题目

Problem

(a) Show that the differential equation

\begin{align*} (x+2)\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2+6x-y\qquad x\ne -2 \end{align*}

can be written in the form

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}+f(x)y=kx \end{align*}

where $f$ is a function to be determined and $k$ is a constant to be found.

(2)

Given that $y=18$ at $x=4$

(b) use the answer to part (a) to determine, in simplest form, the particular solution of the differential equation.

Give the answer in the form $y=g(x)$

(6)

解答

(a)

解法一

思路

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目标形式是一阶线性微分方程。先把 $y$ 移到左边，再除以 $x+2$ 。因为题目给了 $x\ne -2$ ，所以这一步合法。

#### 答题过程

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Starting from

\begin{align*} (x+2)\frac{\mathrm{d}y}{\mathrm{d}x} =&\,3x^2+6x-y\\[4mm] (x+2)\frac{\mathrm{d}y}{\mathrm{d}x}+y =&\,3x^2+6x\\[4mm] =&\,3x(x+2) \end{align*}

Since $x\ne -2$ , divide by $x+2$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{x+2}y =&\,3x \end{align*}

Hence

\begin{align*} f(x)=\frac{1}{x+2}, \qquad k=3 \end{align*}

(b)

解法一

思路

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这是标准一阶线性微分方程。积分因子是 $e^{\int \frac{1}{x+2}\,\mathrm{d}x}$ 。由于初始点 $x=4$ 在 $x>-2$ 的区间内，可以取积分因子为 $x+2$ ，然后两边积分并代入条件。

答题过程

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The integrating factor is

\begin{align*} \mathrm{IF} =&\,\mathrm{e}^{\int \frac{1}{x+2}\,\mathrm{d}x}\\[4mm] =&\,\mathrm{e}^{\ln(x+2)}\\[4mm] =&\,x+2 \end{align*}

Multiplying the differential equation by $x+2$ gives

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\bigl((x+2)y\bigr) =&\,3x(x+2)\\[4mm] =&\,3x^2+6x \end{align*}

Integrating,

\begin{align*} (x+2)y =&\,\int (3x^2+6x)\,\mathrm{d}x\\[4mm] =&\,x^3+3x^2+C \end{align*}

Use $x=4,\ y=18$ :

\begin{align*} 6(18) =&\,4^3+3(4^2)+C\\[4mm] 108 =&\,64+48+C\\[4mm] C =&\,-4 \end{align*}

Therefore

\begin{align*} (x+2)y =&\,x^3+3x^2-4\\[4mm] =&\,(x-1)(x+2)^2 \end{align*}

\begin{align*} y =&\,(x-1)(x+2)\\[4mm] =&\,x^2+x-2 \end{align*}

解法二

思路

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也可以把原方程看成「齐次解 + 特解」。齐次方程给出 $\dfrac{A}{x+2}$ ，再猜一个二次多项式特解。最后用条件会发现齐次部分的常数为 $0$ 。

答题过程

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For the homogeneous equation,

\begin{align*} (x+2)\frac{\mathrm{d}y}{\mathrm{d}x}+y =&\,0\\[4mm] \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\frac{1}{x+2}\\[4mm] \ln y =&\,-\ln(x+2)+C \end{align*}

\begin{align*} y_{\mathrm{c}} =&\,\frac{A}{x+2} \end{align*}

Try a particular solution

\begin{align*} y_{\mathrm{p}}=ax^2+bx+c \end{align*}

Then

\begin{align*} \frac{\mathrm{d}y_{\mathrm{p}}}{\mathrm{d}x} =&\,2ax+b \end{align*}

Substitute into $(x+2)\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=3x^2+6x$ :

\begin{align*} (x+2)(2ax+b)+ax^2+bx+c =&\,3x^2+6x\\[4mm] 3ax^2+(4a+2b)x+(2b+c) =&\,3x^2+6x \end{align*}

Equating coefficients,

\begin{align*} 3a=&\,3,\\[2mm] 4a+2b=&\,6,\\[2mm] 2b+c=&\,0 \end{align*}

Hence

\begin{align*} a=1,\qquad b=1,\qquad c=-2 \end{align*}

The general solution is

\begin{align*} y =&\,x^2+x-2+\frac{A}{x+2} \end{align*}

Use $x=4,\ y=18$ :

\begin{align*} 18 =&\,16+4-2+\frac{A}{6}\\[4mm] \frac{A}{6} =&\,0\\[4mm] A =&\,0 \end{align*}

Therefore

\begin{align*} y=x^2+x-2 \end{align*}

IAL 2025 June A Q2

题目

解答

(a)

解法一

思路

(b)

解法一

思路

答题过程

解法二

思路

答题过程