IAL 2025 June A Q7 | 國際數學站

(a) Show that the substitution $x = e^{u}$ , where $u$ is a function of $x$ , transforms the differential equation

\begin{align*} 2x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} - y = 27x^2 \qquad x > 0 \qquad \text{(I)} \end{align*}

into the differential equation

\begin{align*} 2\frac{\mathrm{d}^2y}{\mathrm{d}u^2} + \frac{\mathrm{d}y}{\mathrm{d}u} - y = 27e^{2u} \qquad \text{(II)} \end{align*}

(4)

(b) By solving differential equation (II), determine the general solution of differential equation (I).

Give the answer in the form $y = f(x)$ where $f$ is a fully simplified function.

(4)

Given that when $x = \dfrac{1}{4}$ , $y = \dfrac{11}{16}$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1$

(c) determine the value of $y$ when $x = \dfrac{1}{8}$ , giving the answer in the form $\dfrac{1}{64}(p\sqrt{2} + q)$ where $p$ and $q$ are integers.

(5)

解答

(a)

Given the substitution $x = \mathrm{e}^u$ , we have:

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}u} = &\,\mathrm{e}^u = x \\[4mm] \implies \frac{\mathrm{d}u}{\mathrm{d}x} = &\,\frac{1}{x} \end{align*}

Using the chain rule for the first derivative:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{\mathrm{d}y}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x}\\[4mm] = &\,\frac{1}{x} \frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

For the second derivative, we differentiate with respect to $x$ using the product rule and chain rule:

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{1}{x} \frac{\mathrm{d}y}{\mathrm{d}u} \right)\\[4mm] = &\,-\frac{1}{x^2} \frac{\mathrm{d}y}{\mathrm{d}u} + \frac{1}{x} \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\mathrm{d}y}{\mathrm{d}u} \right)\\[4mm] = &\,-\frac{1}{x^2} \frac{\mathrm{d}y}{\mathrm{d}u} + \frac{1}{x} \left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} \cdot \frac{\mathrm{d}u}{\mathrm{d}x} \right)\\[4mm] = &\,-\frac{1}{x^2} \frac{\mathrm{d}y}{\mathrm{d}u} + \frac{1}{x^2} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} \end{align*}

Substitute $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ into the original differential equation (I):

\begin{align*} 2x^2 \left( \frac{1}{x^2} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} - \frac{1}{x^2} \frac{\mathrm{d}y}{\mathrm{d}u} \right) + 3x \left( \frac{1}{x} \frac{\mathrm{d}y}{\mathrm{d}u} \right) - y = &\,27x^2\\[4mm] 2 \frac{\mathrm{d}^2y}{\mathrm{d}u^2} - 2 \frac{\mathrm{d}y}{\mathrm{d}u} + 3 \frac{\mathrm{d}y}{\mathrm{d}u} - y = &\,27(\mathrm{e}^u)^2\\[4mm] 2 \frac{\mathrm{d}^2y}{\mathrm{d}u^2} + \frac{\mathrm{d}y}{\mathrm{d}u} - y = &\,27\mathrm{e}^{2u} \quad \text{... (II)} \end{align*}

This completes the proof.

(b)

To solve the differential equation (II), we first find the complementary function (CF) by solving the auxiliary equation:

\begin{align*} 2m^2 + m - 1 = &\,0\\[4mm] (2m - 1)(m + 1) = &\,0\\[4mm] m = &\,\frac{1}{2}, \,-1 \end{align*}

So the complementary function is $y_c = A\mathrm{e}^{\frac{1}{2}u} + B\mathrm{e}^{-u}$ .

For the particular integral (PI), we try $y_p = \lambda \mathrm{e}^{2u}$ :

\begin{align*} y_p = &\,\lambda \mathrm{e}^{2u}\\[4mm] \frac{\mathrm{d}y_p}{\mathrm{d}u} = &\,2\lambda \mathrm{e}^{2u}\\[4mm] \frac{\mathrm{d}^2y_p}{\mathrm{d}u^2} = &\,4\lambda \mathrm{e}^{2u} \end{align*}

Substitute into equation (II):

\begin{align*} 2(4\lambda \mathrm{e}^{2u}) + (2\lambda \mathrm{e}^{2u}) - (\lambda \mathrm{e}^{2u}) = &\,27\mathrm{e}^{2u}\\[4mm] 8\lambda + 2\lambda - \lambda = &\,27\\[4mm] 9\lambda = &\,27\\[4mm] \lambda = &\,3 \end{align*}

So $y_p = 3\mathrm{e}^{2u}$ .

The general solution of (II) is $y = A\mathrm{e}^{\frac{1}{2}u} + B\mathrm{e}^{-u} + 3\mathrm{e}^{2u}$ . Replacing $\mathrm{e}^u$ with $x$ , we get the general solution of (I):

\begin{align*} y = &\,Ax^{\frac{1}{2}} + Bx^{-1} + 3x^2 \end{align*}

(c)

Given $x = \frac{1}{4}$ , $y = \frac{11}{16}$ , and $\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ . First, differentiate our general solution:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{1}{2}Ax^{-\frac{1}{2}} - Bx^{-2} + 6x \end{align*}

Using the condition for $y$ :

\begin{align*} \frac{11}{16} = &\,A\left(\frac{1}{4}\right)^{\frac{1}{2}} + B\left(\frac{1}{4}\right)^{-1} + 3\left(\frac{1}{4}\right)^2\\[4mm] \frac{11}{16} = &\,\frac{1}{2}A + 4B + \frac{3}{16}\\[4mm] \frac{8}{16} = &\,\frac{1}{2}A + 4B\\[4mm] A + 8B = &\,1 \quad \text{... (Eq. 1)} \end{align*}

Using the condition for $\frac{\mathrm{d}y}{\mathrm{d}x}$ :

\begin{align*} 1 = &\,\frac{1}{2}A\left(\frac{1}{4}\right)^{-\frac{1}{2}} - B\left(\frac{1}{4}\right)^{-2} + 6\left(\frac{1}{4}\right)\\[4mm] 1 = &\,\frac{1}{2}A(2) - 16B + \frac{3}{2}\\[4mm] 1 = &\,A - 16B + \frac{3}{2}\\[4mm] A - 16B = &\,-\frac{1}{2} \quad \text{... (Eq. 2)} \end{align*}

Subtract (Eq. 2) from (Eq. 1):

\begin{align*} (A + 8B) - (A - 16B) = &\,1 - \left(-\frac{1}{2}\right)\\[4mm] 24B = &\,\frac{3}{2}\\[4mm] B = &\,\frac{1}{16} \end{align*}

Substitute $B$ into (Eq. 1):

\begin{align*} A + 8\left(\frac{1}{16}\right) = &\,1\\[4mm] A + \frac{1}{2} = &\,1\\[4mm] A = &\,\frac{1}{2} \end{align*}

So the particular solution is:

\begin{align*} y = &\,\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{16}x^{-1} + 3x^2 \end{align*}

Now, find the value of $y$ when $x = \frac{1}{8}$ :

\begin{align*} y = &\,\frac{1}{2}\left(\frac{1}{8}\right)^{\frac{1}{2}} + \frac{1}{16}\left(\frac{1}{8}\right)^{-1} + 3\left(\frac{1}{8}\right)^2\\[4mm] = &\,\frac{1}{2}\left(\frac{1}{2\sqrt{2}}\right) + \frac{1}{16}(8) + 3\left(\frac{1}{64}\right)\\[4mm] = &\,\frac{1}{4\sqrt{2}} + \frac{1}{2} + \frac{3}{64}\\[4mm] = &\,\frac{\sqrt{2}}{8} + \frac{32}{64} + \frac{3}{64}\\[4mm] = &\,\frac{\sqrt{2}}{8} + \frac{35}{64}\\[4mm] = &\,\frac{8\sqrt{2} + 35}{64}\\[4mm] = &\,\frac{1}{64}(8\sqrt{2} + 35) \end{align*}

This is in the required form, where $p=8$ and $q=35$ .