求微分方程

的特解，已知当 $x = 0$ 时 $y = \pi$。答案需写成精确形式。

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。

$y$ 的系数是分母的导数乘上 $-\frac12$ 再除以分母，可直接求出积分因子。乘上它后，配方把右边的积分化为反正弦。

Write $Q=1-2x-x^2=2-(x+1)^2$. On the interval $-1-\sqrt2<x<-1+\sqrt2$ containing $0$, we have $Q>0$. Since $(1+x)/Q=-Q'/(2Q)$, an integrating factor is $Q^{-1/2}$. Indeed,

Hence

In the integral of $1/\sqrt{2-(x+1)^2}$, put $s=(x+1)/\sqrt2$. Then $\mathrm{d}x=\sqrt2\,\mathrm{d}s$, so

Integration therefore gives

At $x=0$, $Q=1$ and $y=\pi$, so $C=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$. Thus