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IAL 2024 Jan Q5

A Level / Edexcel / P4

IAL 2024 Jan Paper · Question 5

题目

Problem

(a) Find

∫x2cos⁡2x dx\int x^2\cos 2x\,dx
(4)

(b) Hence solve the differential equation

dydt=(tcos⁡ty)2\frac{dy}{dt}=\left(\frac{t\cos t}{y}\right)^2

giving your answer in the form yn=f(t)y^n=f(t) where nn is an integer.

(5)
题目中文翻译

(a) 求

∫x2cos⁡2x dx\int x^2\cos 2x\,dx

(b) 进而解微分方程

dydt=(tcos⁡ty)2\frac{dy}{dt}=\left(\frac{t\cos t}{y}\right)^2

并将答案写成 yn=f(t)y^n=f(t) 的形式，其中 nn 为整数。

解答

(a)

解法一

思路

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这是一个典型的两次分部积分。把 x2x^2 看成会越求导越简单的部分，把 cos⁡2x\cos 2x 积分。第一次分部积分后会出现 ∫xsin⁡2x dx\int x\sin 2x\,\mathrm{d}x，再对它做一次分部积分即可。

答题过程

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Using integration by parts,

∫x2cos⁡2x dx= x2⋅12sin⁡2x−∫2x⋅12sin⁡2x dx= 12x2sin⁡2x−∫xsin⁡2x dx.\begin{align*} \int x^2\cos 2x\,\mathrm{d}x =&\, x^2\cdot \frac12\sin 2x -\int 2x\cdot \frac12\sin 2x\,\mathrm{d}x \\[2mm] =&\, \frac12x^2\sin 2x -\int x\sin 2x\,\mathrm{d}x. \end{align*}

For the remaining integral, use integration by parts again:

∫xsin⁡2x dx= x⋅(−12cos⁡2x)−∫1⋅(−12cos⁡2x) dx= −12xcos⁡2x+14sin⁡2x.\begin{align*} \int x\sin 2x\,\mathrm{d}x =&\, x\cdot\left(-\frac12\cos 2x\right) -\int 1\cdot\left(-\frac12\cos 2x\right)\,\mathrm{d}x \\[2mm] =&\, -\frac12x\cos 2x+\frac14\sin 2x. \end{align*}

Therefore

∫x2cos⁡2x dx= 12x2sin⁡2x−(−12xcos⁡2x+14sin⁡2x)= 12x2sin⁡2x+12xcos⁡2x−14sin⁡2x+C.\begin{align*} \int x^2\cos 2x\,\mathrm{d}x =&\, \frac12x^2\sin 2x -\left(-\frac12x\cos 2x+\frac14\sin 2x\right) \\[2mm] =&\, \frac12x^2\sin 2x +\frac12x\cos 2x -\frac14\sin 2x+C. \end{align*}

解法二

思路

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也可以用表格法做分部积分。左边不断对 x2x^2 求导，右边不断对 cos⁡2x\cos 2x 积分，然后按交替符号相乘相加。

答题过程

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Set up the repeated integration by parts table:

DIx2cos⁡2x2x12sin⁡2x2−14cos⁡2x0−18sin⁡2x\begin{array}{c|c} D & I \\ \hline x^2 & \cos 2x\\ 2x & \frac12\sin 2x\\ 2 & -\frac14\cos 2x\\ 0 & -\frac18\sin 2x \end{array}

Hence

∫x2cos⁡2x dx= x2(12sin⁡2x)−2x(−14cos⁡2x)+2(−18sin⁡2x)+C= 12x2sin⁡2x+12xcos⁡2x−14sin⁡2x+C.\begin{align*} \int x^2\cos 2x\,\mathrm{d}x =&\, x^2\left(\frac12\sin 2x\right) -2x\left(-\frac14\cos 2x\right) +2\left(-\frac18\sin 2x\right)+C \\[2mm] =&\, \frac12x^2\sin 2x +\frac12x\cos 2x -\frac14\sin 2x+C. \end{align*}

(b)

解法一

思路

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先把微分方程分离变量，左边只含 yy，右边只含 tt。右边会出现 ∫t2cos⁡2t dt\int t^2\cos^2t\,\mathrm{d}t，用 cos⁡2t=12(1+cos⁡2t)\cos^2t=\frac12(1+\cos 2t)，其中 ∫t2cos⁡2t dt\int t^2\cos 2t\,\mathrm{d}t 正好可以套用 (a) 的结果。

答题过程

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From

dydt=(tcos⁡ty)2,\frac{\mathrm{d}y}{\mathrm{d}t} =\left(\frac{t\cos t}{y}\right)^2,

we have

y2dydt=t2cos⁡2t.y^2\frac{\mathrm{d}y}{\mathrm{d}t}=t^2\cos^2t.

Separate and integrate:

∫y2 dy=∫t2cos⁡2t dt.\int y^2\,\mathrm{d}y =\int t^2\cos^2t\,\mathrm{d}t.

The left hand side is

∫y2 dy=13y3.\int y^2\,\mathrm{d}y=\frac13y^3.

For the right hand side,

∫t2cos⁡2t dt= ∫t2⋅12(1+cos⁡2t) dt= 12∫t2 dt+12∫t2cos⁡2t dt.\begin{align*} \int t^2\cos^2t\,\mathrm{d}t =&\, \int t^2\cdot\frac12(1+\cos 2t)\,\mathrm{d}t \\[2mm] =&\, \frac12\int t^2\,\mathrm{d}t +\frac12\int t^2\cos 2t\,\mathrm{d}t. \end{align*}

Using the result from part (a), with xx replaced by tt,

∫t2cos⁡2t dt=12t2sin⁡2t+12tcos⁡2t−14sin⁡2t.\int t^2\cos 2t\,\mathrm{d}t =\frac12t^2\sin 2t+\frac12t\cos 2t-\frac14\sin 2t.

Therefore

13y3= 16t3+12(12t2sin⁡2t+12tcos⁡2t−14sin⁡2t)+C= 16t3+14t2sin⁡2t+14tcos⁡2t−18sin⁡2t+C.\begin{align*} \frac13y^3 =&\, \frac16t^3 +\frac12\left( \frac12t^2\sin 2t+\frac12t\cos 2t-\frac14\sin 2t \right)+C \\[2mm] =&\, \frac16t^3 +\frac14t^2\sin 2t +\frac14t\cos 2t -\frac18\sin 2t+C. \end{align*}

Multiplying by 33 and absorbing the constant,

y3=12t3+34t2sin⁡2t+34tcos⁡2t−38sin⁡2t+C.y^3=\frac12t^3+\frac34t^2\sin 2t+\frac34t\cos 2t-\frac38\sin 2t+C.

So the solution is

y3=12t3+34t2sin⁡2t+34tcos⁡2t−38sin⁡2t+C.\boxed{y^3=\frac12t^3+\frac34t^2\sin 2t+\frac34t\cos 2t-\frac38\sin 2t+C}.