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FP2 2026 Jan Q3

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 3

题目

Problem

3.

(a) Show that the transformation u=y−4u=y^{-4} transforms the differential equation

4dydx=y−2xy5(I)\begin{align*} 4\frac{\mathrm{d}y}{\mathrm{d}x}=y-2xy^5\qquad\text{(I)} \end{align*}

into the differential equation

dudx+u=2x(II)\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}+u=2x\qquad\text{(II)} \end{align*}
(3)

(b) Solve differential equation (II) to determine a general solution for uu in terms of xx.

(5)

(c) Hence determine the particular solution of differential equation (I) for which y=1y=1 at x=0x=0, giving your answer in the form y4=f(x)y^4=f(x).

(2)
题目中文翻译

(a) 证明代换 u=y−4u=y^{-4} 可将微分方程

4dydx=y−2xy5(I)\begin{align*} 4\frac{\mathrm{d}y}{\mathrm{d}x}=y-2xy^5\qquad\text{(I)} \end{align*}

化为微分方程

dudx+u=2x(II)\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}+u=2x\qquad\text{(II)} \end{align*}
(3)

(b) 解微分方程 (II)，求出以 xx 表示的 uu 的通解。

(5)

(c) 由此求微分方程 (I) 满足 x=0x=0 时 y=1y=1 的特解，将答案写成 y4=f(x)y^4=f(x) 的形式。

(2)

解答

(a)

解法一

思路

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关键是对 u=y−4u=y^{-4} 作链式求导。由于原方程中有 y5y^5，把求导结果乘回去后会自然出现 y−4y^{-4}，也就是 uu。

答题过程

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Since

u=y−4\begin{align*} u=y^{-4} \end{align*}

differentiating with respect to xx gives

dudx= −4y−5dydx\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} =&\, -4y^{-5}\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

From

4dydx=y−2xy5\begin{align*} 4\frac{\mathrm{d}y}{\mathrm{d}x}=y-2xy^5 \end{align*}

we have

dydx= 14(y−2xy5)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac14(y-2xy^5) \end{align*}

Therefore

dudx= −4y−5⋅14(y−2xy5)= −y−4+2x= −u+2x\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} =&\, -4y^{-5}\cdot\frac14(y-2xy^5)\\[3mm] =&\, -y^{-4}+2x\\[3mm] =&\, -u+2x \end{align*}

Hence

dudx+u=2x\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}+u=2x \end{align*}

(b)

解法一

思路

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这是标准一阶线性微分方程。积分因子是 e∫1 dx=ex\mathrm{e}^{\int 1\,\mathrm{d}x}=\mathrm{e}^x，乘上后左边变成乘积求导。

答题过程

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The differential equation is

dudx+u=2x\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}+u=2x \end{align*}

The integrating factor is

e∫1 dx= ex\begin{align*} \mathrm{e}^{\int 1\,\mathrm{d}x} =&\, \mathrm{e}^x \end{align*}

Multiplying by ex\mathrm{e}^x,

exdudx+exu= 2xex\begin{align*} \mathrm{e}^x\frac{\mathrm{d}u}{\mathrm{d}x} +\mathrm{e}^xu =&\, 2x\mathrm{e}^x \end{align*}

So

ddx(uex)= 2xex\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \Big(u\mathrm{e}^x\Big) =&\, 2x\mathrm{e}^x \end{align*}

Integrating,

uex= ∫2xex dx= 2xex−∫2ex dx= 2xex−2ex+C\begin{align*} u\mathrm{e}^x =&\, \int 2x\mathrm{e}^x\,\mathrm{d}x\\[3mm] =&\, 2x\mathrm{e}^x-\int 2\mathrm{e}^x\,\mathrm{d}x\\[3mm] =&\, 2x\mathrm{e}^x-2\mathrm{e}^x+C \end{align*}

Therefore

u=2x−2+Ce−x\begin{align*} u=2x-2+C\mathrm{e}^{-x} \end{align*}

解法二

思路

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先解对应的齐次方程，再试一个一次函数作为特解。因为右边是 2x2x，设特解为 ax+bax+b 即可，通过比较系数求出 aa、bb。

答题过程

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For the homogeneous equation,

ducdx+uc=0\begin{align*} \frac{\mathrm{d}u_{\mathrm{c}}}{\mathrm{d}x}+u_{\mathrm{c}}=0 \end{align*}

Separation and integration give

∫1uc duc= −∫1 dxln⁡∣uc∣= −x+K\begin{align*} \int\frac1{u_{\mathrm{c}}}\,\mathrm{d}u_{\mathrm{c}} =&\,-\int1\,\mathrm{d}x\\[3mm] \ln|u_{\mathrm{c}}| =&\,-x+K \end{align*}

Thus the complementary function is uc=Ce−xu_{\mathrm{c}}=C\mathrm{e}^{-x}, including the zero solution when C=0C=0.

For a particular integral, try up=ax+bu_{\mathrm{p}}=ax+b. Substituting into (II),

a+(ax+b)= 2xax+(a+b)= 2x\begin{align*} a+(ax+b) =&\,2x\\[3mm] ax+(a+b) =&\,2x \end{align*}

Comparing coefficients gives

a= 2,a+b= 0⟹b=−2\begin{align*} a=&\,2,\\[2mm] a+b=&\,0\quad\Longrightarrow\quad b=-2 \end{align*}

Therefore the general solution is

u=uc+up=2x−2+Ce−x\begin{align*} u=u_{\mathrm{c}}+u_{\mathrm{p}}=2x-2+C\mathrm{e}^{-x} \end{align*}

(c)

解法一

思路

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因为 u=y−4u=y^{-4}，条件 y=1y=1 给出 u=1u=1。先求常数，再把 u=y−4u=y^{-4} 转回 y4y^4。

答题过程

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When x=0x=0, y=1y=1, so

u=y−4=1\begin{align*} u=y^{-4}=1 \end{align*}

Using

u=2x−2+Ce−x\begin{align*} u=2x-2+C\mathrm{e}^{-x} \end{align*}

gives

1= 0−2+CC= 3\begin{align*} 1 =&\, 0-2+C\\[3mm] C =&\, 3 \end{align*}

Thus

y−4=2x−2+3e−x\begin{align*} y^{-4}=2x-2+3\mathrm{e}^{-x} \end{align*}

Therefore

y4=12x−2+3e−x\begin{align*} y^4=\frac{1}{2x-2+3\mathrm{e}^{-x}} \end{align*}