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IAL 2025 Jan FP3 Q8

A Level / Edexcel / FP3

IAL 2025 Jan Paper · Question 8

题目

Problem

Given that

y=e3xcosh⁡2xy = e^{3x}\cosh 2x

prove by induction that for n∈Nn \in \mathbb{N}

dnydxn=e3x(5n+12cosh⁡2x+5n−12sinh⁡2x)\frac{d^n y}{dx^n} = e^{3x}\left(\frac{5^n+1}{2}\cosh 2x+\frac{5^n-1}{2}\sinh 2x\right)
(6)
题目中文翻译

已知

y=e3xcosh⁡2xy = e^{3x}\cosh 2x

用数学归纳法证明，对于 n∈Nn \in \mathbb{N}，

dnydxn=e3x(5n+12cosh⁡2x+5n−12sinh⁡2x)\frac{d^n y}{dx^n} = e^{3x}\left(\frac{5^n+1}{2}\cosh 2x+\frac{5^n-1}{2}\sinh 2x\right)

解答

解法一：直接对双曲函数形式作归纳

思路

展开

先验证 n=1n=1。在归纳步骤中，把两个系数分别记作 AkA_k 和 BkB_k；对假设式求导后，乘积法则会把它们变成 3Ak+2Bk3A_k+2B_k 和 2Ak+3Bk2A_k+3B_k。化简这两个新系数，即可得到 n=k+1n=k+1 时的目标形式。

答题过程

For n=1n=1,

dydx=3e3xcosh⁡2x+2e3xsinh⁡2x=e3x(3cosh⁡2x+2sinh⁡2x)=e3x(51+12cosh⁡2x+51−12sinh⁡2x).\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=3e^{3x}\cosh 2x+2e^{3x}\sinh 2x \\ &=e^{3x}\left(3\cosh 2x+2\sinh 2x\right) \\ &=e^{3x}\left( \frac{5^1+1}{2}\cosh 2x +\frac{5^1-1}{2}\sinh 2x \right). \end{aligned}

Therefore, the result is true for n=1n=1.

Assume that the result is true for n=kn=k, where k∈Nk\in\mathbb N. Let

Ak=5k+12andBk=5k−12.A_k=\frac{5^k+1}{2} \quad\text{and}\quad B_k=\frac{5^k-1}{2}.

Then the induction hypothesis is

dkydxk=e3x(Akcosh⁡2x+Bksinh⁡2x).\frac{\mathrm{d}^k y}{\mathrm{d}x^k} =e^{3x}\left(A_k\cosh 2x+B_k\sinh 2x\right).

Differentiating both sides gives

dk+1ydxk+1=3e3x(Akcosh⁡2x+Bksinh⁡2x)+e3x(2Aksinh⁡2x+2Bkcosh⁡2x)=e3x[(3Ak+2Bk)cosh⁡2x+(2Ak+3Bk)sinh⁡2x].\begin{aligned} \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} ={}&3e^{3x}\left(A_k\cosh 2x+B_k\sinh 2x\right) \\ &+e^{3x}\left(2A_k\sinh 2x+2B_k\cosh 2x\right) \\ ={}&e^{3x}\bigl[(3A_k+2B_k)\cosh 2x \\ &\qquad +(2A_k+3B_k)\sinh 2x\bigr]. \end{aligned}

Now

3Ak+2Bk=3(5k+1)+2(5k−1)2=5k+1+12,\begin{aligned} 3A_k+2B_k &=\frac{3(5^k+1)+2(5^k-1)}{2} \\ &=\frac{5^{k+1}+1}{2}, \end{aligned}

and

2Ak+3Bk=2(5k+1)+3(5k−1)2=5k+1−12.\begin{aligned} 2A_k+3B_k &=\frac{2(5^k+1)+3(5^k-1)}{2} \\ &=\frac{5^{k+1}-1}{2}. \end{aligned}

Hence

dk+1ydxk+1=e3x(5k+1+12cosh⁡2x+5k+1−12sinh⁡2x).\frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} =e^{3x}\left( \frac{5^{k+1}+1}{2}\cosh 2x +\frac{5^{k+1}-1}{2}\sinh 2x \right).

Thus, if the result is true for n=kn=k, it is true for n=k+1n=k+1. Since it is true for n=1n=1, the result follows by induction for all n∈Nn\in\mathbb N.

解法二：先改写成指数函数

思路

展开

利用 cosh⁡2x=12(e2x+e−2x)\cosh 2x=\frac12(e^{2x}+e^{-2x})，先把 yy 化为两个指数函数之和。它的 nn 阶导数形式很容易用归纳法证明；最后再将结果重新组合成 cosh⁡2x\cosh 2x 与 sinh⁡2x\sinh 2x，便得到题目所要求的式子。

答题过程

Using the exponential form of cosh⁡2x\cosh 2x,

y=e3x(e2x+e−2x2)=e5x+ex2.y=e^{3x}\left(\frac{e^{2x}+e^{-2x}}{2}\right) =\frac{e^{5x}+e^x}{2}.

We first prove by induction that

dnydxn=5ne5x+ex2.\frac{\mathrm{d}^n y}{\mathrm{d}x^n} =\frac{5^n e^{5x}+e^x}{2}.

For n=1n=1,

dydx=5e5x+ex2=51e5x+ex2,\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{5e^{5x}+e^x}{2} =\frac{5^1e^{5x}+e^x}{2},

so the result is true for n=1n=1.

Assume that it is true for n=kn=k. Then

dkydxk=5ke5x+ex2.\frac{\mathrm{d}^k y}{\mathrm{d}x^k} =\frac{5^k e^{5x}+e^x}{2}.

Differentiating both sides,

dk+1ydxk+1=5(5ke5x)+ex2=5k+1e5x+ex2.\begin{aligned} \frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} &=\frac{5\left(5^k e^{5x}\right)+e^x}{2} \\ &=\frac{5^{k+1}e^{5x}+e^x}{2}. \end{aligned}

Therefore, by induction,

dnydxn=5ne5x+ex2\frac{\mathrm{d}^n y}{\mathrm{d}x^n} =\frac{5^n e^{5x}+e^x}{2}

for all n∈Nn\in\mathbb N.

Finally,

dnydxn=e3x2(5ne2x+e−2x)=e3x4[(5n+1)(e2x+e−2x)+(5n−1)(e2x−e−2x)]=e3x(5n+12cosh⁡2x+5n−12sinh⁡2x),\begin{aligned} \frac{\mathrm{d}^n y}{\mathrm{d}x^n} &=\frac{e^{3x}}{2}\left(5^n e^{2x}+e^{-2x}\right) \\ &=\frac{e^{3x}}{4}\bigl[ (5^n+1)(e^{2x}+e^{-2x}) \\ &\hspace{5.5em}+(5^n-1)(e^{2x}-e^{-2x}) \bigr] \\ &=e^{3x}\left( \frac{5^n+1}{2}\cosh 2x +\frac{5^n-1}{2}\sinh 2x \right), \end{aligned}

as required.