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IAL 2025 June A Q7

A Level / Edexcel / FP2

IAL 2025 June A Paper · Question 7

题目

Problem

(a) Show that the substitution x=eux=e^u, where uu is a function of xx, transforms the differential equation

2x2d2ydx2+3xdydx−y=27x2x>0(I)\begin{align*} 2x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +3x\frac{\mathrm{d}y}{\mathrm{d}x} -y=27x^2\qquad x>0\qquad \text{(I)} \end{align*}

into the differential equation

2d2ydu2+dydu−y=27e2u(II)\begin{align*} 2\frac{\mathrm{d}^2y}{\mathrm{d}u^2} +\frac{\mathrm{d}y}{\mathrm{d}u} -y=27e^{2u}\qquad \text{(II)} \end{align*}
(4)

(b) By solving differential equation (II), determine the general solution of differential equation (I).

Give the answer in the form y=f(x)y=f(x) where ff is a fully simplified function.

(4)

Given that when x=14x=\dfrac14, y=1116y=\dfrac{11}{16} and dydx=1\dfrac{\mathrm{d}y}{\mathrm{d}x}=1

(c) determine the value of yy when x=18x=\dfrac18, giving the answer in the form 164(p2+q)\dfrac{1}{64}(p\sqrt2+q) where pp and qq are integers.

(5)

解答

(a)

解法一

思路

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关键是把对 xx 的导数全部换成对 uu 的导数。由 x=eux=e^u 得 dudx=1x\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac1x，然后连续使用链式法则。

#### 答题过程
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Since x=eux=e^u,

dxdu=eu=x⇒dudx=1x\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}u}=e^u=x \quad \Rightarrow \quad \frac{\mathrm{d}u}{\mathrm{d}x}=\frac1x \end{align*}

So

dydx= dydududx= 1xdydu\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x}\\[4mm] =&\,\frac1x\frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

For the second derivative,

d2ydx2= ddx(1xdydu)= −1x2dydu+1xddx(dydu)= −1x2dydu+1x(d2ydu2dudx)= 1x2d2ydu2−1x2dydu\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac1x\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[4mm] =&\,-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} +\frac1x \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[4mm] =&\,-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} +\frac1x \left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} \frac{\mathrm{d}u}{\mathrm{d}x} \right)\\[4mm] =&\,\frac{1}{x^2}\frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

Substitute into (I):

2x2(1x2d2ydu2−1x2dydu)+3x(1xdydu)−y= 27x22d2ydu2−2dydu+3dydu−y= 27e2u2d2ydu2+dydu−y= 27e2u\begin{align*} 2x^2\left( \frac{1}{x^2}\frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} \right) +3x\left(\frac1x\frac{\mathrm{d}y}{\mathrm{d}u}\right)-y =&\,27x^2\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}u^2} -2\frac{\mathrm{d}y}{\mathrm{d}u} +3\frac{\mathrm{d}y}{\mathrm{d}u}-y =&\,27e^{2u}\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}u^2} +\frac{\mathrm{d}y}{\mathrm{d}u} -y =&\,27e^{2u} \end{align*}

This is equation (II).

解法二

思路

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利用算子关系。由 x=eux = \mathrm{e}^u 可得 dudx=1x\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{x}，即 xdydx=dydux \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u}。随后对 uu 再次求导，利用乘积求导法则与链式法则，可以更直观地导出 x2d2ydx2=d2ydu2−dydux^2 \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - \dfrac{\mathrm{d}y}{\mathrm{d}u} 的关系，最后代入原方程化简。

答题过程

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Since x=eux=\mathrm{e}^u, we have:

dudx=  1x\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} =&\,\, \frac1x \end{align*}

Using the chain rule:

dydx=  dydududx=  1xdydu\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, \frac{\mathrm{d}y}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\\[4mm] =&\,\, \frac1x \frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

which can be written as:

xdydx=  dydu\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, \frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

Differentiate both sides with respect to uu:

d2ydu2=  ddu(xdydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} =&\,\, \frac{\mathrm{d}}{\mathrm{d}u}\left( x\frac{\mathrm{d}y}{\mathrm{d}x} \right) \end{align*}

Using the Product Rule on the right-hand side, noting that x=eu  ⟹  dxdu=xx = \mathrm{e}^u \implies \dfrac{\mathrm{d}x}{\mathrm{d}u} = x, and ddu=xddx\dfrac{\mathrm{d}}{\mathrm{d}u} = x\dfrac{\mathrm{d}}{\mathrm{d}x}:

d2ydu2=  dxdudydx+xddu(dydx)=  xdydx+x(xddx(dydx))=  xdydx+x2d2ydx2\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} =&\,\, \frac{\mathrm{d}x}{\mathrm{d}u}\frac{\mathrm{d}y}{\mathrm{d}x} + x\frac{\mathrm{d}}{\mathrm{d}u}\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)\\[4mm] =&\,\, x\frac{\mathrm{d}y}{\mathrm{d}x} + x\left( x\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) \right)\\[4mm] =&\,\, x\frac{\mathrm{d}y}{\mathrm{d}x} + x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

Substitute xdydx=dydux\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u}:

d2ydu2=  dydu+x2d2ydx2\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} =&\,\, \frac{\mathrm{d}y}{\mathrm{d}u} + x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

Rearranging this gives:

x2d2ydx2=  d2ydu2−dydu\begin{align*} x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\, \frac{\mathrm{d}^2y}{\mathrm{d}u^2} - \frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

Substitute xdydx=dydux\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} and x2d2ydx2=d2ydu2−dydux^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - \dfrac{\mathrm{d}y}{\mathrm{d}u} into equation (I):

2x2d2ydx2+3xdydx−y=  27x22(d2ydu2−dydu)+3dydu−y=  27e2u2d2ydu2+dydu−y=  27e2u\begin{align*} 2x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} - y =&\,\, 27x^2\\[4mm] 2\left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} - \frac{\mathrm{d}y}{\mathrm{d}u} \right) + 3\frac{\mathrm{d}y}{\mathrm{d}u} - y =&\,\, 27\mathrm{e}^{2u}\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}u^2} + \frac{\mathrm{d}y}{\mathrm{d}u} - y =&\,\, 27\mathrm{e}^{2u} \end{align*}

(b)

解法一

思路

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(II) 是常系数二阶非齐次微分方程。先解辅助方程得到互补函数，再用 3e2u3e^{2u} 型的特解，最后把 eu=xe^u=x 代回去。

答题过程

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For the complementary function,

2m2+m−1= 0(2m−1)(m+1)= 0m= 12, −1\begin{align*} 2m^2+m-1 =&\,0\\[4mm] (2m-1)(m+1) =&\,0\\[4mm] m =&\,\frac12,\ -1 \end{align*}

So

yc=Ae12u+Be−u\begin{align*} y_{\mathrm{c}} =Ae^{\frac12u}+Be^{-u} \end{align*}

Try

yp=λe2u\begin{align*} y_{\mathrm{p}}=\lambda e^{2u} \end{align*}

Then

yp′= 2λe2u,yp′′= 4λe2u\begin{align*} y_{\mathrm{p}}'=&\,2\lambda e^{2u},\\[2mm] y_{\mathrm{p}}''=&\,4\lambda e^{2u} \end{align*}

Substitute into (II):

2(4λe2u)+2λe2u−λe2u= 27e2u9λ= 27λ= 3\begin{align*} 2(4\lambda e^{2u})+2\lambda e^{2u}-\lambda e^{2u} =&\,27e^{2u}\\[4mm] 9\lambda =&\,27\\[4mm] \lambda =&\,3 \end{align*}

Thus

y= Ae12u+Be−u+3e2u\begin{align*} y =&\,Ae^{\frac12u}+Be^{-u}+3e^{2u} \end{align*}

Since eu=xe^u=x,

y=Ax12+Bx−1+3x2\begin{align*} y=Ax^{\frac12}+Bx^{-1}+3x^2 \end{align*}

(c)

解法一

思路

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先用 x=14x=\frac14 的两个条件求 A,BA,B。注意导数条件是对 xx 的导数，所以要先把 (b) 的答案对 xx 微分。

答题过程

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From part (b),

y=Ax12+Bx−1+3x2\begin{align*} y=Ax^{\frac12}+Bx^{-1}+3x^2 \end{align*}

Differentiate:

dydx= 12Ax−12−Bx−2+6x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac12Ax^{-\frac12}-Bx^{-2}+6x \end{align*}

Use x=14, y=1116x=\frac14,\ y=\frac{11}{16}:

1116= 12A+4B+31612A+4B= 12\begin{align*} \frac{11}{16} =&\,\frac12A+4B+\frac{3}{16}\\[4mm] \frac12A+4B =&\,\frac12 \end{align*}

Use x=14, dydx=1x=\frac14,\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=1:

1= A−16B+32A−16B= −12\begin{align*} 1 =&\,A-16B+\frac32\\[4mm] A-16B =&\,-\frac12 \end{align*}

From the first equation,

A+8B=1\begin{align*} A+8B=1 \end{align*}

Solve

A+8B= 1,A−16B= −12\begin{align*} A+8B=&\,1,\\[2mm] A-16B=&\,-\frac12 \end{align*}

Subtracting gives

24B= 32B= 116\begin{align*} 24B =&\,\frac32\\[4mm] B =&\,\frac{1}{16} \end{align*}

Then

A= 12\begin{align*} A =&\,\frac12 \end{align*}

So

y= 12x12+116x−1+3x2\begin{align*} y =&\,\frac12x^{\frac12}+\frac{1}{16}x^{-1}+3x^2 \end{align*}

At x=18x=\frac18,

y= 12(18)12+116(18)−1+3(18)2= 142+12+364= 28+3564= 164(82+35)\begin{align*} y =&\,\frac12\left(\frac18\right)^{\frac12} +\frac{1}{16}\left(\frac18\right)^{-1} +3\left(\frac18\right)^2\\[4mm] =&\,\frac{1}{4\sqrt2}+\frac12+\frac{3}{64}\\[4mm] =&\,\frac{\sqrt2}{8}+\frac{35}{64}\\[4mm] =&\,\frac{1}{64}(8\sqrt2+35) \end{align*}