# 2011年考研数二第18题解析：导数、三角函数、对数、二阶微分方程

## 解析

$$\tan \alpha = \frac{dy}{dx}.$$

$$(\sec^{2} \alpha) \cdot (\frac{d \alpha}{dx}) = \frac{d^{2}y}{dx^{2}}$$

$$(\tan x)^{‘} = \sec^{2} x = 1 + \tan^{2} x.$$

$$(1+ \tan^{2} \alpha) (\frac{d \alpha}{dx}) = \frac{d^{2}y}{dx^{2}}.$$

$$\frac{d \alpha}{dx} = \frac{dy}{dx} = \tan \alpha.$$

$$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx}[1+(\frac{dy}{dx})^{2}].$$

$$p = \frac{dy}{dx}.$$

$$\frac{d^{2}y}{dx^{2}} = p(1+p^{2}). (1)$$

$\frac{d^{2}y}{dx^{2}} \neq (\frac{dy}{dx})(\frac{dy}{dx})$,

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{dy}{dx}) \Rightarrow$$

$$\frac{dp}{dx} = \frac{dy}{dx} \cdot \frac{dp}{dy} = p \cdot \frac{dp}{dy}.$$

$$p \cdot \frac{dp}{dy} = p(1+p^{2}).$$

$$\frac{dy}{dx} \neq 0 \Rightarrow$$

$$p \neq 0.$$

$$\frac{dp}{dy} = 1 + p^{2} \Rightarrow$$

$$\frac{dp}{1+p^{2}} = dy \Rightarrow$$

$$\int \frac{dp}{1+p^{2}} = \int dy \Rightarrow$$

$$\arctan (p) = y + C_{1}.$$

$$y(0) = 0, y^{‘}(0) = \frac{dy}{dx} = p = 1.$$

$$\arctan(1) = 0 + C_{1} \Rightarrow$$

$$C_{1} = \frac{\pi}{4}.$$

$$\arctan (p) = y + \frac{\pi}{4} \Rightarrow$$

$$\tan[\arctan(p)] = \tan (y + \frac{\pi}{4}) \Rightarrow$$

$$p = \tan (y + \frac{\pi}{4}) \Rightarrow$$

$$\frac{dy}{dx} = \tan (y + \frac{\pi}{4}) \Rightarrow$$

$$\frac{dy}{\tan (y + \frac{\pi}{4})} = dx \Rightarrow$$

$$\int \frac{dy}{\tan (y + \frac{\pi}{4})} = \int dx \Rightarrow$$

$[\ln \sin(y + \frac{\pi}{4})]^{‘} =$
$\frac{[\sin(y + \frac{\pi}{4})]^{‘}}{\sin(y + \frac{\pi}{4})} =$
$\frac{\cos(y + \frac{\pi}{4})}{\sin(y + \frac{\pi}{4})} =$
$\frac{1}{\tan (y + \frac{\pi}{4})}.$

$$\ln \sin (y + \frac{\pi}{4}) = x + \ln C_{2} (2)$$

$(2)$ 式中的 $”\ln C_{2}”$ 表示这是一个未知常数，这里当然也可以单独用 $”C_{2}”$ 表示未知常数，不过，为了和全式保持一致，并在之后的运算中消去 $”\ln”$, 我们这里选择使用 $”\ln C_{2}”$ 表示未知常数。

$$\log_{e}^{e^{x}} = x \Rightarrow$$

$$x = \ln e^{x}.$$

$$\ln \sin (y + \frac{\pi}{4}) = \ln e^{x} + \ln C_{2} \Rightarrow$$

$$\ln \sin (y + \frac{\pi}{4}) = \ln(C_{2} e^{x}) \Rightarrow$$

$\ln \sin (y + \frac{\pi}{4}) = \ln e^{x} + \ln C_{2} \nRightarrow$
$\sin (y + \frac{\pi}{4}) = e^{x} + C_{2}.$

$$\sin (y + \frac{\pi}{4}) = C_{2}e^{x}.$$

$$x = 0 时，y(0) = 0.$$

$$\sin (\frac{\pi}{4}) = C_{2} \cdot 1 \Rightarrow$$

$$C_{2} = \frac{\sqrt{2}}{2}.$$

$$\sin (y + \frac{\pi}{4}) = \frac{\sqrt{2}}{2}e^{x} \Rightarrow$$

$$\arcsin[\sin (y + \frac{\pi}{4})] = \arcsin(\frac{\sqrt{2}}{2}e^{x}) \Rightarrow$$

$$y + \frac{\pi}{4} = \arcsin(\frac{\sqrt{2}}{2}e^{x}) \Rightarrow$$

$$y = \arcsin(\frac{\sqrt{2}}{2}e^{x}) – \frac{\pi}{4}.$$