(a) 用代换 $u=1-(\theta-1)^2$ 求

(b) 求微分方程

的解，其中 $0<\theta<2$，且当 $\theta=1$ 时 $y=1$。将答案写成 $y=\mathrm{f}(\theta)$ 的形式。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

代换后，$(\theta-1)\,\mathrm{d}\theta=-\frac12\mathrm{d}u$，被积函数化为 $u^{-1/2}$ 的倍数。

With $u=1-(\theta-1)^2$, we have $\mathrm{d}u=-2(\theta-1)\,\mathrm{d}\theta$. Therefore

除以 $\theta$ 后，用积分因子 $\theta^{-1}$。右边先平移变量，再分部积分，余项可用 (a)。

On $0<\theta<2$, the standard form is

Since $\theta>0$, the integrating factor is $e^{\int-\theta^{-1}\,\mathrm{d}\theta}=e^{-\ln\theta}=\theta^{-1}$. Multiplying by it gives $(y/\theta)'=\arcsin(\theta-1)$. Put $s=\theta-1$, so $\mathrm{d}s=\mathrm{d}\theta$. For integration by parts, take $u=\arcsin s$ and $\mathrm{d}v=\mathrm{d}s$, so $\mathrm{d}u=\mathrm{d}s/\sqrt{1-s^2}$ and $v=s$. The remaining integral is exactly (a) with $s=\theta-1$. Thus

At $\theta=1$, $s=0$ and $y=1$, so $C=0$. Thus