IAL 2025 June A Q2 | 國際數學站

(a) Show that the differential equation

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

Given the differential equation:

\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(x+2) \end{align*}

Divide through by $(x+2)$ since $x \neq -2$ :

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

This is in the form $\frac{\mathrm{d}y}{\mathrm{d}x} + f(x)y = kx$ , where $f(x) = \frac{1}{x+2}$ and $k = 3$ .

(b)

The integrating factor (IF) is given by:

\begin{align*} \text{IF} = &\,\mathrm{e}^{\int f(x)\,\mathrm{d}x}\\[4mm] = &\,\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 the integrating factor gives:

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

Integrating both sides with respect to $x$ :

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

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

\begin{align*} (4+2)(18) = &\,(4)^3 + 3(4)^2 + c\\[4mm] 108 = &\,64 + 48 + c\\[4mm] 108 = &\,112 + c\\[4mm] c = &\,-4 \end{align*}

Substitute $c = -4$ back into the general solution:

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

To simplify, we factorise the cubic $x^3+3x^2-4$ . By the factor theorem, let $P(x) = x^3+3x^2-4$ . $P(1) = 1+3-4=0$ , so $(x-1)$ is a factor.

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

Therefore:

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