IAL 2025 Jan Q3 | 國際數學站

\begin{align*} 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} - 6xy = 0 \end{align*}

Given that $\frac{\mathrm{d}y}{\mathrm{d}x} = 3$ and $y = \frac{1}{2}$ at $x = 2$

(a) determine the value of $\frac{\mathrm{d}^3y}{\mathrm{d}x^3}$ at $x = 2$

(6)

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

(2)

解答

(a)

Substitute $x = 2$ , $y = \frac{1}{2}$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 3$ into the given differential equation to find the value of $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ at $x = 2$ :

\begin{align*} 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y\frac{\mathrm{d}y}{\mathrm{d}x} - 6xy =&\, 0\\[4mm] 2(2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{1}{2}\right)(3) - 6(2)\left(\frac{1}{2}\right) =&\, 0\\[4mm] 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{3}{2} - 6 =&\, 0\\[4mm] 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{9}{2}\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{9}{8} \end{align*}

Differentiate the given differential equation with respect to $x$ , using the product rule:

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

Substitute the known values $x = 2$ , $y = \frac{1}{2}$ , $\frac{\mathrm{d}y}{\mathrm{d}x} = 3$ , and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{9}{8}$ into this differentiated equation:

\begin{align*} 2\left(\frac{9}{8}\right) + 2(2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + (3)^2 + \left(\frac{1}{2}\right)\left(\frac{9}{8}\right) - 6\left(\frac{1}{2}\right) - 6(2)(3) =&\, 0\\[4mm] \frac{9}{4} + 4\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 9 + \frac{9}{16} - 3 - 36 =&\, 0\\[4mm] 4\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + \frac{45}{16} - 30 =&\, 0\\[4mm] 4\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, 30 - \frac{45}{16}\\[4mm] 4\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, \frac{480 - 45}{16}\\[4mm] 4\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, \frac{435}{16}\\[4mm] \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, \frac{435}{64} \end{align*}

(b)

The Taylor series expansion for $y$ about $x = 2$ 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] =&\, \frac{1}{2} + 3(x - 2) + \frac{\frac{9}{8}}{2}(x - 2)^2 + \frac{\frac{435}{64}}{6}(x - 2)^3\\[4mm] =&\, \frac{1}{2} + 3(x - 2) + \frac{9}{16}(x - 2)^2 + \frac{435}{384}(x - 2)^3\\[4mm] =&\, \frac{1}{2} + 3(x - 2) + \frac{9}{16}(x - 2)^2 + \frac{145}{128}(x - 2)^3 \end{align*}

解答

(a)

\begin{align*} \frac{r+4}{r(r+1)(r+2)} = &\,\frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}\\[4mm] r+4 = &\,A(r+1)(r+2) + Br(r+2)\\[4mm] &\,\hspace{2pt}+ Cr(r+1) \end{align*}

By substituting values for $r$ :

\begin{align*} r = 0 \implies &\,4 = 2A \implies A = 2\\[4mm] r = -1 \implies &\,3 = -B \implies B = -3\\[4mm] r = -2 \implies &\,2 = 2C \implies C = 1 \end{align*}

Hence,

\begin{align*} \frac{r+4}{r(r+1)(r+2)} = &\,\frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+2} \end{align*}

(b)

\begin{align*} &\, \sum_{r=1}^{n}\frac{r+4}{r(r+1)(r+2)}\\[4mm] = &\, \sum_{r=1}^{n}\left(\frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+2}\right)\\[4mm] = &\,\left(\frac{2}{1} - \frac{3}{2} + \frac{1}{3}\right)\\[4mm] &\,\hspace{2pt}+ \left(\frac{2}{2} - \frac{3}{3} + \frac{1}{4}\right)\\[4mm] &\,\hspace{2pt}+ \left(\frac{2}{3} - \frac{3}{4} + \frac{1}{5}\right)\\[4mm] &\,\hspace{2pt}+ \cdots\\[4mm] &\,\hspace{2pt}+ \left(\frac{2}{n-1} - \frac{3}{n} + \frac{1}{n+1}\right)\\[4mm] &\,\hspace{2pt}+ \left(\frac{2}{n} - \frac{3}{n+1} + \frac{1}{n+2}\right)\\[4mm] &\,\hspace{2pt}\colorbox{aqua}{middle terms cancel}\\[4mm] = &\,\left(\frac{2}{1} - \frac{1}{2}\right) + \left(-\frac{2}{n+1} + \frac{1}{n+2}\right)\\[4mm] = &\,\frac{3}{2} - \frac{2}{n+1} + \frac{1}{n+2}\\[4mm] = &\,\frac{3(n+1)(n+2) - 4(n+2) + 2(n+1)}{2(n+1)(n+2)}\\[4mm] = &\,\frac{3(n^2+3n+2) - 4n - 8 + 2n + 2}{2(n+1)(n+2)}\\[4mm] = &\,\frac{3n^2 + 9n + 6 - 2n - 6}{2(n+1)(n+2)}\\[4mm] = &\,\frac{3n^2 + 7n}{2(n+1)(n+2)}\\[4mm] = &\,\frac{n(3n+7)}{2(n+1)(n+2)} \end{align*}

where $P=3$ , $Q=7$ , $R=1$ , and $S=2$ .