Section04_高阶线性微分方程

定义

1. $n$ 阶齐次线性微分方程 $(*)$ $$y^{(n)} + a_{1}(x) y^{(n-1)} +\cdots +a_{n-1}(x)y'+ a_{n}(x)y = 0 \quad (*)$$
2. $n$ 阶非齐次线性微分方程 $(**)$ $$y^{(n)} + a_{1}(x) y^{(n-1)} +\cdots +a_{n-1}(x)y'+ a_{n}(x)y = f(x)\quad (**)$$
   • 若 $f(x) = f_{1}(x) + f_{2}(x)$ $$\begin{split} y^{(n)} + a_{1}(x) y^{(n-1)} +\cdots +a_{n-1}(x)y'+ a_{n}(x)y = f_{1}(x)\quad (**') \\ y^{(n)} + a_{1}(x) y^{(n-1)} +\cdots +a_{n-1}(x)y'+ a_{n}(x)y = f_{2}(x)\quad (**'') \end{split}$$

解的结构

1. $\phi_{1}(x), \cdots,\phi_{s}(x)$ 为 $(*)$ 的解，则 $y = k_{1}\phi_{1}(x)+\cdots +k_{s}\phi_{s}(x)$ 还为 $(*)$ 的解
2. $\phi_{1}(x), \cdots,\phi_{s}(x)$ 为 $(**)$ 的解
   1. 若 $y = k_{1}\phi_{1}(x)+\cdots +k_{s}\phi_{s}(x)$ 为 $(*)$ 的解，则 $k_{1} + \cdots + k_{s} = 0$
   2. 若 $y = k_{1}\phi_{1}(x)+\cdots +k_{s}\phi_{s}(x)$ 为 $(**)$ 的解，则 $k_{1} + \cdots + k_{s} = 1$
3. $\phi_{1}(x),\phi_{2}(x)$ 分别为 $(*), (**)$ 的解，则 $\phi_{1}(x)+\phi_{2}(x)$ 为 $(**)$ 的解
4. $\phi_{1}(x),\phi_{2}(x)$ 为 $(**)$ 的解，则 $\phi_{1}(x) - \phi_{2}(x)$ 为 $(*)$ 的解 （2.1的特殊情况）
5. $\phi_{1}(x),\phi_{2}(x)$ 分别为 $(**'), (**'')$ 的解，则 $\phi_{1}(x)+\phi_{2}(x)$ 为 $(**)$ 的解

特殊形式

Notes 一元二次方程的解 $ax^{2} + bx + c =0$ $$\begin{array}{c}\displaystyle\Delta = b^{2} - 4ac\\\displaystyle x = -\frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a} \end{array}$$

特一 二阶常系数齐次线性微分方程

$$y'' + py' +qy = 0\quad (*)$$

通解

1. 转化为特征方程 $\lambda^{2} + p\lambda + q = 0$
2. 解出 $\lambda_{1},\lambda_{2}$
   1. $\Delta > 0,\lambda_{1}\ne \lambda_{2}$ $$y = C_{1}e^{\lambda_{1} x} + C_{2}e^{\lambda_{2} x}$$
   2. $\Delta =0,\lambda_{1} = \lambda{2}$ $$y = (C_{1} + C_{2}x)e^{\lambda x}$$
   3. $\lambda <0, \lambda_{1,2} = \alpha \pm \beta i$ $$y = e^{\alpha x}(C_{1}\cos \beta x + C_{2}\sin \beta x)$$

例题

1. $y'' + y' -2y = 0$，求通解 $$\begin{array}{ll} & \text{特征方程为} \lambda^{2} + \lambda -2 =0 \\ \therefore & \begin{cases} \lambda_{1} = -2 \\ \lambda_{2} = 1 \end{cases} \\ \therefore & \text{通解为} y = C_{1}e^{-2x} + C_{2} e^{x} \\ \end{array}$$
2. $y'' - 4y' + 4y =0$，求通解 $$\begin{array}{ll} & \text{特征方程为} \lambda^{2} - 4\lambda + 4 =0 \\ \therefore & \lambda_{1} =\lambda_{2} = 2 \\ \therefore & \text{通解为} y = (C_{1} + C_{2}x)e^{2x} \\ \end{array}$$
3. $y'' -2y' +2y =0$，求通解 $$\begin{array}{ll} & \text{特征方程为} \lambda^{2} - 2\lambda + 2 =0 \\ \therefore & \lambda_{1,2} = 1 \pm i \\ \therefore & \text{通解为} y = e^{x}(C_{1}\cos x + C_{2}\sin x) \\ \end{array}$$

特二 二阶常系数非齐次线性微分方程

$$y'' + py' + qy = f(x)\quad (**)$$

通解

• $\text{通解} = (*)\text{的通解} + (**)\text{的特解}$

型一 $f(x) = P_{n}(x) e^{kx}$

1. $y''+y'-2y = (2x+1)e^{2x}$，求通解 $\color{#D0104C}{\lambda_{1}\ne\lambda_{2}\ne k}$ $$\begin{array}{ll} & \text{特征方程为} \lambda^{2} + \lambda - 2 =0 \\ \therefore & \begin{cases} \lambda_{1} =-2 \\ \lambda_{2}=1 \end{cases} \\ \therefore & y''+y'-2y=0 \text{的通解为} y = C_{1}e^{-2x} + C_{2}e^{x} \\ & \text{令} C(x) = (ax+b)e^{2x} \text{为其的一个特解} \\ \therefore & C'(x) = ae^{2x} + 2(ax + b)e^{2x} \\ & C''(x) = 2ae^{2x} + e^{2x}(2a +4ax + 4b) \\ \therefore & 2a e^{2x} + e^{2x}(2a+4ax+4b) + ae^{2x} + 2(ax+b)e^{2x} -2(ax+b)e^{2x} = (2x+1)e^{2x} \\ & (5a +4b +4ax)e^{2x} = (2x+1)e^{2x} \\ \therefore & \begin{cases} a=\frac{1}{2} \\ b= -\frac{3}{8} \end{cases} \\ \therefore & \text{原方程的通解} C_{1}e^{-2x} + C_{2}e^{x} + (2x+1)e^{2x} \\ \end{array}$$

2. $y'' - 3y' + 2y = (2x-1)e^{x}$，求通解 $\color{#D0104C}{\lambda_{1}\ne \lambda_{2} = k}$ $$\begin{array}{ll} & \text{特征方程为} \lambda^{2} - 3\lambda + 2 =0 \\ \therefore & \begin{cases} \lambda_{1} =2 \\ \lambda_{2} =1 \end{cases} \\ \therefore & y''+y'-2y=0 \text{的通解为} y = C_{1}e^{2x} + C_{2}e^{x} \\ & \text{令} y_{0}(x) = x(ax+b)e^{x} = (ax^{2} + bx)e^{x} \text{为其的一个特解} \\ \therefore & y_{0}'(x) = (2ax + b + ax^{2}+bx)e^{x} = (ax^{2} + 2ax + bx +b)e^{x} \\ & y_{0}''(x) = (2a +2ax +b +2ax + b + ax^{2} + bx) e^{x} = (ax^{2} + bx + 4ax +2a + 2b)e^{x} \\ \therefore & (ax^{2} + bx + 4ax +2a + 2b)e^{x} - 3(ax^{2} + 2ax + bx +b)e^{x} + 2(ax^{2} + bx)e^{x}= (2x-1)e^{x} \\ & (-2ax +2a-b) e^{x} = (2x-1)e^{x} \\ \therefore & \begin{cases} a=-1 \\ b= -1 \end{cases} \\ \therefore & \text{原方程的通解} C_{1}e^{2x} + C_{2}e^{x} + (-x^{2}-x)e^{x} \\ \end{array}$$

3. $y'' - 4y' + 4y = (3x-1)e^{2x}$ 求通解 $\color{#D0104C}{\lambda_{1} = \lambda_{2} = k}$ $$\begin{array}{ll} & \text{特征方程为} \lambda^{2} - 4\lambda + 4 =0 \\ \therefore & \lambda_{1} = \lambda_{2} = 2 \\ \therefore & y''+y'-2y=0 \text{的通解为} y = (C_{1} + C_{2}x)e^{2x} \\ & \text{令} y_{0}(x) = x^{2}(ax+b)e^{2x} = (ax^{3} + bx^{2})e^{2x} \text{为其的一个特解} \\ \therefore & (ax^{3} + bx^{2})'' = 6ax + 2b = 3x -1 \\ \therefore & \begin{cases} a = \frac{1}{2} \\ b = -\frac{1}{2} \end{cases} \\ \therefore & \text{通解为} y = (C_{1} + C_{2}x)e^{2x} + (\frac{1}{2}x^{3}-\frac{1}{2}x^{2})e^{2x} \\ \end{array}$$

Notes 对于 $\lambda_{1} = \lambda_{2} = k$ 的情形，特解 $y_{0}(x) = Q(x)e^{kx}$ 则 $$Q(x)'' = (ax^{3} + bx^{2})'' = P_{n}(x)$$