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IAL 2023 Jan Q3

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 3

题目

Problem

(a) Show that the transformation y=1zy = \frac{1}{z} 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)} \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)} \end{align*}
(3)

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

(4)

(c) Hence determine the particular solution of differential equation (I) for which y=−38y=-\frac{3}{8} at x=3x=3

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

(2)

解答

(a)

解法一

思路

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由 y=1z=z−1y=\frac{1}{z}=z^{-1}，先求 dydx\frac{\mathrm{d}y}{\mathrm{d}x}，再代入原方程。最后把整式乘开，整理成关于 zz 的一阶线性方程。

答题过程

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Since

y=1z=z−1,\begin{align*} y=\frac{1}{z}=z^{-1}, \end{align*}

we have

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

Substitute into 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\\[2mm] -\frac{x^2}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x} +\frac{x}{z} =&\,\frac{2}{z^2}. \end{align*}

Multiply by −z2x2-\dfrac{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 is equation (II).

(b)

解法一

思路

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这是关于 zz 的一阶线性微分方程。积分因子为

e∫−1x dx=1x.\begin{align*} \mathrm{e}^{\int -\frac{1}{x}\,\mathrm{d}x} =\frac{1}{x}. \end{align*}

乘上积分因子后，左边变成 ddx(zx)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x}\right)。

答题过程

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Equation (II) is

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

The integrating factor is

e∫−1x dx= e−ln⁡x= 1x.\begin{align*} \mathrm{e}^{\int -\frac{1}{x}\,\mathrm{d}x} =&\,\mathrm{e}^{-\ln x}\\[2mm] =&\,\frac{1}{x}. \end{align*}

Multiplying by 1x\frac{1}{x},

1xdzdx−zx2= −2x3ddx(zx)= −2x−3.\begin{align*} \frac{1}{x}\frac{\mathrm{d}z}{\mathrm{d}x} -\frac{z}{x^2} =&\,-\frac{2}{x^3}\\[2mm] \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x}\right) =&\,-2x^{-3}. \end{align*}

Integrate:

zx= ∫−2x−3 dx= x−2+C.\begin{align*} \frac{z}{x} =&\,\int -2x^{-3}\,\mathrm{d}x\\[2mm] =&\,x^{-2}+C. \end{align*}

Therefore

z=1x+Cx.\begin{align*} \boxed{z=\frac{1}{x}+Cx}. \end{align*}

(c)

解法一

思路

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由 y=1zy=\frac{1}{z}，所以 z=1yz=\frac{1}{y}。把 x=3x=3、y=−38y=-\frac38 代入 (b) 的结果求常数，再反解出 yy。

答题过程

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Since

z=1y,\begin{align*} z=\frac{1}{y}, \end{align*}

and y=−38y=-\frac{3}{8} when x=3x=3,

z=−83.\begin{align*} z=-\frac{8}{3}. \end{align*}

Using

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

we get

−83= 13+3C−3= 3CC= −1.\begin{align*} -\frac{8}{3} =&\,\frac{1}{3}+3C\\[2mm] -3=&\,3C\\[2mm] C=&\,-1. \end{align*}

Thus

1y= 1x−x= 1−x2x.\begin{align*} \frac{1}{y} =&\,\frac{1}{x}-x\\[2mm] =&\,\frac{1-x^2}{x}. \end{align*}

Therefore

y=x1−x2.\begin{align*} \boxed{y=\frac{x}{1-x^2}}. \end{align*}