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IAL 2023 Jan Q3

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 3

(a) Show that the transformation

y= 1z\begin{align*} y =\,& \frac{1}{z}\\[2mm] \end{align*}

transforms the differential equation

x2dydx+xy= 2y2(I)\begin{align*} x^2\frac{\mathrm{d}y}{\mathrm{d}x} + xy =\,& 2y^2 \qquad \text{(I)}\\[2mm] \end{align*}

into the differential equation

dzdx−zx= −2x2(II)\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{z}{x} =\,& -\frac{2}{x^2} \qquad \text{(II)}\\[2mm] \end{align*}
(4)

(b) Solve differential equation (II) to determine zz in terms of xx .

(3)

(c) Hence determine the particular solution of differential equation (I) for which at x=3x = 3

y= −38\begin{align*} y =\,& -\frac{3}{8}\\[2mm] \end{align*}

Give your answer in the form y=f(x)y = \mathrm{f}(x) .

(2)
解答

(a)

Given the transformation y=1z=z−1y = \frac{1}{z} = z^{-1}, we differentiate with respect to xx:

dydx= −z−2dzdx=−1z2dzdx\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,-z^{-2}\frac{\mathrm{d}z}{\mathrm{d}x} = -\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x} \end{align*}

Substitute yy and dydx\frac{\mathrm{d}y}{\mathrm{d}x} into differential equation (I):

x2(−1z2dzdx)+x(1z)= 2(1z)2−x2z2dzdx+xz= 2z2\begin{align*} x^2\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right) + x\left(\frac{1}{z}\right) = &\,2\left(\frac{1}{z}\right)^2\\[4mm] -\frac{x^2}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x} + \frac{x}{z} = &\,\frac{2}{z^2} \end{align*}

Multiply the entire equation by −z2x2-\frac{z^2}{x^2}:

dzdx−zx= −2x2\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{z}{x} = &\,-\frac{2}{x^2} \end{align*}

This matches differential equation (II).

(b)

Differential equation (II) is a first-order linear ODE. The integrating factor (IF) is:

IF= e∫−1x dx= e−ln⁡x= 1x\begin{align*} \text{IF} = &\,\mathrm{e}^{\int -\frac{1}{x}\,\mathrm{d}x}\\[4mm] = &\,\mathrm{e}^{-\ln x}\\[4mm] = &\,\frac{1}{x} \end{align*}

Multiply equation (II) by the integrating factor:

1xdzdx−1x2z= −2x3ddx(zx)= −2x3\begin{align*} \frac{1}{x}\frac{\mathrm{d}z}{\mathrm{d}x} - \frac{1}{x^2}z = &\,-\frac{2}{x^3}\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x}\right) = &\,-\frac{2}{x^3} \end{align*}

Integrate both sides with respect to xx:

zx= ∫−2x−3 dxzx= x−2+czx= 1x2+c\begin{align*} \frac{z}{x} = &\,\int -2x^{-3}\,\mathrm{d}x\\[4mm] \frac{z}{x} = &\,x^{-2} + c\\[4mm] \frac{z}{x} = &\,\frac{1}{x^2} + c \end{align*}

Multiply by xx to find zz:

z= 1x+cx\begin{align*} z = &\,\frac{1}{x} + cx \end{align*}

(c)

Reverse the substitution y=1zy = \frac{1}{z}, so z=1yz = \frac{1}{y}:

1y= 1x+cx\begin{align*} \frac{1}{y} = &\,\frac{1}{x} + cx \end{align*}

Use the condition y=−38y = -\frac{3}{8} when x=3x = 3:

−83= 13+3c−93= 3c−3= 3cc= −1\begin{align*} -\frac{8}{3} = &\,\frac{1}{3} + 3c\\[4mm] -\frac{9}{3} = &\,3c\\[4mm] -3 = &\,3c\\[4mm] c = &\,-1 \end{align*}

Substitute c=−1c = -1 back into the equation:

1y= 1x−x1y= 1−x2xy= x1−x2\begin{align*} \frac{1}{y} = &\,\frac{1}{x} - x\\[4mm] \frac{1}{y} = &\,\frac{1 - x^2}{x}\\[4mm] y = &\,\frac{x}{1 - x^2} \end{align*}