IAL 2025 June Q2 | 國際數學站

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\frac{\mathrm{d}y}{\mathrm{d}x} - 5y^2 = \mathrm{e}^{x^2 - 9} \end{align*}

Given that $y = 2$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = -1$ at $x = 3$ , determine a Taylor series for $y$ in ascending powers of $(x - 3)$ , up to and including the term in $(x - 3)^3$

(5)

解答

Given the differential equation:

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\frac{\mathrm{d}y}{\mathrm{d}x} - 5y^2 = &\,\mathrm{e}^{x^2-9} \end{align*}

Substitute $x = 3$ , $y = 2$ , and $y' = -1$ :

\begin{align*} y''(3) - 4(-1) - 5(2)^2 = &\,\mathrm{e}^{3^2-9}\\[4mm] y''(3) + 4 - 20 = &\,\mathrm{e}^0\\[4mm] y''(3) - 16 = &\,1\\[4mm] y''(3) = &\,17 \end{align*}

Differentiate the equation with respect to $x$ :

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 10y\frac{\mathrm{d}y}{\mathrm{d}x} = &\,2x\mathrm{e}^{x^2-9} \end{align*}

Substitute $x = 3$ :

\begin{align*} y'''(3) - 4(17) - 10(2)(-1) = &\,2(3)\mathrm{e}^{3^2-9}\\[4mm] y'''(3) - 68 + 20 = &\,6(1)\\[4mm] y'''(3) - 48 = &\,6\\[4mm] y'''(3) = &\,54 \end{align*}

Using the Taylor series expansion about $x = 3$ :

\begin{align*} y(x) = &\,y(3) + y'(3)(x - 3) + \frac{y''(3)}{2!}(x - 3)^2\\[4mm] &\,\hspace{2pt}+ \frac{y'''(3)}{3!}(x - 3)^3 + \cdots\\[4mm] = &\,2 - 1(x - 3) + \frac{17}{2}(x - 3)^2 + \frac{54}{6}(x - 3)^3\\[4mm] = &\,2 - (x - 3) + \frac{17}{2}(x - 3)^2 + 9(x - 3)^3 \end{align*}