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IAL 2023 Jan Q4

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 4

题目

Problem

dydx= y2−x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =\,& y^2 - x \end{align*}

(a) Show that

d4ydx4= Ayd3ydx3+Bdydxd2ydx2\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} =\,& Ay\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + B\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

where AA and BB are integers to be determined.

(4)

Given that y=1y = 1 at x=−1x = -1

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

(3)

解答

(a)

解法一

思路

展开

从 y′=y2−xy'=y^2-x 开始连续求导。每次对含 yy 的式子求导时，都要记得 yy 是 xx 的函数。

答题过程

展开

Given

dydx=y2−x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=y^2-x. \end{align*}

Differentiate:

d2ydx2= 2ydydx−1.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,2y\frac{\mathrm{d}y}{\mathrm{d}x}-1. \end{align*}

Differentiate again:

d3ydx3= 2yd2ydx2+2(dydx)2.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,2y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2. \end{align*}

Now differentiate once more:

d4ydx4= 2yd3ydx3+2dydxd2ydx2 +4dydxd2ydx2= 2yd3ydx3+6dydxd2ydx2.\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} =&\,2y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} +2\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] &\,\hspace{2pt}+4\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] =&\,2y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} +6\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Therefore

A=2,B=6.\begin{align*} \boxed{A=2,\qquad B=6}. \end{align*}

(b)

解法一

思路

展开

展开中心是 x=−1x=-1，所以 Taylor series 要写成 (x+1)(x+1) 的幂。先用微分方程和 (a) 的结果求出在 x=−1x=-1 时的各阶导数。

答题过程

展开

At x=−1x=-1,

y=1.\begin{align*} y=1. \end{align*}

First,

y′= y2−xy′(−1)= 12−(−1)=2.\begin{align*} y'=&\,y^2-x\\[2mm] y'(-1)=&\,1^2-(-1)=2. \end{align*}

Also,

y′′= 2yy′−1,y′′(−1)= 2(1)(2)−1=3.\begin{align*} y''=&\,2yy'-1,\\[2mm] y''(-1)=&\,2(1)(2)-1=3. \end{align*}

Using

y′′′=2yy′′+2(y′)2,\begin{align*} y'''=2yy''+2(y')^2, \end{align*}

we get

y′′′(−1)= 2(1)(3)+2(2)2= 6+8= 14.\begin{align*} y'''(-1)=&\,2(1)(3)+2(2)^2\\[2mm] =&\,6+8\\[2mm] =&\,14. \end{align*}

Using part (a),

y(4)=2yy′′′+6y′y′′,\begin{align*} y^{(4)} =2yy'''+6y'y'', \end{align*}

so

y(4)(−1)= 2(1)(14)+6(2)(3)= 28+36= 64.\begin{align*} y^{(4)}(-1) =&\,2(1)(14)+6(2)(3)\\[2mm] =&\,28+36\\[2mm] =&\,64. \end{align*}

Therefore

y= y(−1)+y′(−1)(x+1)+y′′(−1)2!(x+1)2 +y′′′(−1)3!(x+1)3+y(4)(−1)4!(x+1)4+⋯= 1+2(x+1)+32(x+1)2 +146(x+1)3+6424(x+1)4+⋯= 1+2(x+1)+32(x+1)2 +73(x+1)3+83(x+1)4+⋯ .\begin{align*} y=&\,y(-1)+y'(-1)(x+1) +\frac{y''(-1)}{2!}(x+1)^2\\[2mm] &\,\hspace{2pt}+\frac{y'''(-1)}{3!}(x+1)^3 +\frac{y^{(4)}(-1)}{4!}(x+1)^4+\cdots\\[2mm] =&\,1+2(x+1)+\frac{3}{2}(x+1)^2\\[2mm] &\,\hspace{2pt}+\frac{14}{6}(x+1)^3+\frac{64}{24}(x+1)^4+\cdots\\[2mm] =&\,1+2(x+1)+\frac{3}{2}(x+1)^2\\[2mm] &\,\hspace{2pt}+\frac{7}{3}(x+1)^3+\frac{8}{3}(x+1)^4+\cdots. \end{align*}

Thus

y=1+2(x+1)+32(x+1)2+73(x+1)3+83(x+1)4+⋯.\begin{align*} \boxed{ y=1+2(x+1)+\frac{3}{2}(x+1)^2 +\frac{7}{3}(x+1)^3+\frac{8}{3}(x+1)^4+\cdots }. \end{align*}