IAL 2020 Oct Q1 | 國際數學站

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} = 2\cos x \end{align*}

(a) Express $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}$ in terms of $x$ , $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$

(3)

At $x = 0$ , $y = 2$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 5$

(b) Determine the value of $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}$ at $x = 0$

(1)

(3)

解答

Note. Dashed notation is acceptable throughout this question.

(a)

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 3\frac{\mathrm{d}y}{\mathrm{d}x} + 3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2\sin x \end{align*}

Hence

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -2\sin x - 3\frac{\mathrm{d}y}{\mathrm{d}x} - 3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

(b)

At $x = 0$ ,

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -3 \times 5 = -15 \end{align*}

(c)

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -3 \times 0 \times 5 + 2 = 2 \end{align*}

Using the Taylor expansion about $x = 0$ ,

\begin{align*} y = 2 + 5x + \frac{2}{2!}x^2 + \frac{-15}{3!}x^3 \end{align*}

Hence

\begin{align*} y = 2 + 5x + x^2 - \frac{5}{2}x^3 \end{align*}