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IAL 2023 June Q7

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 7

题目

Problem

(a) Show that the substitution z=y−2z = y^{-2} transforms the differential equation

xdydx+y+4x2y3ln⁡x=0x>0(I)\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} + y + 4x^2y^3\ln x ={}& 0 \qquad x > 0 \qquad \text{(I)} \end{align*}

into the differential equation

dzdx−2zx=8xln⁡xx>0(II)\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{2z}{x} ={}& 8x\ln x \qquad x > 0 \qquad \text{(II)} \end{align*}
(5)

(b) By solving differential equation (II), determine the general solution of differential equation (I), giving your answer in the form y2=f(x)y^2 = \mathrm{f}(x)

(6)

解答

(a)

解法一

思路

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代换 z=y−2z=y^{-2} 的关键是把 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 改写成含 dzdx\frac{\mathrm{d}z}{\mathrm{d}x} 的形式。原方程里有 y3y^3，所以把方程除以 y3y^3 后会自然出现 y−2=zy^{-2}=z 和 y−3dydxy^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}。

答题过程

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Given

z=y−2\begin{align*} z=y^{-2} \end{align*}

differentiate with respect to xx:

dzdx=−2y−3dydx\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} ={}& -2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

So

y−3dydx=−12dzdx\begin{align*} y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} ={}& -\frac12\frac{\mathrm{d}z}{\mathrm{d}x} \end{align*}

Starting from differential equation (I),

xdydx+y+4x2y3ln⁡x=0\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} +y+4x^2y^3\ln x ={}&0 \end{align*}

divide by y3y^3:

xy−3dydx+y−2+4x2ln⁡x=0\begin{align*} xy^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} +y^{-2} +4x^2\ln x ={}&0 \end{align*}

Substitute y−2=zy^{-2}=z and y−3dydx=−12dzdxy^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac12\frac{\mathrm{d}z}{\mathrm{d}x}:

−x2dzdx+z+4x2ln⁡x=0−x2dzdx+z=−4x2ln⁡x\begin{align*} -\frac{x}{2}\frac{\mathrm{d}z}{\mathrm{d}x} +z+4x^2\ln x ={}&0\\[4mm] -\frac{x}{2}\frac{\mathrm{d}z}{\mathrm{d}x} +z ={}& -4x^2\ln x \end{align*}

Multiply by −2x-\frac{2}{x}, using x>0x>0:

dzdx−2zx=8xln⁡x\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} -\frac{2z}{x} ={}& 8x\ln x \end{align*}

This is the required differential equation (II).

(b)

解法一

思路

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(II) 是一阶线性微分方程。先找 integrating factor，然后把左边写成一个乘积的导数。积分 ∫x−1ln⁡x dx\int x^{-1}\ln x\,\mathrm{d}x 时，可以令 u=ln⁡xu=\ln x。

答题过程

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The differential equation is

dzdx−2xz=8xln⁡x\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-\frac{2}{x}z=8x\ln x \end{align*}

The integrating factor is

I.F.=e∫−2x dx=e−2ln⁡x=x−2\begin{align*} \mathrm{I.F.} ={}& \mathrm{e}^{\int -\frac{2}{x}\,\mathrm{d}x}\\[4mm] ={}& \mathrm{e}^{-2\ln x}\\[4mm] ={}& x^{-2} \end{align*}

Multiply the equation by x−2x^{-2}:

x−2dzdx−2x−3z=8x−1ln⁡xddx(x−2z)=8x−1ln⁡x\begin{align*} x^{-2}\frac{\mathrm{d}z}{\mathrm{d}x} -2x^{-3}z ={}& 8x^{-1}\ln x\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}(x^{-2}z) ={}& 8x^{-1}\ln x \end{align*}

Integrate both sides:

x−2z=∫8x−1ln⁡x dx\begin{align*} x^{-2}z ={}& \int 8x^{-1}\ln x\,\mathrm{d}x \end{align*}

Let u=ln⁡xu=\ln x, so du=x−1 dx\mathrm{d}u=x^{-1}\,\mathrm{d}x. Then

∫8x−1ln⁡x dx=∫8u du=4u2+C=4(ln⁡x)2+C\begin{align*} \int 8x^{-1}\ln x\,\mathrm{d}x ={}& \int 8u\,\mathrm{d}u\\[4mm] ={}& 4u^2+C\\[4mm] ={}& 4(\ln x)^2+C \end{align*}

Therefore

x−2z=4(ln⁡x)2+Cz=x2(4(ln⁡x)2+C)\begin{align*} x^{-2}z ={}& 4(\ln x)^2+C\\[4mm] z ={}& x^2\left(4(\ln x)^2+C\right) \end{align*}

Since z=y−2z=y^{-2},

y−2=x2(4(ln⁡x)2+C)y2=1x2(4(ln⁡x)2+C)\begin{align*} y^{-2} ={}& x^2\left(4(\ln x)^2+C\right)\\[4mm] y^2 ={}& \frac{1}{x^2\left(4(\ln x)^2+C\right)} \end{align*}

Thus the general solution is

y2=1x2(4(ln⁡x)2+C)\begin{align*} \boxed{ y^2=\frac{1}{x^2\left(4(\ln x)^2+C\right)} } \end{align*}