题目

Problem

Use the substitution $z=x+y$ to find the solution of the differential equation

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+3x+3y}{3x+3y-1}

for which $y=0$ when $x=1$ . Give your answer in the form $a\ln(x+y)+b(x-y)+c=0$ , where $a$ , $b$ and $c$ are constants to be determined.

(7)

题目中文翻译

用代换 $z=x+y$ ，求微分方程

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+3x+3y}{3x+3y-1}

满足当 $x=1$ 时 $y=0$ 的解。将答案写成 $a\ln(x+y)+b(x-y)+c=0$ 的形式，其中 $a$ 、 $b$ 和 $c$ 是待求的常数。

(7)

原卷通用作答要求

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。

解答

解法一

思路

展开

令 $z=x+y$ ，把右边改写成只含 $z$ 的式子。得到可分离方程后，对 $z$ 积分，再代初值和 $z=x+y$ 。

答题过程

展开

Since $z=x+y$ , we have

\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} =&\,1+\frac{\mathrm{d}y}{\mathrm{d}x}\\ =&\,1+\frac{1+3z}{3z-1}\\ =&\,\frac{6z}{3z-1}. \end{align*}

On the branch containing the initial value $z=1$ , we can separate variables and integrate:

\begin{align*} \frac{3z-1}{6z}\,\mathrm{d}z=&\,\mathrm{d}x,\\ \int\bigg(\frac12-\frac1{6z}\bigg)\,\mathrm{d}z=&\,\int\mathrm{d}x,\\ x=&\,\frac z2-\frac16\ln z+C. \end{align*}

At $x=1$ , $y=0$ gives $z=1$ , so $C=\frac12$ . Substituting $z=x+y$ and multiplying by $6$ gives $6x=3(x+y)-\ln(x+y)+3$ . Rearranging,

\boxed{\ln(x+y)+3(x-y)-3=0}.

This is the branch $x+y>1/3$ containing the initial point, and one choice is $(a,b,c)=(1,3,-3)$ .

CIE 9231 2023 June Paper 21 Q2

题目

解答

解法一

思路

答题过程