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IAL 2024 June Q8

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 8

题目

Problem

(a) Given that t=ln⁡xt = \ln x , where x>0x > 0 , show that

d2ydx2= e−2t(d2ydt2−dydt)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \mathrm{e}^{-2t} \left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t}\right) \end{align*}
(3)

(b) Hence show that the transformation t=ln⁡xt = \ln x , where x>0x > 0 , transforms the differential equation

x2d2ydx2−2y= 1+4ln⁡x−2(ln⁡x)2(I)\begin{align*} x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2y =&\, 1 + 4\ln x - 2(\ln x)^2 \qquad \text{(I)} \end{align*}

into the differential equation

d2ydt2−dydt−2y= 1+4t−2t2(II)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t} - 2y =&\, 1 + 4t - 2t^2 \qquad \text{(II)} \end{align*}
(1)

(c) Solve differential equation (II) to determine yy in terms of tt.

(5)

(d) Hence determine the general solution of differential equation (I).

(1)

解答

(a)

解法一

思路

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t=ln⁡xt=\ln x 给出 dtdx=1x=e−t\frac{\mathrm{d}t}{\mathrm{d}x}=\frac1x=\mathrm{e}^{-t}。先求 dydx\frac{\mathrm{d}y}{\mathrm{d}x}，再对它关于 xx 求导。第二次求导时要把 ddx\frac{\mathrm{d}}{\mathrm{d}x} 看成 dtdxddt\frac{\mathrm{d}t}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}t}。

答题过程

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Since t=ln⁡xt=\ln x, x=etx=\mathrm{e}^t and

dtdx= 1x=e−t\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x} =&\, \frac1x=\mathrm{e}^{-t} \end{align*}

By the chain rule,

dydx= dydtdtdx= e−tdydt\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac{\mathrm{d}y}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}x}\\[2mm] =&\, \mathrm{e}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t} \end{align*}

Differentiate again:

d2ydx2= ddx(e−tdydt)= dtdxddt(e−tdydt)= e−t(−e−tdydt+e−td2ydt2)= e−2t(d2ydt2−dydt)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{\mathrm{d}}{\mathrm{d}x} \left( \mathrm{e}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t} \right)\\[4mm] =&\, \frac{\mathrm{d}t}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathrm{e}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t} \right)\\[4mm] =&\, \mathrm{e}^{-t} \left( -\mathrm{e}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t} +\mathrm{e}^{-t}\frac{\mathrm{d}^2y}{\mathrm{d}t^2} \right)\\[4mm] =&\, \mathrm{e}^{-2t} \left( \frac{\mathrm{d}^2y}{\mathrm{d}t^2} -\frac{\mathrm{d}y}{\mathrm{d}t} \right) \end{align*}

(b)

解法一

思路

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因为 x=etx=\mathrm{e}^t，所以 x2e−2t=1x^2\mathrm{e}^{-2t}=1。右边的 ln⁡x\ln x 直接换成 tt。

答题过程

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Substitute the result from part (a) into (I):

x2d2ydx2−2y= 1+4ln⁡x−2(ln⁡x)2e2te−2t(d2ydt2−dydt)−2y= 1+4t−2t2\begin{align*} x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-2y =&\, 1+4\ln x-2(\ln x)^2\\[4mm] \mathrm{e}^{2t}\mathrm{e}^{-2t} \left( \frac{\mathrm{d}^2y}{\mathrm{d}t^2} -\frac{\mathrm{d}y}{\mathrm{d}t} \right)-2y =&\, 1+4t-2t^2 \end{align*}

Therefore

d2ydt2−dydt−2y= 1+4t−2t2\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} -\frac{\mathrm{d}y}{\mathrm{d}t} -2y =&\, 1+4t-2t^2 \end{align*}

(c)

解法一

思路

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(II) 是二阶常系数非齐次微分方程。右边是二次多项式，所以 particular integral 设为 at2+bt+cat^2+bt+c。

答题过程

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The auxiliary equation is

m2−m−2=0\begin{align*} m^2-m-2=0 \end{align*}

Hence

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

So the complementary function is

yc=Ae2t+Be−t\begin{align*} y_c=A\mathrm{e}^{2t}+B\mathrm{e}^{-t} \end{align*}

For a particular integral, let

yp=at2+bt+c\begin{align*} y_p=at^2+bt+c \end{align*}

Then

yp′= 2at+byp′′= 2a\begin{align*} y_p'=&\,2at+b\\ y_p''=&\,2a \end{align*}

Substitute into (II):

2a−(2at+b)−2(at2+bt+c)= 1+4t−2t2−2at2+(−2a−2b)t+(2a−b−2c)= −2t2+4t+1\begin{align*} 2a-(2at+b)-2(at^2+bt+c) =&\, 1+4t-2t^2\\[4mm] -2at^2+(-2a-2b)t+(2a-b-2c) =&\, -2t^2+4t+1 \end{align*}

Equating coefficients:

−2a= −2⇒a=1−2a−2b= 4⇒b=−32a−b−2c= 1⇒c=2\begin{align*} -2a=&\,-2 &&\Rightarrow a=1\\ -2a-2b=&\,4 &&\Rightarrow b=-3\\ 2a-b-2c=&\,1 &&\Rightarrow c=2 \end{align*}

Therefore

y=Ae2t+Be−t+t2−3t+2\begin{align*} \boxed{ y=A\mathrm{e}^{2t}+B\mathrm{e}^{-t}+t^2-3t+2 } \end{align*}

(d)

解法一

思路

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把 t=ln⁡xt=\ln x 代回，且 e2t=x2\mathrm{e}^{2t}=x^2，e−t=x−1\mathrm{e}^{-t}=x^{-1}。

答题过程

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Since t=ln⁡xt=\ln x,

e2t=x2,e−t=x−1\begin{align*} \mathrm{e}^{2t}=x^2,\qquad \mathrm{e}^{-t}=x^{-1} \end{align*}

Therefore

y=Ax2+Bx−1+(ln⁡x)2−3ln⁡x+2\begin{align*} \boxed{ y=Ax^2+Bx^{-1}+(\ln x)^2-3\ln x+2 } \end{align*}