Second-Order ODE

Second-Order ODE | 二阶常微分方程

Definition | 定义

A second-order ordinary differential equation (ODE) is a differential equation of the form:

二阶常微分方程的形式为：

$${\mathbf{y}}^{\prime \prime }(x)+p(x){\mathbf{y}}^{\prime }(x)+q(x)\mathbf{y}(x)=g(x)$$

where ${\mathbf{y}}^{\prime \prime }(x)$ is the second derivative of $\mathbf{y}(x)$ with respect to $x$, ${\mathbf{y}}^{\prime }(x)$ is the first derivative, $p(x)$ and $q(x)$ are given functions, and $g(x)$ is a known function.

其中 ${\mathbf{y}}^{\prime \prime }(x)$ 是 $\mathbf{y}(x)$ 对 $x$ 的二阶导数， ${\mathbf{y}}^{\prime }(x)$ 是一阶导数， $p(x)$ 和 $q(x)$ 是已知函数， $g(x)$ 是已知函数。

Types of Second-Order ODEs | 二阶常微分方程的类型

1. Homogeneous Second-Order ODE | 齐次二阶常微分方程

   If $g(x)=0$, the ODE is called homogeneous:

   若 $g(x)=0$，则该方程称为齐次方程：

   $${\mathbf{y}}^{\prime \prime }(x)+p(x){\mathbf{y}}^{\prime }(x)+q(x)\mathbf{y}(x)=0$$

2. **Non-Homogeneous Second-Order ODE | 非齐次二阶常微分方程** If $g(x) \neq 0$, the ODE is called non-homogeneous: 若 $g(x) \neq 0$，则该方程称为非齐次方程：

   \mathbf{y}''(x) + p(x)\mathbf{y}'(x) + q(x)\mathbf{y}(x) = g(x)

$$

Solution Methods | 解法

1. Homogeneous Equations | 齐次方程

For homogeneous second-order ODEs:

对于齐次二阶常微分方程：

$${\mathbf{y}}^{\prime \prime }+p(x){\mathbf{y}}^{\prime }+q(x)\mathbf{y}=0$$

Characteristic Equation | 特征方程

1. Assume a solution of the form $\mathbf{y}=e^{\mathbf{r}x}$ 设解形式为 $\mathbf{y}=e^{\mathbf{r}x}$

2. Substitute $\mathbf{y}$ into the ODE to get the characteristic equation: 将 $\mathbf{y}$ 代入方程得到特征方程：

$${\mathbf{r}}^2+p\mathbf{r}+q=0$$

3. Solve for $\mathbf{r}$: 解 $\mathbf{r}$：
   • If ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$ are real and distinct: 若 ${\mathbf{r}}_1$ 和 ${\mathbf{r}}_2$ 为实且不同：

$$\mathbf{y}(x)=C_1{e}^{{\mathbf{r}}_{1}x}+C_2{e}^{{\mathbf{r}}_{2}x}$$

• If ${\mathbf{r}}_1={\mathbf{r}}_2=\mathbf{r}$ 若 ${\mathbf{r}}_1={\mathbf{r}}_2=\mathbf{r}$：

$$\mathbf{y}(x)=(C_1+C_{2}x)e^{\mathbf{r}x}$$

• If ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$ are complex: 若 ${\mathbf{r}}_1$ 和 ${\mathbf{r}}_2$ 为复数：

$${\mathbf{r}}_1,{\mathbf{r}}_2=\alpha \pm \beta i=\alpha e^{\pm i\beta }$$

$$

 \mathbf{y}(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))

$$

2. Non-Homogeneous Equations | 非齐次方程

For non-homogeneous second-order ODEs:

对于非齐次二阶常微分方程：

$${\mathbf{y}}^{\prime \prime }+p(x){\mathbf{y}}^{\prime }+q(x)\mathbf{y}=g(x)$$

Particular Solutions of Non-Homogeneous Second-Order ODEs | 非齐次二阶常微分方程的特解

When solving a non-homogeneous second-order ODE, the particular solution ${\mathbf{y}}_p(x)$ can be found using different methods depending on the form of $g(x)$. Here, we discuss some common approaches:

在求解非齐次二阶常微分方程时，特解 ${\mathbf{y}}_p(x)$ 的求法取决于 $g(x)$ 的形式。以下是一些常见的方法：

Method of Undetermined Coefficients | 未定系数法

This method is applicable when $g(x)$ is a polynomial, exponential function, sine or cosine, or a combination of these.

当 $g(x)$ 是多项式、指数函数、正弦或余弦函数，或这些的组合时，可以使用未定系数法。

Steps | 步骤

1. Identify the form of $g(x)$ | 确定 $g(x)$ 的形式

$$g(x)=P(x)e^{\alpha x}\cos (\beta x)+Q(x)e^{\alpha x}\sin (\beta x)$$

where $P(x)$ and $Q(x)$ are polynomials.

其中 $P(x)$ 和 $Q(x)$ 是多项式。

1. Assume a particular solution ${\mathbf{y}}_p(x)$ with undetermined coefficients | 假设具有未定系数的特解 ${\mathbf{y}}_p(x)$

   • For a polynomial $g(x)=a{x}^n+b{x}^{n-1}+\cdots +c$:

$${\mathbf{y}}_p(x)=A_n{x}^n+A_{n-1}{x}^{n-1}+\cdots +A_0$$

• For an exponential function $g(x)=e^{\alpha x}$:

$${\mathbf{y}}_p(x)=A{e}^{\alpha x}$$

• For a sine or cosine function $g(x)=\cos (\beta x)$ or $g(x)=\sin (\beta x)$:

$${\mathbf{y}}_p(x)=A\cos (\beta x)+B\sin (\beta x)$$

• For a combination like $g(x)=P(x)e^{\alpha x}\cos (\beta x)+Q(x)e^{\alpha x}\sin (\beta x)$:

$${\mathbf{y}}_p(x)=e^{\alpha x}(A\cos (\beta x)+B\sin (\beta x))$$

1. Substitute ${\mathbf{y}}_p(x)$ into the non-homogeneous ODE | 将 ${\mathbf{y}}_p(x)$ 代入非齐次方程

2. Solve for the undetermined coefficients | 解出未定系数

Example | 示例

Given:

$${\mathbf{y}}^{\prime \prime }-3{\mathbf{y}}^{\prime }+2\mathbf{y}=4e^{2x}$$

Assume a particular solution:

$${\mathbf{y}}_p(x)=A{e}^{2x}$$

Substitute into the ODE:

$$(4A-6A+2A)e^{2x}=4e^{2x}$$ $$0=4e^{2x}$$

Clearly, the assumption must be modified since it yields a contradiction. Instead, assume:

$${\mathbf{y}}_p(x)=Ax{e}^{2x}$$

Substitute and solve for $A$:

$$(4Ax+2A-6Ax+2Ax)e^{2x}=4e^{2x}$$ $$2A{e}^{2x}=4e^{2x}$$ $$A=2$$

Thus, the particular solution is:

$${\mathbf{y}}_p(x)=2x{e}^{2x}$$

Variation of Parameters | 参数变换法

This method is applicable when $g(x)$ is not suitable for the method of undetermined coefficients, or when the form of $g(x)$ is more complicated.

当 $g(x)$ 不适用于未定系数法，或 $g(x)$ 的形式更复杂时，可以使用参数变换法。

Steps | 步骤

1. Solve the corresponding homogeneous equation | 解对应的齐次方程

$${\mathbf{y}}_h(x{)}^{\prime \prime }+p(x){\mathbf{y}}_h^{\prime }(x)+q(x){\mathbf{y}}_h(x)=0$$

1. Find the fundamental solutions ${\mathbf{y}}_1(x)$ and ${\mathbf{y}}_2(x)$ | 找到基本解 ${\mathbf{y}}_1(x)$ 和 ${\mathbf{y}}_2(x)$

2. Form the particular solution as | 特解形式为

$${\mathbf{y}}_p(x)={\mathbf{y}}_1(x){\mathbf{u}}_1(x)+{\mathbf{y}}_2(x){\mathbf{u}}_2(x)$$

1. Determine ${\mathbf{u}}_1(x)$ and ${\mathbf{u}}_2(x)$ by solving | 通过解以下方程确定 ${\mathbf{u}}_1(x)$ 和 ${\mathbf{u}}_2(x)$

$$\{\begin{array}{l}{\mathbf{y}}_1(x){\mathbf{u}}_1^{\prime }(x)+{\mathbf{y}}_2(x){\mathbf{u}}_2^{\prime }(x)=0\\ {\mathbf{y}}_1^{\prime }(x){\mathbf{u}}_1^{\prime }(x)+{\mathbf{y}}_2^{\prime }(x){\mathbf{u}}_2^{\prime }(x)=g(x)\end{array}$$

1. Integrate to find ${\mathbf{u}}_1(x)$ and ${\mathbf{u}}_2(x)$ | 积分求 ${\mathbf{u}}_1(x)$ 和 ${\mathbf{u}}_2(x)$

$${\mathbf{u}}_1(x)=\int -\frac{{\mathbf{y}}_2(x)g(x)}{\mathbf{W}({\mathbf{y}}_1,{\mathbf{y}}_2)}\mathrm{d}x$$ $${\mathbf{u}}_2(x)=\int \frac{{\mathbf{y}}_1(x)g(x)}{\mathbf{W}({\mathbf{y}}_1,{\mathbf{y}}_2)}\mathrm{d}x、$$

1. Combine ${\mathbf{u}}_1(x)$ and ${\mathbf{u}}_2(x)$ with ${\mathbf{y}}_1(x)$ and ${\mathbf{y}}_2(x)$ to get ${\mathbf{y}}_p(x)$ | 将 ${\mathbf{u}}_1(x)$ 和 ${\mathbf{u}}_2(x)$ 与 ${\mathbf{y}}_1(x)$ 和 ${\mathbf{y}}_2(x)$ 结合得到 ${\mathbf{y}}_p(x)$

Example | 示例

Given:

$${\mathbf{y}}^{\prime \prime }+\mathbf{y}=\tan (x)$$

Solve the homogeneous equation:

$${\mathbf{y}}_h^{\prime \prime }+{\mathbf{y}}_h=0$$

The fundamental solutions are:

$${\mathbf{y}}_1(x)=\cos (x),{\mathbf{y}}_2(x)=\sin (x)$$

Particular solution:

$${\mathbf{y}}_p(x)=\cos (x){\mathbf{u}}_1(x)+\sin (x){\mathbf{u}}_2(x)$$

Determine ${\mathbf{u}}_1(x)$ and ${\mathbf{u}}_2(x)$:

$$\{\begin{array}{l}\cos (x){\mathbf{u}}_1^{\prime }(x)+\sin (x){\mathbf{u}}_2^{\prime }(x)=0\\ -\sin (x){\mathbf{u}}_1^{\prime }(x)+\cos (x){\mathbf{u}}_2^{\prime }(x)=\tan (x)\end{array}$$

Solve for ${\mathbf{u}}_1^{\prime }(x)$ and ${\mathbf{u}}_2^{\prime }(x)$:

$${\mathbf{u}}_1^{\prime }(x)=-\sin (x)\tan (x),{\mathbf{u}}_2^{\prime }(x)=\cos (x)\tan (x)$$

Integrate to find ${\mathbf{u}}_1(x)$ and ${\mathbf{u}}_2(x)$:

$${\mathbf{u}}_1(x)=-\int \sin (x)\tan (x)\mathrm{d}x=-\int \sin (x)\frac{\sin (x)}{\cos (x)}\mathrm{d}x=-\int \frac{{\sin }^2(x)}{\cos (x)}\mathrm{d}x$$ $${\mathbf{u}}_2(x)=\int \cos (x)\tan (x)\mathrm{d}x=\int \sin (x)\mathrm{d}x=-\cos (x)$$

Thus, the particular solution is:

$${\mathbf{y}}_p(x)=\cos (x)(-\frac{1}{2}\ln |\cos (x)|)+\sin (x)(-\cos (x))$$