IAL 2024 June Q2 | 國際數學站

\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} - y^3 = 4 \end{align*}

(a) Show that

\begin{align*} x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = ay\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + (by^2 + c)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

where $a$ , $b$ and $c$ are integers to be determined.

(4)

Given that $y = 1$ at $x = 2$

(b) determine the Taylor series expansion for $y$ in ascending powers of $(x - 2)$ , up to and including the term in $(x - 2)^3$ , giving each coefficient in simplest form.

(3)

解答

(a)

Given the equation:

\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} - y^3 = &\,4 \end{align*}

Differentiate both sides with respect to $x$ using the product rule on the first term:

\begin{align*} \left( 1 \cdot \frac{\mathrm{d}y}{\mathrm{d}x} + x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right) - 3y^2\frac{\mathrm{d}y}{\mathrm{d}x} = &\,0 \end{align*}

Differentiate again with respect to $x$ :

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( 1 \cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} \right) - \left( 6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right) = &\,0\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - 3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,0\\[4mm] x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + (3y^2 - 2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

This is in the required form, with $a = 6$ , $b = 3$ , and $c = -2$ .

(b)

We are given $x = 2$ and $y = 1$ . Substitute these into the original equation to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ :

\begin{align*} 2\frac{\mathrm{d}y}{\mathrm{d}x} - 1^3 = &\,4\\[4mm] 2\frac{\mathrm{d}y}{\mathrm{d}x} = &\,5\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{5}{2} \end{align*}

Substitute into the first derivative equation to find $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ :

\begin{align*} \frac{5}{2} + 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 3(1)^2\left(\frac{5}{2}\right) = &\,0\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,\frac{15}{2} - \frac{5}{2}\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,5\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,\frac{5}{2} \end{align*}

Substitute into the second derivative equation to find $\frac{\mathrm{d}^3y}{\mathrm{d}x^3}$ :

\begin{align*} 2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,6(1)\left(\frac{5}{2}\right)^2 + (3(1)^2 - 2)\left(\frac{5}{2}\right)\\[4mm] 2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,6\left(\frac{25}{4}\right) + (1)\left(\frac{5}{2}\right)\\[4mm] 2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,\frac{75}{2} + \frac{5}{2}\\[4mm] 2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,40\\[4mm] \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,20 \end{align*}

The Taylor series expansion in ascending powers of $(x-2)$ is given by:

\begin{align*} y = &\,y(2) + y'(2)(x-2) + \frac{y''(2)}{2!}(x-2)^2 + \frac{y'''(2)}{3!}(x-2)^3 + \dots\\[4mm] = &\,1 + \frac{5}{2}(x-2) + \frac{5/2}{2}(x-2)^2 + \frac{20}{6}(x-2)^3 + \dots\\[4mm] = &\,1 + \frac{5}{2}(x - 2) + \frac{5}{4}(x - 2)^2 + \frac{10}{3}(x - 2)^3 \end{align*}