习题7.5 可降阶的高阶微分方程
1. 求下列微分方程的通解
【题目】
$y'' = x + \sin x;$
$xy'' + y' = 0;$
$yy'' + 2y'^{2} = 0;$
$y'' = y'^{2} + 1.$
【解析】
连续积分两次：
$$y' = \int (x + \sin x)dx = \frac{x^2}{2} - \cos x + C_1$$$$y = \int \left(\frac{x^2}{2} - \cos x + C_1\right)dx = \frac{x^3}{6} - \sin x + C_1 x + C_2$$不显含 $y$ 型，令 $p = y' \implies y'' = p'$：
$$x p' + p = 0 \implies \frac{dp}{p} = -\frac{dx}{x} \implies \ln|p| = -\ln|x| + \ln|C_1| \implies p = \frac{C_1}{x}$$再积分：
$$y = \int \frac{C_1}{x}dx = C_1\ln|x| + C_2$$不显含 $x$ 型，令 $p = y' \implies y'' = p\frac{dp}{dy}$：
$$y p \frac{dp}{dy} + 2p^2 = 0 \implies p\left(y\frac{dp}{dy} + 2p\right) = 0$$除以 $p$（$p \neq 0$）：
$$\frac{dp}{p} = -\frac{2}{y}dy \implies \ln|p| = -2\ln|y| + \ln|C_1| \implies p = \frac{C_1}{y^2}$$由于 $p = \frac{dy}{dx}$：
$$y^2 dy = C_1 dx \implies \frac{y^3}{3} = C_1 x + C_2' \implies y^3 = C_1 x + C_2$$令 $p = y' \implies y'' = p'$：
$$p' = p^2 + 1 \implies \frac{dp}{p^2 + 1} = dx \implies \arctan p = x + C_1 \implies p = \tan(x + C_1)$$再积分：
$$y = \int \tan(x + C_1)dx = -\ln|\cos(x + C_1)| + C_2 = \ln|\sec(x + C_1)| + C_2$$
2. 特解求解
【题目】
$$ y y ^ {\prime \prime} = 2 \left(y ^ {2} - y ^ {\prime}\right), y (0) = 1, y ^ {\prime} (0) = 2. $$【解析】
不显含 $x$ 型，令 $p = y' \implies y'' = p\frac{dp}{dy}$。原方程为：
$$y p \frac{dp}{dy} = 2(y^2 - p)$$解得：
$$\arctan y = x + \frac{\pi}{4}$$