1991 年考研数二真题解析

三、解答题 (本题满分 25 分, 每小题 5 分)

(1) 设 $\{\begin{array}{l}x=t\cos t,\\ y=t\sin t,\end{array}$ 求 $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$.

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\cdot \frac{\mathrm{d}t}{\mathrm{d}x}\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}t}=\sin t+t\cos t$$

$$\frac{\mathrm{d}x}{\mathrm{d}t}=\cos t-t\sin t\Rightarrow$$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sin t+t\cos t}{\cos t-t\sin t}\Rightarrow \frac{d^{2}y}{\mathrm{d}{x}^2}=\frac{d}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\cdot \frac{\mathrm{d}t}{\mathrm{d}x}\Rightarrow$$

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{t^2+2}{(\cos t-t\sin t{)}^2}$$

$$\frac{d^{2}y}{\mathrm{d}{x}^2}=\frac{t^2+2}{(\cos t-t\sin t{)}^3}$$

(2) 计算 ${\int }_1^4\frac{\mathrm{d}x}{x(1+\sqrt{x})}$.

$$t=\sqrt{x}\Rightarrow t^2=x\Rightarrow t\in (1,2)\Rightarrow \mathrm{d}x=2t\mathrm{d}t\Rightarrow$$

$${\int }_1^4\frac{\mathrm{d}x}{x(1+\sqrt{x})}=2{\int }_1^2\frac{t\mathrm{d}t}{t^2(1+t)}=$$

$$2{\int }_1^2\frac{\mathrm{d}t}{t(1+t)}\Rightarrow 2{\int }_1^2(\frac{1}{t}-\frac{1}{t+1})\mathrm{d}t=$$

$${2(\ln t-\ln (t+1))|}_1^2=2[\ln 2-\ln 3-(0-\ln 2)]=$$

$$2(2\ln 2-\ln 3)=2(\ln 4-\ln 3)=2\ln \frac{4}{3}$$

(3) 求 $\underset{x\to 0}{lim}\frac{x-\sin x}{x^2({\mathrm{e}}^x-1)}$.

$$x\to 0\Rightarrow \frac{\frac{1}{6}{x}^3}{x^2\cdot x}=\frac{1}{6}$$

(4) 求 $\int x{\sin }^{2}x\mathrm{d}x$.

错误解法（这是不定积分，不能用定积分的公式计算）：

$$\int x{\sin }^{2}x\mathrm{d}x=\frac{\pi }{2}\int {\sin }^{2}x\mathrm{d}x=$$

$$\frac{\pi }{2}\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{{\pi }^2}{8}$$

正确解法（降幂）：

$$\cos 2\alpha =1-2{\sin }^2\alpha \Rightarrow$$

$${\sin }^2\alpha =\frac{1}{2}(1-\cos 2\alpha )\Rightarrow$$

$${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$$

$$\int x{\sin }^{2}x\mathrm{d}x=\frac{1}{2}\int x(1-\cos 2x)\mathrm{d}x=$$

$$\frac{1}{2}\int x\mathrm{d}x-\frac{1}{2}\int x\cos 2x\mathrm{d}x=$$

$$\frac{1}{4}{x}^2-\frac{1}{4}\int x\mathrm{d}(\sin 2x)=$$

$$\frac{1}{4}{x}^2-\frac{1}{4}x\sin 2x+\frac{1}{4}\cdot \frac{1}{2}\int \sin 2x\mathrm{d}(2x)=$$

$$\frac{1}{4}{x}^2-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+C$$

(5) 求微分方程 $x{y}^{\prime }+y=x{\mathrm{e}}^x$ 满足 $y(1)=1$ 的特解.

$$x{y}^{\prime }+y=x{e}^x\Rightarrow y^{\prime }+\frac{1}{x}y=e^x\Rightarrow$$

$$y=[\int e^x\cdot e^{\int \frac{1}{x}\mathrm{d}x}\mathrm{d}x+c]e^{-\int \frac{1}{x}\mathrm{d}x}\Rightarrow$$

$$y=[[x{e}^x\mathrm{d}x+c]\cdot \frac{1}{x}\Rightarrow$$

$$y=[x{e}^x-e^x+c]\cdot \frac{1}{x}\Rightarrow x=1,y=1\Rightarrow$$

$$1=(e-e+c)\Rightarrow c=1\Rightarrow$$

$$y=(x{e}^x-e^x+1)\cdot \frac{1}{x}$$