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IAL 2025 Jan Q1

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 1

题目

Problem

dydx−3ytan⁡x=sec⁡2x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} - 3y\tan x = \sec^2 x \end{align*}

(a) Show that an integrating factor for this differential equation is given by

p(x)=cos⁡3x\begin{align*} p(x)=\cos^3 x \end{align*}
(2)

Given that y=4y=4 when x=π4x=\frac{\pi}{4}

(b) determine the particular solution of the differential equation.

Give your answer in the form y=f(x)y=f(x).

(4)

解答

(a)

解法一

思路

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一阶线性方程 dydx+P(x)y=Q(x)\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x) 的积分因子是 e∫P(x) dx\mathrm{e}^{\int P(x)\,\mathrm{d}x}。这里 P(x)=−3tan⁡xP(x)=-3\tan x。

答题过程

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The differential equation is in the form

dydx+P(x)y=Q(x)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x) \end{align*}

where

P(x)=−3tan⁡x\begin{align*} P(x)=-3\tan x \end{align*}

Therefore an integrating factor is

I.F.=exp⁡(∫−3tan⁡x dx)=exp⁡(3ln⁡(cos⁡x))=cos⁡3x\begin{align*} \mathrm{I.F.} ={}& \exp\left(\int -3\tan x\,\mathrm{d}x\right)\\[3mm] ={}& \exp\left(3\ln(\cos x)\right)\\[3mm] ={}& \cos^3 x \end{align*}

So

p(x)=cos⁡3x\begin{align*} p(x)=\cos^3x \end{align*}

解法二

思路

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也可以直接验证。若乘上 cos⁡3x\cos^3x 后，左边正好是 ddx(ycos⁡3x)\frac{\mathrm{d}}{\mathrm{d}x}(y\cos^3x)，那么 cos⁡3x\cos^3x 就是积分因子。

答题过程

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Differentiate ycos⁡3xy\cos^3x:

ddx(ycos⁡3x)=cos⁡3xdydxy⋅3cos⁡2x(−sin⁡x)=cos⁡3xdydx−3ysin⁡xcos⁡2x=cos⁡3x(dydx−3ytan⁡x)\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(y\cos^3x\right) ={}& \cos^3x\frac{\mathrm{d}y}{\mathrm{d}x} y\cdot 3\cos^2x(-\sin x)\\[3mm] ={}& \cos^3x\frac{\mathrm{d}y}{\mathrm{d}x} -3y\sin x\cos^2x\\[3mm] ={}& \cos^3x\left( \frac{\mathrm{d}y}{\mathrm{d}x}-3y\tan x \right) \end{align*}

Thus multiplying the differential equation by cos⁡3x\cos^3x makes the left hand side an exact derivative. Hence

p(x)=cos⁡3x\begin{align*} p(x)=\cos^3x \end{align*}

(b)

解法一

思路

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用 (a) 的积分因子乘整条方程。左边变成乘积的导数，右边 sec⁡2xcos⁡3x\sec^2x\cos^3x 化简成 cos⁡x\cos x，所以积分很短。最后代入 x=π4,y=4x=\frac{\pi}{4},y=4 求常数。

答题过程

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Multiply the differential equation by cos⁡3x\cos^3x:

cos⁡3xdydx−3ycos⁡3xtan⁡x=sec⁡2xcos⁡3xddx(ycos⁡3x)=cos⁡x\begin{align*} \cos^3x\frac{\mathrm{d}y}{\mathrm{d}x} -3y\cos^3x\tan x ={}& \sec^2x\cos^3x\\[3mm] \frac{\mathrm{d}}{\mathrm{d}x}\left(y\cos^3x\right) ={}& \cos x \end{align*}

Integrate:

ycos⁡3x=∫cos⁡x dx=sin⁡x+C\begin{align*} y\cos^3x ={}& \int \cos x\,\mathrm{d}x\\[2mm] ={}& \sin x+C \end{align*}

Use y=4y=4 when x=π4x=\frac{\pi}{4}:

4cos⁡3π4=sin⁡π4+C4(22)3=22+C2=22+C\begin{align*} 4\cos^3\frac{\pi}{4} ={}& \sin\frac{\pi}{4}+C\\[3mm] 4\left(\frac{\sqrt2}{2}\right)^3 ={}& \frac{\sqrt2}{2}+C\\[3mm] \sqrt2 ={}& \frac{\sqrt2}{2}+C \end{align*}

Hence

C=22\begin{align*} C=\frac{\sqrt2}{2} \end{align*}

So

ycos⁡3x=sin⁡x+22y=sin⁡x+22cos⁡3x\begin{align*} y\cos^3x ={}& \sin x+\frac{\sqrt2}{2}\\[3mm] \boxed{ y=\frac{\sin x+\frac{\sqrt2}{2}}{\cos^3x} } \end{align*}

Equivalently,

y=tan⁡xsec⁡2x+22sec⁡3x\begin{align*} \boxed{ y=\tan x\sec^2x+\frac{\sqrt2}{2}\sec^3x } \end{align*}