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IAL 2023 Jan Q4

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 4

(a) Show that

dydx= y2−x  ⟹  d4ydx4= Ayd3ydx3+Bdydxd2ydx2\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =\,& y^2 - x\\[4mm] \implies \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)

Given dydx=y2−x\frac{\mathrm{d}y}{\mathrm{d}x} = y^2 - x. Differentiate with respect to xx using implicit differentiation and the product rule:

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 a second time:

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*}

Differentiate a third time:

d4ydx4= (2yd3ydx3+2dydxd2ydx2)+4dydxd2ydx2= 2yd3ydx3+6dydxd2ydx2\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} = &\,\left( 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} \right) + 4\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] = &\,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*}

This is in the required form with A=2A = 2 and B=6B = 6.

(b)

Evaluate yy and its derivatives at x=−1x = -1. We are given y=1y = 1.

y′(−1)= (1)2−(−1)=2y′′(−1)= 2(1)(2)−1=3y′′′(−1)= 2(1)(3)+2(2)2=6+8=14y(4)(−1)= 2(1)(14)+6(2)(3)=28+36=64\begin{align*} y'(-1) = &\,(1)^2 - (-1) = 2\\[4mm] y''(-1) = &\,2(1)(2) - 1 = 3\\[4mm] y'''(-1) = &\,2(1)(3) + 2(2)^2 = 6 + 8 = 14\\[4mm] y^{(4)}(-1) = &\,2(1)(14) + 6(2)(3) = 28 + 36 = 64 \end{align*}

The Taylor series expansion in ascending powers of (x+1)(x+1) is given by:

y(x)≈ 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(x) \approx &\,y(-1) + y'(-1)(x+1) + \frac{y''(-1)}{2!}(x+1)^2 + \frac{y'''(-1)}{3!}(x+1)^3 + \frac{y^{(4)}(-1)}{4!}(x+1)^4\\[4mm] \approx &\,1 + 2(x+1) + \frac{3}{2}(x+1)^2 + \frac{14}{6}(x+1)^3 + \frac{64}{24}(x+1)^4\\[4mm] \approx &\,1 + 2(x+1) + \frac{3}{2}(x+1)^2 + \frac{7}{3}(x+1)^3 + \frac{8}{3}(x+1)^4 \end{align*}