# 2014年考研数二第16题解析：一阶线性微分方程求极值、求导

## 解析

$$x^{2}+y^{2}y^{‘} = 1-y^{‘} \Rightarrow$$

$$x^{2} + y^{2} \frac{dy}{dx} = 1 – \frac{dy}{dx} \Rightarrow$$

$$y^{2} \frac{dy}{dx} + \frac{dy}{dx} = 1 – x^{2} \Rightarrow$$

$$(y^{2}+1)\frac{dy}{dx} = 1-x^{2} \Rightarrow$$

$$(y^{2} + 1)dy = (1-x^{2}) dx \Rightarrow$$

$$\int (y^{2}+1) dy = \int (1-x^{2}) dx \Rightarrow$$

$$\frac{1}{3} y^{3} + y = x – \frac{1}{3} x^{3} + C.$$

$$0=2-\frac{8}{3} + C \Rightarrow$$

$$\frac{-2}{3} + C = 0 \Rightarrow$$

$$C = \frac{2}{3}.$$

$$\frac{1}{3} y^{3} + y = x – \frac{1}{3} x^{3} + \frac{2}{3}.$$

$$y^{2} y^{‘} + y^{‘} = 1-x^{2} \Rightarrow$$

$$(y^{2}+1)y^{‘} = 1-x^{2} \Rightarrow$$

$$y^{‘} = \frac{1-x^{2}}{1+y^{2}}\Rightarrow$$

$$y^{”} = \frac{-2x(1+y^{2}) – 2yy^{‘}(1-x^{2})}{(1+y^{2})^{2}}.$$

$$1-x^{2} = 0 \Rightarrow$$

$$x = \pm 1.$$

$$\frac{1}{3} y^{3}(1) + y(1) = 1-\frac{1}{3} + \frac{2}{3} \Rightarrow$$

$$\frac{1}{3} y^{3}(1) + y(1) = \frac{4}{3} \Rightarrow$$

$$y(1) = 1.$$

$$\frac{1}{3} y^{3}(-1) + y(-1) = -1+\frac{1}{3} + \frac{2}{3} \Rightarrow$$

$$\frac{1}{3} y^{3}(-1) + y(-1) = 0 \Rightarrow$$

$$y(-1) = 0.$$

$$y^{”}(1) = \frac{-2(1+1)-0}{(1+1)^{2}} \Rightarrow$$

$$y^{”}(1) = \frac{-4}{4} = -1 < 0.$$

$$y^{”}(-1) = \frac{2(1+0)}{1^{2}} \Rightarrow$$

$$y^{”}(-1) = 2 > 0$$