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IAL 2022 Jan Q8

A Level / Edexcel / FP2

IAL 2022 Jan Paper · Question 8

题目

Problem

(a) Show that the transformation v=y−2xv = y - 2x transforms the differential equation

dydx+2yx(y−4x)=2−8x3(I)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + 2yx(y - 4x) = 2 - 8x^3 \qquad \text{(I)} \end{align*}

into the differential equation

dvdx=−2xv2(II)\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x} = -2xv^2 \qquad \text{(II)} \end{align*}
(4)

(b) Solve the differential equation (II) to determine vv as a function of xx

(4)

(c) Hence obtain the general solution of the differential equation (I).

(1)

(d) Sketch the solution curve that passes through the point (−1,−1)(-1, -1).

On your sketch show clearly the equation of any horizontal or vertical asymptotes.

You do not need to find the coordinates of any intercepts with the coordinate axes or the coordinates of any stationary points.

(5)

解答

(a)

解法一

思路

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由

v=y−2x\begin{align*} v=y-2x \end{align*}

得 y=v+2xy=v+2x，且

dydx=dvdx+2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}v}{\mathrm{d}x}+2. \end{align*}

把这些代入原方程后，利用

y−4x=v−2x\begin{align*} y-4x=v-2x \end{align*}

化简。

答题过程

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Given

v=y−2x,\begin{align*} v=y-2x, \end{align*}

we have

y=v+2x.\begin{align*} y=v+2x. \end{align*}

Differentiate:

dvdx= dydx−2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x} =&\,\frac{\mathrm{d}y}{\mathrm{d}x}-2. \end{align*}

So

dydx= dvdx+2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}v}{\mathrm{d}x}+2. \end{align*}

Substitute into (I):

dvdx+2+2x(v+2x)(v+2x−4x)= 2−8x3.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}+2 +2x(v+2x)\left(v+2x-4x\right) =&\,2-8x^3. \end{align*}

Since

v+2x−4x=v−2x,\begin{align*} v+2x-4x=v-2x, \end{align*}

we get

dvdx+2+2x(v+2x)(v−2x)= 2−8x3.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}+2 +2x(v+2x)(v-2x) =&\,2-8x^3. \end{align*}

Use difference of squares:

(v+2x)(v−2x)=v2−4x2.\begin{align*} (v+2x)(v-2x)=v^2-4x^2. \end{align*}

Therefore

dvdx+2+2x(v2−4x2)= 2−8x3.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}+2 +2x(v^2-4x^2) =&\,2-8x^3. \end{align*}

Expand:

dvdx+2+2xv2−8x3= 2−8x3.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}+2 +2xv^2-8x^3 =&\,2-8x^3. \end{align*}

Cancel 2−8x32-8x^3 from both sides:

dvdx+2xv2=0.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}+2xv^2=0. \end{align*}

Hence

dvdx=−2xv2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}=-2xv^2. \end{align*}

(b)

解法一

思路

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(II) 是 separable differential equation：

dvdx=−2xv2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}=-2xv^2. \end{align*}

把 vv 放一边、xx 放一边，然后积分。

答题过程

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From (II),

dvdx=−2xv2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}=-2xv^2. \end{align*}

Separate the variables:

1v2dvdx=−2x.\begin{align*} \frac{1}{v^2}\frac{\mathrm{d}v}{\mathrm{d}x}=-2x. \end{align*}

So

∫v−2 dv= ∫−2x dx.\begin{align*} \int v^{-2}\,\mathrm{d}v =&\,\int -2x\,\mathrm{d}x. \end{align*}

Integrate:

−v−1= −x2+C.\begin{align*} -v^{-1} =&\,-x^2+C. \end{align*}

Multiply by −1-1:

1v=x2+C.\begin{align*} \frac1v=x^2+C. \end{align*}

Therefore

v=1x2+C.\begin{align*} v=\frac{1}{x^2+C}. \end{align*}

(c)

解法一

思路

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由 v=y−2xv=y-2x 得 y=v+2xy=v+2x。把 (b) 的 vv 代入即可。

答题过程

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Since

v=y−2x,\begin{align*} v=y-2x, \end{align*}

we have

y=v+2x.\begin{align*} y=v+2x. \end{align*}

Using

v=1x2+C,\begin{align*} v=\frac{1}{x^2+C}, \end{align*}

the general solution of (I) is

y=2x+1x2+C.\begin{align*} y=2x+\frac{1}{x^2+C}. \end{align*}

(d)

解法一

思路

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先用点 (−1,−1)(-1,-1) 求出常数 CC。得到具体曲线后，再判断渐近线。若

y=2x+1x2,\begin{align*} y=2x+\frac1{x^2}, \end{align*}

则 x=0x=0 是 vertical asymptote；当 ∣x∣|x| 很大时，1x2→0\dfrac1{x^2}\to0，所以曲线接近斜线 y=2xy=2x。题目只要求 horizontal 或 vertical asymptotes，因此需要标出 x=0x=0。

答题过程

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From part (c),

y=2x+1x2+C.\begin{align*} y=2x+\frac{1}{x^2+C}. \end{align*}

The curve passes through (−1,−1)(-1,-1), so

−1= 2(−1)+1(−1)2+C= −2+11+C.\begin{align*} -1 =&\,2(-1)+\frac{1}{(-1)^2+C}\\[2mm] =&\,-2+\frac{1}{1+C}. \end{align*}

Thus

1=11+C.\begin{align*} 1=\frac{1}{1+C}. \end{align*}

So

C=0.\begin{align*} C=0. \end{align*}

Hence the required solution curve is

y=2x+1x2.\begin{align*} y=2x+\frac1{x^2}. \end{align*}

The vertical asymptote is

x=0.\begin{align*} x=0. \end{align*}

There is no horizontal asymptote. As x→±∞x\to\pm\infty,

y−2x=1x2→0,\begin{align*} y-2x=\frac1{x^2}\to0, \end{align*}

so the curve approaches the oblique asymptote y=2xy=2x.

For the sketch:

  • one branch lies for x<0x<0, passes through (−1,−1)(-1,-1), and tends to +∞+\infty as x→0−x\to0^-;
  • the other branch lies for x>0x>0, tends to +∞+\infty as x→0+x\to0^+, then has a minimum in the first quadrant before rising again;
  • mark the vertical asymptote x=0x=0.