CalcGospel 國際數學圖譜
返回

IAL 2024 June Q4

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 4

题目

Problem

(a) Show that the substitution y2=1ty^2 = \frac{1}{t} transforms the differential equation

dydx+y=xy3(I)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + y = xy^3 \qquad \text{(I)} \end{align*}

into the differential equation

dtdx−2t=−2x(II)\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x} - 2t = -2x \qquad \text{(II)} \end{align*}
(3)

(b) Solve differential equation (II) and determine y2y^2 in terms of xx.

(6)

解答

(a)

解法一

思路

展开

由 y2=1ty^2=\frac1t 可写成 t=y−2t=y^{-2} 或 y=t−1/2y=t^{-1/2}。这里用 y=t−1/2y=t^{-1/2} 求导比较直接，再代入原方程。

答题过程

展开

Since

y2=1t,\begin{align*} y^2=\frac1t, \end{align*}

we can write

y=t−12\begin{align*} y=t^{-\frac12} \end{align*}

Differentiate:

dydx= −12t−32dtdx\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, -\frac12t^{-\frac32}\frac{\mathrm{d}t}{\mathrm{d}x} \end{align*}

Substitute into (I):

−12t−32dtdx+t−12= xt−32\begin{align*} -\frac12t^{-\frac32}\frac{\mathrm{d}t}{\mathrm{d}x} +t^{-\frac12} =&\, x t^{-\frac32} \end{align*}

Multiply by −2t32-2t^{\frac32}:

dtdx−2t= −2x\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x}-2t =&\, -2x \end{align*}

This is (II).

(b)

解法一

思路

展开

(II) 是一阶线性微分方程。积分因子是 e−2x\mathrm{e}^{-2x}。右边积分 ∫−2xe−2x dx\int -2x\mathrm{e}^{-2x}\,\mathrm{d}x 用分部积分。

答题过程

展开

The differential equation is

dtdx−2t=−2x\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x}-2t=-2x \end{align*}

The integrating factor is

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

Multiplying by e−2x\mathrm{e}^{-2x} gives

ddx(te−2x)= −2xe−2x\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(t\mathrm{e}^{-2x}\right) =&\, -2x\mathrm{e}^{-2x} \end{align*}

Integrate:

te−2x= ∫−2xe−2x dx\begin{align*} t\mathrm{e}^{-2x} =&\, \int -2x\mathrm{e}^{-2x}\,\mathrm{d}x \end{align*}

Using integration by parts with u=xu=x and dv=−2e−2x dx\mathrm{d}v=-2\mathrm{e}^{-2x}\,\mathrm{d}x,

∫−2xe−2x dx= xe−2x−∫e−2x dx= xe−2x+12e−2x+C\begin{align*} \int -2x\mathrm{e}^{-2x}\,\mathrm{d}x =&\, x\mathrm{e}^{-2x}-\int \mathrm{e}^{-2x}\,\mathrm{d}x\\[4mm] =&\, x\mathrm{e}^{-2x}+\frac12\mathrm{e}^{-2x}+C \end{align*}

So

te−2x= xe−2x+12e−2x+Ct= x+12+Ce2x\begin{align*} t\mathrm{e}^{-2x} =&\, x\mathrm{e}^{-2x}+\frac12\mathrm{e}^{-2x}+C\\[4mm] t =&\, x+\frac12+C\mathrm{e}^{2x} \end{align*}

Since y2=1ty^2=\frac1t,

y2=1x+12+Ce2x\begin{align*} \boxed{ y^2=\frac{1}{x+\frac12+C\mathrm{e}^{2x}} } \end{align*}

解法二

思路

展开

这条方程也可以当作常系数线性方程来解。齐次部分给出 e2x\mathrm{e}^{2x}，右边是一次式，所以特解可以设成 ax+bax+b。

答题过程

展开

Solve

dtdx−2t=−2x\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x}-2t=-2x \end{align*}

The complementary function satisfies

dtdx−2t=0\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x}-2t=0 \end{align*}

so

tc=Ae2x\begin{align*} t_{\mathrm{c}}=A\mathrm{e}^{2x} \end{align*}

For a particular integral, try

tp=ax+b\begin{align*} t_{\mathrm{p}}=ax+b \end{align*}

Then

dtpdx−2tp= a−2(ax+b)= −2ax+(a−2b)\begin{align*} \frac{\mathrm{d}t_{\mathrm{p}}}{\mathrm{d}x}-2t_{\mathrm{p}} =&\, a-2(ax+b)\\[2mm] =&\, -2ax+(a-2b) \end{align*}

Compare with −2x-2x:

−2a= −2a−2b= 0\begin{align*} -2a=&\,-2\\[2mm] a-2b=&\,0 \end{align*}

Hence

a= 1,b= 12\begin{align*} a=&\,1,\\[2mm] b=&\,\frac12 \end{align*}

Therefore

t= Ae2x+x+12\begin{align*} t =&\, A\mathrm{e}^{2x}+x+\frac12 \end{align*}

Since y2=1ty^2=\frac1t,

y2=1x+12+Ae2x\begin{align*} \boxed{ y^2=\frac{1}{x+\frac12+A\mathrm{e}^{2x}} } \end{align*}