常数变易法求解非齐次线性微分方程

常数变易法求解一阶非齐次线性微分方程

对于一阶非齐次线性微分方程

$$y' + p(x)y = q(x)$$

先用分离变量法求解对应的齐次方程

$$\begin{aligned} & y' + p(x)y = 0 \\ \Rightarrow & y = C e^{- \int p(x)dx} \end{aligned}$$

将 $C$ 改为 $C(x)$，令 $y = C(x) e^{- \int p(x)dx}$，代入原非齐次方程得

$$\begin{aligned} & \left[ C'(x) e^{- \int p(x)dx} - p(x) e^{- \int p(x)dx} \right] + p(x) e^{- \int p(x)dx} = q(x) \\ \Rightarrow & C'(x) e^{- \int p(x)dx} = q(x) \\ \Rightarrow & C(x) = \int q(x) e^{\int p(x)dx} dx + C \end{aligned}$$

所以一阶非齐次线性微分方程的通解为

$$y = e^{- \int p(x)dx} \left( \int q(x) e^{\int p(x)dx} dx + C \right)$$

常数变易法求解二阶非齐次线性微分方程

对于二阶非齐次线性微分方程

$$y''+p(x)y'+q(x)y=f(x)$$

设对应齐次方程的两个线性无关解为 $y_1,y_2$，则其通解为

$$y = C_1 y_1 + C_2 y_2（C_1,C_2为任意常数）$$

因此可设非齐次方程的特解为

$$y^* = C_1(x) y_1 + C_2(x) y_2$$

为确定函数 $C_1(x),C_2(x)$，可对上式进行求导得

$$\begin{aligned} (y^*)' &= [C_1'(x) y_1 + C_1(x) y_1'] + [C_2'(x) y_2 + C_2(x) y_2'] \\ &= [C_1'(x) y_1 + C_2'(x) y_2] + [C_1(x) y_1' + C_2(x) y_2'] \end{aligned}$$

接下来对上式再进行一次求导，不过在此之前，为了使得 $y''$ 中不含 $C_1''(x),C_2''(x)$，可令 $C_1'(x) y_1 + C_2'(x) y_2 = 0$，现在对上式求导得

$$\begin{aligned} (y^*)'' &= [C_1(x) y_1' + C_2(x) y_2']' \\ &= [C_1'(x) y_1' + C_1(x) y_1''] + [C_2'(x) y_2' + C_2(x) y_2''] \end{aligned}$$

将 $y,y',y''$ 代入原非齐次方程得

$$\begin{aligned} & [C_1'(x) y_1' + C_1(x) y_1''] + [C_2'(x) y_2' + C_2(x) y_2''] + p(x)[C_1(x) y_1' + C_2(x) y_2'] + q(x) [C_1(x) y_1 + C_2(x) y_2] = f(x) \\ \Rightarrow & C_1(x) [y_1''+p(x)y_1'+q(x)y_1] + C_2(x) [y_2''+p(x)y_2'+q(x)y_2] + [C_1'(x) y_1' + C_2'(x) y_2'] = f(x) \\ \Rightarrow & C_1'(x) y_1' + C_2'(x) y_2' = f(x) \end{aligned}$$

联立两个方程

$$\begin{cases} C_1'(x) y_1 + C_2'(x) y_2 = 0 \\ C_1'(x) y_1' + C_2'(x) y_2' = f(x) \\ \end{cases}$$

即可求得 $C_1'(x),C_2'(x)$，最后进行积分得到 $C_1(x),C_2(x)$。

【注】常数变易法在同济七版高等数学中有介绍，适用于求解任意二阶非齐次常系数线性微分方程（提醒：在考研范围内，非齐次项的形式是固定的，而非任意形式）。

例题

【例 1】求解微分方程 $y'' + 3y' + 2y = x^2$

【解】先求对应齐次通解：$y = C_1 e^{-x} + C_2 e^{-2x}$，所以 $y_1 = e^{-x}, y_2 = e^{-2x}$，解方程组

$$\begin{cases} C_1'(x) e^{-x} + C_2'(x) e^{-2x} = 0 \\ -C_1'(x) e^{-x} - 2C_2'(x) e^{-2x} = x^2 \\ \end{cases}$$

可求得

$$\begin{cases} C_1'(x) = \begin{vmatrix} 0 & e^{-2x} \\ x^2 & -2e^{-2x} \end{vmatrix} / \begin{vmatrix} e^{-x} & e^{-2x} \\ -e^{-x} & -2e^{-2x} \end{vmatrix} = \frac{-x^2e^{-2x}}{-e^{-3x}} = x^2 e^x\\ C_2'(x) = \begin{vmatrix} e^{-x} & 0 \\ -e^{-x} & x^2 \end{vmatrix} / \begin{vmatrix} e^{-x} & e^{-2x} \\ -e^{-x} & -2e^{-2x} \end{vmatrix} = \frac{x^2e^{-x}}{-e^{-3x}} = -x^2 e^{2x} \end{cases}$$

于是

$$\begin{cases} C_1(x) = (x^2-2x+2) e^x\\ C_2(x) = -\frac{1}{4}(2x^2 - 2x + 1) e^{2x} \end{cases}$$

所以特解为

$$\begin{aligned} y^* &= C_1(x) y_1 + C_2(x) y_2 \\ &= (x^2-2x+2) - \frac{1}{4}(2x^2 - 2x + 1) \\ &= \frac{1}{2}x^2 - \frac{3}{2}x + \frac{7}{4} \end{aligned}$$

【例 2】求解微分方程 $y'' + 3y' + 2y = \sin x$

【解】先求对应齐次通解：$y = C_1 e^{-x} + C_2 e^{-2x}$，所以 $y_1 = e^{-x}, y_2 = e^{-2x}$，解方程组

$$\begin{cases} C_1'(x) e^{-x} + C_2'(x) e^{-2x} = 0 \\ -C_1'(x) e^{-x} - 2C_2'(x) e^{-2x} = \sin x \\ \end{cases}$$

可求得

$$\begin{cases} C_1'(x) = \begin{vmatrix} 0 & e^{-2x} \\ \sin x & -2e^{-2x} \end{vmatrix} / \begin{vmatrix} e^{-x} & e^{-2x} \\ -e^{-x} & -2e^{-2x} \end{vmatrix} = \frac{e^{-2x}\sin x}{-e^{-3x}} = e^x \sin x\\ C_2'(x) = \begin{vmatrix} e^{-x} & 0 \\ -e^{-x} & \sin x \end{vmatrix} / \begin{vmatrix} e^{-x} & e^{-2x} \\ -e^{-x} & -2e^{-2x} \end{vmatrix} = \frac{e^{-x}\sin x}{-e^{-3x}} = -e^{2x} \sin x \end{cases}$$

于是

$$\begin{cases} C_1(x) = \frac{1}{2} (\sin x - \cos x) e^x\\ C_2(x) = -\frac{4}{5}(\frac{1}{2} \sin x - \frac{1}{4} \cos x) e^{2x} \end{cases}$$

所以特解为

$$\begin{aligned} y^* &= C_1(x) y_1 + C_2(x) y_2 \\ &= \frac{1}{2} (\sin x - \cos x) -\frac{4}{5}(\frac{1}{2} \sin x - \frac{1}{4} \cos x) \\ &= \frac{1}{10} \sin x - \frac{3}{10} \cos x \end{aligned}$$

【例 3】求解微分方程 $y'' + 4y = \cos 2x$

【解】先求对应齐次通解：$y = C_1 \cos 2x + C_2 \sin 2x$，所以 $y_1 = \cos 2x, y_2 = \sin 2x$，解方程组

$$\begin{cases} C_1'(x) \cos 2x + C_2'(x) \sin 2x = 0 \\ -2C_1'(x) \sin 2x + 2C_2'(x) \cos 2x = \cos 2x \\ \end{cases}$$

可求得

$$\begin{cases} C_1'(x) = \begin{vmatrix} 0 & \sin 2x \\ \cos 2x & 2\cos 2x \end{vmatrix} / \begin{vmatrix} \cos 2x & \sin 2x \\ -2\sin 2x & 2\cos 2x \end{vmatrix} = -\frac{\sin 2x \cos 2x}{2\cos^2 2x + 2\sin^2 2x} = -\frac{1}{4} \sin 4x \\ C_2'(x) = \begin{vmatrix} \cos 2x & 0 \\ -2\sin 2x & \cos 2x \end{vmatrix} / \begin{vmatrix} \cos 2x & \sin 2x \\ -2\sin 2x & 2\cos 2x \end{vmatrix} = \frac{\cos^2 2x}{2\cos^2 2x + 2\sin^2 2x} = \frac{1}{4} (\cos 4x + 1) \end{cases}$$

于是

$$\begin{cases} C_1(x) = \frac{1}{16} \cos 4x \\ C_2(x) = \frac{1}{4} x + \frac{1}{16}\sin 4x \end{cases}$$

所以特解为

$$\begin{aligned} y^* &= C_1(x) y_1 + C_2(x) y_2 \\ &= \frac{1}{16} \cos 4x \cos 2x + (\frac{1}{4} x + \frac{1}{16}\sin 4x) \sin 2x \\ &= \frac{1}{16} (\cos 4x \cos 2x + \sin 4x \sin 2x) + \frac{1}{4} x \sin 2x \\ &= \frac{1}{16} \cos 2x + \frac{1}{4} x \sin 2x \end{aligned}$$

由于方程的通解为

$$\begin{aligned} y &= (C_1 + \frac{1}{16}) \cos 2x + C_2 \sin 2x + \frac{1}{4} x \sin 2x \\ &= C_3 \cos 2x + C_2 \sin 2x + \frac{1}{4} x \sin 2x \end{aligned}$$

所以特解应为

$$y^* = \frac{1}{4} x \sin 2x$$

【例 4】求解微分方程 $y'' - 2y' + y = xe^x$

【解】先求对应齐次通解：$y = C_1 e^{x} + C_2 xe^{x}$，所以 $y_1 = e^{x}, y_2 = xe^{x}$，解方程组

$$\begin{cases} C_1'(x) e^{x} + C_2'(x) xe^{x} = 0 \\ C_1'(x) e^{x} + C_2'(x) (x+1)e^{x} = xe^x \\ \end{cases}$$

可求得

$$\begin{cases} C_1'(x) = \begin{vmatrix} 0 & xe^{x} \\ xe^{x} & (x+1)e^{x} \end{vmatrix} / \begin{vmatrix} e^{x} & xe^{x} \\ e^{x} & (x+1)e^{x} \end{vmatrix} = \frac{-x^2e^{2x}}{e^{2x}} = -x^2 \\ C_2'(x) = \begin{vmatrix} e^{x} & 0 \\ e^{x} & xe^{x} \end{vmatrix} / \begin{vmatrix} e^{x} & xe^{x} \\ e^{x} & (x+1)e^{x} \end{vmatrix} = \frac{xe^{2x}}{e^{2x}} = x \end{cases}$$

于是

$$\begin{cases} C_1(x) = -\frac{1}{3} x^3 \\ C_2(x) = \frac{1}{2} x^2 \end{cases}$$

所以特解为

$$\begin{aligned} y^* &= C_1(x) y_1 + C_2(x) y_2 \\ &= -\frac{1}{3} x^3 \cdot e^{x} + \frac{1}{2} x^2 \cdot xe^{x} \\ &= \frac{1}{6} x^3 e^{x} \end{aligned}$$