First-Order ODE

一阶常微分方程 (First-Order Ordinary Differential Equations)

定义 (Definition)

一阶常微分方程是只包含一个自变量及其一阶导数的方程。

A first-order ordinary differential equation is an equation that contains only one independent variable and its first derivative.

一般形式 (General form):

$F(x,y,\frac{dy}{dx})=0$ 或 (or) $\frac{dy}{dx}=f(x,y)$

其中 $y=y(x)$ 是未知函数，$x$ 是自变量。

Where $y=y(x)$ is the unknown function and $x$ is the independent variable.

主要类型 (Main Types)

1. 可分离变量方程 (Separable Equations)

定义 (Definition):

可以将方程改写成 $g(y)dy=h(x)dx$ 形式的方程。

Equations that can be rewritten in the form $g(y)dy=h(x)dx$.

解法 (Solution method):

1. 分离变量 Separate the variables
2. 两边积分 Integrate both sides
3. 解出 $y$ Solve for $y$

例子 (Example):

$\frac{dy}{dx}=xy$

解 (Solution):

$$\frac{dy}{y}=xdx$$ $$\int \frac{dy}{y}=\int xdx$$ $$\ln |y|=\frac{1}{2}{x}^2+C$$ $$y=\pm e^{\frac{1}{2}{x}^2+C}=C{e}^{\frac{1}{2}{x}^2}$$

2. 线性方程 (Linear Equations)

定义 (Definition):

形如 $\frac{dy}{dx}+P(x)y=Q(x)$ 的方程。

Equations of the form $\frac{dy}{dx}+P(x)y=Q(x)$.

解法 (Solution method):

1. 求积分因子 $\mu (x)=e^{\int P(x)dx}$ Find the integrating factor $\mu (x)=e^{\int P(x)dx}$
2. 两边乘以积分因子 Multiply both sides by the integrating factor
3. 整理并积分 Simplify and integrate

详细步骤 (Detailed steps):

1. 将方程改写成 Rewrite the equation as

$$\mu (x)\frac{dy}{dx}+\mu (x)P(x)y=y^{\prime }{e}^{\int P(x)dx}+P(x)y=\mu (x)Q(x)$$

1. 左边为 The left side becomes

$$y^{\prime }(e^{\int P(x)dx})+(e^{\int P(x)dx}{)}^{\prime }y=(y{e}^{\int P(x)dx}{)}^{\prime }$$

1. 解出 $\mu (x)y=\int \mu (x)Q(x)dx$ Solve for $\mu (x)y=\int \mu (x)Q(x)dx$

例子 (Example):

$\frac{dy}{dx}+2xy=x$

解 (Solution):

积分因子 (Integrating factor): $\mu (x)=e^{\int 2xdx}=e^{x^2}$

$$e^{x^2}\frac{dy}{dx}+2x{e}^{x^2}y=x{e}^{x^2}$$ $$\frac{d}{dx}(e^{x^2}y)=x{e}^{x^2}$$ $$e^{x^2}y=\int x{e}^{x^2}dx=\frac{1}{2}{e}^{x^2}+C$$ $$y=\frac{1}{2}+C{e}^{-x^2}$$

3. 伯努利方程 (Bernoulli Equations)

定义 (Definition):

形如 $\frac{dy}{dx}+P(x)y=Q(x)y^n$ 的方程，其中 $n\ne 0,1$。

Equations of the form $\frac{dy}{dx}+P(x)y=Q(x)y^n$, where $n\ne 0,1$.

解法 (Solution method):

1. 令 $v=y^{1-n}$，将方程转化为线性方程 Let $v=y^{1-n}$, transform the equation into a linear equation
2. 解转化后的线性方程 Solve the resulting linear equation
3. 代回 $y=v^{\frac{1}{1-n}}$ 得到原方程的解 Substitute back $y=v^{\frac{1}{1-n}}$ to get the solution of the original equation

例子 (Example):

$\frac{dy}{dx}-2xy=x^3{y}^3$

解 (Solution):

令 (Let) $v=y^{-2}$, $\frac{dv}{dx}=-2y^{-3}\frac{dy}{dx}$, 则 (Then) $y=v^{-\frac{1}{2}}$, $\frac{dy}{dx}=-\frac{1}{2}{v}^{-\frac{3}{2}}\frac{dv}{dx}$

$$-\frac{1}{2}{v}^{-\frac{3}{2}}\frac{dv}{dx}-2x{v}^{-\frac{1}{2}}=x^3{v}^3$$ $$\frac{dv}{dx}+4xv=-2x^3$$

这是一个线性方程，可以用上面的方法解决。

This is a linear equation that can be solved using the method above.

4. 精确方程 (Exact Equations)

定义 (Definition):

形如 $M(x,y)dx+N(x,y)dy=0$ 的方程，其中 $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$。

Equations of the form $M(x,y)dx+N(x,y)dy=0$, where $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

解法 (Solution method):

1. 验证方程是否为精确方程 Verify if the equation is exact
2. 找到函数 $\varphi (x,y)$ 使得 $\frac{\partial \varphi }{\partial x}=M$ 和 $\frac{\partial \varphi }{\partial y}=N$ Find a function $\varphi (x,y)$ such that $\frac{\partial \varphi }{\partial x}=M$ and $\frac{\partial \varphi }{\partial y}=N$
3. 解方程 $\varphi (x,y)=C$ Solve the equation $\varphi (x,y)=C$

例子 (Example):

$(2xy+y^2)dx+(x^2+2xy)dy=0$

解 (Solution):

验证 (Verify): $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=2x+2y$

$$\varphi (x,y)=\int (2xy+y^2)dx=x^{2}y+x{y}^2+f(y)$$ $$\frac{\partial \varphi }{\partial y}=x^2+2xy+f^{\prime }(y)=x^2+2xy$$

因此 (Therefore), $f^{\prime }(y)=0$, $f(y)=C$

最终解 (Final solution): $x^{2}y+x{y}^2=C$

注意事项 (Important notes):

• 识别方程类型是解题的关键第一步 Identifying the equation type is the crucial first step in solving
• 有些方程可能需要先进行变量替换或其他转化才能归类 Some equations may require variable substitution or other transformations before classification
• 在考试中，熟练应用这些方法可以大大提高解题效率 In exams, proficiency in applying these methods can greatly improve solving efficiency
• 解的验证很重要，可以通过将解代入原方程来检查 Solution verification is important, which can be done by substituting the solution back into the original equation