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IAL 2025 June Q2

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 2

题目

Problem

d2ydx2−4dydx−5y2=ex2−9\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=2y=2 and dydx=−1\frac{\mathrm{d}y}{\mathrm{d}x}=-1 at x=3x=3, determine a Taylor series for yy in ascending powers of (x−3)(x-3), up to and including the term in (x−3)3(x-3)^3

(5)

解答

解法一

思路

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Taylor series 到三次项需要 y(3)y(3)、y′(3)y'(3)、y′′(3)y''(3)、y′′′(3)y'''(3)。前两个已给出，先代入原方程求 y′′(3)y''(3)，再把原方程对 xx 求导来求 y′′′(3)y'''(3)。

答题过程

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Given

y(3)=2,y′(3)=−1\begin{align*} y(3)=2,\qquad y'(3)=-1 \end{align*}

Substitute into the differential equation:

y′′(3)−4(−1)−5(2)2=e32−9y′′(3)+4−20=1y′′(3)=17\begin{align*} y''(3)-4(-1)-5(2)^2 ={}& \mathrm{e}^{3^2-9}\\[3mm] y''(3)+4-20 ={}& 1\\[3mm] y''(3) ={}& 17 \end{align*}

Differentiate the differential equation:

y′′′−4y′′−10yy′=2xex2−9\begin{align*} y'''-4y''-10yy' ={}& 2x\mathrm{e}^{x^2-9} \end{align*}

At x=3x=3,

y′′′(3)−4(17)−10(2)(−1)=2(3)e32−9y′′′(3)−68+20=6y′′′(3)=54\begin{align*} y'''(3)-4(17)-10(2)(-1) ={}& 2(3)\mathrm{e}^{3^2-9}\\[3mm] y'''(3)-68+20 ={}& 6\\[3mm] y'''(3) ={}& 54 \end{align*}

Using Taylor’s formula about x=3x=3,

y=y(3)+y′(3)(x−3)+y′′(3)2!(x−3)2 +y′′′(3)3!(x−3)3+⋯\begin{align*} y ={}& y(3)+y'(3)(x-3) +\frac{y''(3)}{2!}(x-3)^2\\[2mm] &\,\hspace{2pt}+\frac{y'''(3)}{3!}(x-3)^3+\cdots \end{align*}

Therefore

y=2−(x−3)+172(x−3)2+546(x−3)3+⋯=2−(x−3)+172(x−3)2+9(x−3)3+⋯\begin{align*} y ={}& 2-(x-3) +\frac{17}{2}(x-3)^2 +\frac{54}{6}(x-3)^3+\cdots\\[3mm] ={}& \boxed{ 2-(x-3)+\frac{17}{2}(x-3)^2+9(x-3)^3+\cdots } \end{align*}