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IAL 2026 Jan Q4

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 4

题目

Problem

d2ydx2=x2y+dydx\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

(a) Show that

d4ydx4=d3ydx3+Ax2d2ydx2+Bxdydx+Cy\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} = \frac{\mathrm{d}^3y}{\mathrm{d}x^3} +Ax^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +Bx\frac{\mathrm{d}y}{\mathrm{d}x} +Cy \end{align*}

where AA, BB and CC are integers to be determined.

(4)

Given that dydx=3\dfrac{\mathrm{d}y}{\mathrm{d}x}=3 and y=1y=1 at x=1x=1

(b) determine the Taylor series solution for yy in ascending powers of (x1)(x-1), up to and including the term in (x1)4(x-1)^4, giving each coefficient in simplest form.

(4)

解答

(a)

解法一

思路

展开

从给定的二阶导数开始连续求导。每次对 x2yx^2y 求导时都要用乘积法则。

答题过程

展开

Given

d2ydx2=x2y+dydx\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

differentiating gives

d3ydx3=2xy+x2dydx+d2ydx2\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} ={}& 2xy+x^2\frac{\mathrm{d}y}{\mathrm{d}x} +\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

Differentiate again:

d4ydx4=2y+2xdydx+2xdydx+x2d2ydx2+d3ydx3=d3ydx3+x2d2ydx2+4xdydx+2y\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} ={}& 2y+2x\frac{\mathrm{d}y}{\mathrm{d}x} +2x\frac{\mathrm{d}y}{\mathrm{d}x}\\[3mm] &\,\hspace{2pt}+x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[3mm] ={}& \frac{\mathrm{d}^3y}{\mathrm{d}x^3} +x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +4x\frac{\mathrm{d}y}{\mathrm{d}x} +2y \end{align*}

Therefore

A=1,B=4,C=2\begin{align*} A=1,\qquad B=4,\qquad C=2 \end{align*}

(b)

解法一

思路

展开

Taylor 展开需要 y(1)y(1)y(1)y'(1)y(1)y''(1)y(1)y'''(1)y(1)y''''(1)。前两个已给,后面三个用微分方程逐个算。

答题过程

展开

At x=1x=1,

y=1,dydx=3\begin{align*} y=1,\qquad \frac{\mathrm{d}y}{\mathrm{d}x}=3 \end{align*}

Using the differential equation,

d2ydx2=x2y+dydx\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

so

y(1)=12(1)+3=4\begin{align*} y''(1)=1^2(1)+3=4 \end{align*}

Also,

y(1)=2(1)(1)+12(3)+4=9\begin{align*} y'''(1) ={}& 2(1)(1)+1^2(3)+4\\[3mm] ={}& 9 \end{align*}

and, using part (a),

y(1)=9+12(4)+4(1)(3)+2(1)=27\begin{align*} y''''(1) ={}& 9+1^2(4)+4(1)(3)+2(1)\\[3mm] ={}& 27 \end{align*}

Therefore

y=1+3(x1)+42!(x1)2+93!(x1)3+274!(x1)4=1+3(x1)+2(x1)2+32(x1)3+98(x1)4\begin{align*} y ={}& 1+3(x-1)+\frac{4}{2!}(x-1)^2 +\frac{9}{3!}(x-1)^3\\[3mm] &\,\hspace{2pt}+\frac{27}{4!}(x-1)^4\\[3mm] ={}& 1+3(x-1)+2(x-1)^2 +\frac32(x-1)^3 +\frac98(x-1)^4 \end{align*}