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FP2 2023 June Q7

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 7

题目

Problem

7.

(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 = f(x)

(6)
题目中文翻译

(a) 证明代换 z=y−2z=y^{-2} 将微分方程

xdydx+y+4x2y3ln⁡x= 0,x>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*}

变为微分方程

dzdx−2zx= 8xln⁡x,x>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) 通过求解微分方程 (II)，求微分方程 (I) 的通解，将答案写成 y2=f(x)y^2=f(x) 的形式。

(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\bigg(4(\ln x)^2+C\bigg) \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\bigg(4(\ln x)^2+C\bigg)\\[4mm] y^2 =&\, \frac{1}{x^2\bigg(4(\ln x)^2+C\bigg)} \end{align*}

Thus the general solution is

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

解法二

思路

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仍先用积分因子将方程化为乘积的导数。对于积分 ∫ln⁡xx dx\int\frac{\ln x}{x}\,\mathrm{d}x，选 u=ln⁡xu=\ln x、dv=1x dx\mathrm{d}v=\frac1x\,\mathrm{d}x 作分部积分；右边会出现同一个积分，移项后便能求出。

答题过程

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For equation (II), the integrating factor is

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

Multiplying equation (II) by x−2x^{-2},

ddx(x−2z)= 8ln⁡xx.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(x^{-2}z) =&\,\frac{8\ln x}{x}. \end{align*}

To integrate, put

J= ∫ln⁡xx dx.\begin{align*} J=&\,\int\frac{\ln x}{x}\,\mathrm{d}x. \end{align*}

Use integration by parts with

u= ln⁡x,dv= 1x dx,du= 1x dx,v= ln⁡x.\begin{align*} u=&\,\ln x,&\mathrm{d}v=&\,\frac1x\,\mathrm{d}x,\\[2mm] \mathrm{d}u=&\,\frac1x\,\mathrm{d}x,&v=&\,\ln x. \end{align*}

Then

J= (ln⁡x)2−∫ln⁡xx dx+C0= (ln⁡x)2−J+C0,2J= (ln⁡x)2+C0,J= 12(ln⁡x)2+C1.\begin{align*} J=&\,(\ln x)^2-\int\frac{\ln x}{x}\,\mathrm{d}x+C_0\\[2mm] =&\,(\ln x)^2-J+C_0,\\[2mm] 2J=&\,(\ln x)^2+C_0,\\[2mm] J=&\,\frac12(\ln x)^2+C_1. \end{align*}

Therefore, combining the arbitrary constants,

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

Since z=y−2z=y^{-2},

y2=1x2(4(ln⁡x)2+C).\begin{align*} \boxed{y^2=\frac1{x^2\big(4(\ln x)^2+C\big)}}. \end{align*}