\mathrm{d} t} + p_{0}\left(t\right)\right] \phi_{\alpha}\left(t\right) = 0, \alpha = 1, \cdots, n. \end{aligned} \end{equation} $$ $\mathcal{W}\left(t\right)$ 为朗斯基行列式, $$ \begin{equation}\label{eq:朗斯基行列式} \begin{aligned} \mathcal{W}\left(t\right) \equiv {} & \mathrm{det} \Phi\left(t\right) \\ & \\ = {} & \begin{vmatrix} \phi_{1}\left(t\right) & \cdots & \phi_{n}\left(t\right) \\ \dfrac{\mathrm{d}}{\mathrm{d} t} \phi_{1}\left(t\right) & \cdots & \dfrac{\mathrm{d}}{\mathrm{d} t} \phi_{n}\left(t\right) \\ \vdots & \ddots & \vdots \\ \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} \phi_{1}\left(t\right) & \cdots & \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} \phi_{n}\left(t\right) \end{vmatrix}, \end{aligned} \end{equation} $$ $\mathcal{W}_{\alpha}\left(t\right)$ 为将 $\mathcal{W}\left(t\right)$ 中第 $\alpha$ 列换成 $Q\left(t\right)$ 后所得之行列式, $$ \begin{equation}\label{eq:替换列后之朗斯基行列式} \begin{aligned} \mathcal{W}_{\alpha}\left(t\right) \equiv \begin{vmatrix} \phi_{1}\left(t\right) & \cdots & \phi_{\alpha – 1}\left(t\right) & 0 & \phi_{\alpha + 1}\left(t\right) & \cdots & \phi_{n}\left(t\right) \\ \dfrac{\mathrm{d}}{\mathrm{d} t} \phi_{1}\left(t\right) & \cdots & \dfrac{\mathrm{d}}{\mathrm{d} t} \phi_{\alpha – 1}\left(t\right) & 0 & \dfrac{\mathrm{d}}{\mathrm{d} t} \phi_{\alpha + 1}\left(t\right) & \cdots & \dfrac{\mathrm{d}}{\mathrm{d} t} \phi_{n}\left(t\right) \\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} \phi_{1}\left(t\right) & \cdots & \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} \phi_{\alpha – 1}\left(t\right) & q\left(t\right) & \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} \phi_{\alpha + 1}\left(t\right) & \cdots& \dfrac{\mathrm{d}^{n – 1}}{\mathrm{d} t^{n – 1}} \phi_{n}\left(t\right) \end{vmatrix}. \end{aligned} \end{equation} $$