习题7.2 可分离变量的微分方程
1. 求解下列可分离变量微分方程
【题目】
$xy' - y \ln y = 0;$
$y' = \tan x \tan y;$
$(x y + x ^ {3} y) d y = \left(1 + y ^ {2}\right) d x;$
$y' - xy' = 2(y^{2} + y').$
【解析】
分离变量：
$$x \frac{dy}{dx} = y\ln y \implies \frac{dy}{y\ln y} = \frac{dx}{x}$$两边积分：
$$\int \frac{d(\ln y)}{\ln y} = \int \frac{dx}{x} \implies \ln|\ln y| = \ln|x| + C_1 \implies \ln y = C x \implies y = e^{Cx}$$分离变量：
$$\frac{dy}{dx} = \tan x \tan y \implies \cot y dy = \tan x dx$$两边积分：
$$\int \frac{\cos y}{\sin y}dy = \int \frac{\sin x}{\cos x}dx \implies \ln|\sin y| = -\ln|\cos x| + C_1 \implies \ln|\sin y\cos x| = C_1 \implies \cos x\sin y = C$$分离变量：
$$x(1 + x^2) y dy = (1 + y^2) dx \implies \frac{y}{1 + y^2}dy = \frac{1}{x(1 + x^2)}dx = \left(\frac{1}{x} - \frac{x}{1 + x^2}\right)dx$$两边积分：
$$\frac{1}{2}\ln(1 + y^2) = \ln|x| - \frac{1}{2}\ln(1 + x^2) + C_1 \implies \ln(1 + y^2) + \ln(1 + x^2) - 2\ln|x| = 2C_1$$整理得：
$$(1 + y^2)(1 + x^2) = C x^2$$整理关于 $y'$ 的同类项：
$$y'(1 - x - 2) = 2y^2 \implies -(x + 1)y' = 2y^2 \implies -(x + 1)\frac{dy}{dx} = 2y^2$$分离变量：
$$-\frac{dy}{y^2} = \frac{2}{x + 1}dx$$两边积分：
$$\frac{1}{y} = 2\ln|x + 1| + C = \ln(x + 1)^2 + C$$
2. 求微分方程的特解
【题目】
$y' = e^{2x - y}, \ y(0) = 0;$
$e^x\cos ydx + (1 + e^x)\sin ydy = 0, \ y(0) = \frac{\pi}{4}.$
【解析】
分离变量：
$$\frac{dy}{dx} = \frac{e^{2x}}{e^y} \implies e^y dy = e^{2x}dx$$两边积分：
$$e^y = \frac{1}{2}e^{2x} + C$$代入初值条件 $y(0) = 0$：
$$e^0 = \frac{1}{2}e^0 + C \implies 1 = \frac{1}{2} + C \implies C = \frac{1}{2}$$特解为：
$$e^y = \frac{1}{2}(e^{2x} + 1) \implies y = \ln(e^{2x} + 1) - \ln 2$$分离变量：
$$(1 + e^x)\sin y dy = -e^x\cos y dx \implies \frac{\sin y}{\cos y}dy = -\frac{e^x}{1 + e^x}dx \implies \tan y dy = -\frac{e^x}{1 + e^x}dx$$两边积分：
$$-\ln|\cos y| = -\ln(1 + e^x) + C_1 \implies \ln|\sec y| = -\ln(1 + e^x) + C_1 \implies (1 + e^x)\sec y = C$$代入初值条件 $x = 0, y = \frac{\pi}{4}$：
$$(1 + e^0)\sec\frac{\pi}{4} = 2 \cdot \sqrt{2} = 2\sqrt{2} = C$$特解为：
$$(1 + e^x)\sec y = 2\sqrt{2}$$