变量 $x$ 和 $y$ 满足当 $x=0$ 时 $y=0$，且

(a) 证明当 $x=0$ 时，$\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{3}{4}$。

(b) 求 $y$ 的麦克劳林级数，展开至并包括 $x^2$ 项。

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。

对 $(x+1)y$ 用乘积法则，对三次项用链式法则。最后代入 $x=y=0$。

Differentiating implicitly gives

At $x=y=0$, this is $y'+3(1+y')=0$, so $y'=-3/4$.

对 (a) 再求一次导数。将零点处的 $y,y'$ 代入求 $y''$，再应用 Maclaurin 公式。

The derivative of $y+(x+1)y'$ is $2y'+(x+1)y''$. By the product and chain rules, the derivative of $3(x+y+1)^2(1+y')$ is $6(x+y+1)(1+y')^2+3(x+y+1)^2y''$. Thus differentiating the equation from (a) gives

At $x=y=0$, with $y'=-3/4$, this becomes

so $4y''=3/2-3/8=9/8$, giving $y''(0)=9/32$. Using $y=y(0)+y'(0)x+\frac12y''(0)x^2+O(x^3)$ gives

也可将待求的级数直接代回原方程。用 (a) 的一次项，再比较 $x^2$ 的系数。

Write $y=px+qx^2+O(x^3)$, with $p=-3/4$ from (a). Then

and

The quadratic coefficient on the left must vanish, so $p+4q+3(1+p)^2=0$. Substituting $p=-3/4$ gives $4q=3/4-3/16=9/16$, so $q=9/64$. Hence