Section02_一阶微分方程

可分离变量的微分方程

定义

• 设 $\displaystyle \frac{dy}{dx} = f(x,y)$ 若 $\displaystyle f(x,y) = \phi_{2}(x)\phi_{2}(y)$ 则称为可分离变量的微分方程

解法

$$\frac{dy}{dx} = \phi_{1}(x)\phi_{2}(y) \Rightarrow \int \frac{dy}{\phi_{2}(y)} = \int \phi_{1}(x)\cdot dx + C$$

例题

1. 求 $\displaystyle \frac{dy}{dx} = 1 + x + y^{2} + xy^{2}$ 的通解 $$\begin{array}{ll} & \displaystyle \frac{dy}{dx} = 1 + x + y^{2} + xy^{2} = (1+x)(1+y^{2}) \\ \because & 1+y^{2} > 0 \\ \therefore & \displaystyle \frac{dy}{1+y^{2}} = (1+x)\cdot dx \\ & \displaystyle \int_{}^{}\frac{dy}{1+y^{2}} = \int_{}^{} (1+x) \cdot dx + C \\ \therefore & \arctan y = x + \frac{1}{2} x^{2} + C\\ & y = \tan(x+\frac{1}{2}x^{2} +C) \end{array}$$
2. $\displaystyle \frac{dy}{dx} = 2xy$ 求通解 $$\begin{array}{ll} & \text{When } y = 0 \\ \therefore & y = 0 \text{为该方程的一个解} \\ & \text{When } y \neq 0 \\ \therefore & \displaystyle \frac{dy}{y} = 2x\cdot dx \\ & \displaystyle \int_{}^{}\frac{dy}{y} = \int_{}^{} 2x\cdot dx + C_{0} \\ \therefore & \ln \vert y \vert = x^{2} + C_{0} \\ & \vert y \vert = e^{C_{0}}e^{x^{2}} \\ \therefore & y = (\pm e^{C_{0}})e^{x^{2}} \\ \because & y \ne 0 \\ \therefore & \pm e^{C_{0}} \in (-\infty, 0)\cup(0,+\infty) \\ \therefore & \text{令} \pm e^{C_{0}} = C \\ \therefore & \text{通解为} y = \begin{cases} Ce^{x^{2}}\quad C\in(-\infty, 0)\cup(0,+\infty) \\ 0\\ \end{cases} \\ & \text{即} y = Ce^{x^{2}} \quad C\in (-\infty, +\infty) \end{array}$$

齐次微分方程

定义

• 设 $\displaystyle \frac{dy}{dx} = f(x,y)$ 若 $\displaystyle f(x,y) = \phi(\frac{y}{x})$ 则为齐次微分方程

解法

$$\begin{array}{ll} & \text{令} u = \frac{y}{x}, \text{则} y = ux \\ \therefore & \displaystyle \frac{d(ux)}{dx} = u + \frac{xdu}{dx} = f(u) \\ \therefore & \displaystyle \frac{du}{f(u)-u} = \frac{dx}{x} \\ & \displaystyle \int_{}^{} \frac{du}{f(u) - u} = \int_{}^{} \frac{dx}{x} +C \\ & \text{解出} u = u(x), \text{再将}u =\frac{y}{x} \text{代入} \end{array}$$

例题

1. $\displaystyle \frac{dy}{dx} = 2\frac{y}{x} - 1$ $$\begin{array}{ll} & \text{令} u = \frac{y}{x}, \text{则} y = ux \\ \therefore & \displaystyle u + x \frac{du}{dx} = 2u - 1\\ & \displaystyle \frac{du}{u-1} = \frac{dx}{x} \\ \therefore & \displaystyle \int_{}^{} \frac{du}{u-1} = \int_{}^{} \frac{dx}{x} + \ln C \\ & \ln(u-1) = \ln Cx \\ \therefore & u-1 = Cx \\ & \frac{y}{x} - 2 = Cx \\ & y = Cx^{2} + 2x \end{array}$$
2. $\displaystyle xdy - (y+\sqrt{x^{2}+y^{2}})\cdot dx = 0\quad (x>0)$，求通解 $$\begin{array}{ll} & \displaystyle \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \bigg(\frac{y}{x}\bigg)^{2}} \\ & \displaystyle \text{令}\frac{y}{x} = u \\ \therefore & \displaystyle u + x\frac{du}{dx} = u + \sqrt{1+u^{2}}\\ & \displaystyle \frac{du}{\sqrt{1+u^{2}}} = \frac{dx}{x} \\ \therefore & \displaystyle \int_{}^{} \frac{du}{\sqrt{1+u^{2}}} = \int_{}^{} \frac{dx}{x} + \ln C\\ & \ln (u + \sqrt{1+u^{2}}) = \ln Cx \\ \therefore & y + \sqrt{y^{2}+ x^{2}} = Cx^{2}\\ \end{array}$$

一阶齐次线性微分方程 重点

定义

• 形如 $\displaystyle \frac{dy}{dx}+p(x)y = 0$ 称为一阶齐次线性微分方程

解法

$$\begin{array}{ll} & \displaystyle \frac{dy}{dx} + p(x) y = 0 \\ & \displaystyle \frac{dy}{y} = -p(x)dx \\ & \displaystyle \int_{}^{} \frac{dy}{y} = \int_{}^{}-p(x)\cdot dx + \ln C \\\\ & \displaystyle \color{#D0104C}{y = C e^{-\int_{}^{}p(x)\cdot dx}} \end{array}$$

例题

1. $\displaystyle \frac{dy}{dx} = 2xy$ $$\begin{array}{ll} & \displaystyle \frac{dy}{dx} = 2xy \\ & \displaystyle \frac{dy}{dx} - 2xy = 0 \\ \therefore & p(x) = -2x \\ & \displaystyle y = Ce^{-\int_{}^{}-2x\cdot dx} = C e^{x^{2}} \end{array}$$
2. $\displaystyle xy'-2y =0$ $$\begin{array}{ll} & xy' - 2y = 0\\ & y' - \frac{2}{x} y = 0 \\ \therefore & p(x) = -\frac{2}{x} \\ \therefore & y = Ce^{-\int_{}^{}-\frac{2}{x}\cdot dx} \\ & y = Ce^{2\ln x} = C x^{2} \end{array}$$
3. $\displaystyle \frac{dy}{dx} - xy = 0$，求满足 $y(0) = \pi$ 的特解 $$\begin{array}{ll} & \frac{dy}{dx} - xy = 0 \\ \therefore & p(x) = -x \\ \therefore & y = Ce^{-\int_{}^{}-x\cdot dx} = C e^{\frac{1}{2}x^{2}} \\ \because & y(0) = \pi \\ \therefore & \pi = C \\ \therefore & \text{特解为} y = \pi e^{\frac{1}{2}x^{2}} \\ \end{array}$$

一阶非齐次线性微分方程

定义

• 形如 $\displaystyle \frac{dy}{dx} + p(x)y = Q(x)$

解法：常数变易法

$$\begin{array}{ll} & \text{对于} \frac{dy}{dx} + p(x) = 0 \text{，通解为} y = Ce^{-\int p(x)\cdot dx} \\ & \text{对于} \frac{dy}{dx} + p(x) = Q(x) \text{，通解为} y = C(x)e^{-\int p(x)\cdot dx} \\ \therefore & y' = C'(x)e^{-\int_{}^{}p(x)\cdot dx} - C(x)p(x)e^{-\int_{}^{}p(x)\cdot dx} \\ \therefore & C'(x)e^{-\int_{}^{}p(x)\cdot dx} - C(x)p(x)e^{-\int_{}^{}p(x)\cdot dx} + p(x)C(x)e^{-\int_{}^{}p(x)\cdot dx} = Q(x) \\ \therefore & C'(x) = Q(x)e^{\int_{}^{}p(x)\cdot dx} \\ \therefore & C(x) = \int_{}^{}Q(x)e^{\int_{}^{}p(x)\cdot dx} \cdot dx +C\\ \end{array}$$

$$\color{#D0104C}{y = \bigg[\int Q(x)e^{\int p(x)\cdot dx}\cdot dx + C\bigg]e^{-\int p(x)\cdot dx}}$$

例题

1. $\displaystyle \frac{dy}{dx} = 2\frac{y}{x}-1$ $$\begin{array}{ll} & \frac{dy}{dx} - 2 \frac{1}{x}y = -1 \\ \therefore & p(x) = -\frac{2}{x}, Q(x) = -1 \\ \therefore & y = [\int_{}^{} -1 e ^{\int_{}^{}-\frac{2}{x}\cdot dx}\cdot dx + C] e ^{- \int_{}^{}-\frac{2}{x}\cdot dx}\\ & = [x^{-1} + C] x^{2} = Cx^{2} +x \end{array}$$
2. $\displaystyle y' + y\tan x = \cos x$ $$\begin{array}{ll} \therefore & p(x) = \tan{x}, Q(x) = \cos x \\ \therefore & y = [\int_{}^{} \cos x e^{\int_{}^{}\tan x\cdot dx} \cdot dx + C] e^{- \int_{}^{}\tan x\cdot dx} \\ & y = [\int_{}^{}\cos x e^{-\ln (\cos x)}\cdot dx + C] e^{\ln (\cos x)} \\ & y = [\int_{}^{}1 \cdot dx + C]\cos x \\ & y = x\cos x + C\cos x \end{array}$$