题目

Problem

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

(a) Show that

\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} = \frac{\mathrm{d}^3y}{\mathrm{d}x^3} +Ax^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +Bx\frac{\mathrm{d}y}{\mathrm{d}x} +Cy \end{align*}

where $A$ , $B$ and $C$ are integers to be determined.

(4)

Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3$ and $y=1$ at $x=1$

(b) determine the Taylor series solution for $y$ in ascending powers of $(x-1)$ , up to and including the term in $(x-1)^4$ , giving each coefficient in simplest form.

(4)

题目中文翻译

已知

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

(a) 证明

\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} =&\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}+Ax^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] &\,\hspace{2pt}+Bx\frac{\mathrm{d}y}{\mathrm{d}x}+Cy \end{align*}

并求出整数 $A$ 、 $B$ 和 $C$ 。

(4)

已知当 $x=1$ 时， $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3$ 且 $y=1$ 。

(b) 求 $y$ 按 $(x-1)$ 的升幂排列的泰勒级数解，保留到 $(x-1)^4$ 项，并将各项系数化为最简形式。

(4)

解答

(a)

解法一

思路

展开

从给定的二阶导数开始连续求导。每次对 $x^2y$ 求导时都要用乘积法则。

答题过程

展开

Given

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

differentiating gives

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, 2xy+x^2\frac{\mathrm{d}y}{\mathrm{d}x} +\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

Differentiate again:

\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} =&\, 2y+2x\frac{\mathrm{d}y}{\mathrm{d}x} +2x\frac{\mathrm{d}y}{\mathrm{d}x}\\[3mm] &\,\hspace{2pt}+x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[3mm] =&\, \frac{\mathrm{d}^3y}{\mathrm{d}x^3} +x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +4x\frac{\mathrm{d}y}{\mathrm{d}x} +2y \end{align*}

Therefore

\begin{align*} A=1,\qquad B=4,\qquad C=2 \end{align*}

(b)

解法一

思路

展开

Taylor 展开需要 $y(1)$ 、 $y'(1)$ 、 $y''(1)$ 、 $y'''(1)$ 、 $y''''(1)$ 。前两个已给，后面三个用微分方程逐个算。

答题过程

展开

At $x=1$ ,

\begin{align*} y=1,\qquad \frac{\mathrm{d}y}{\mathrm{d}x}=3 \end{align*}

Using the differential equation,

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, x^2y+\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

\begin{align*} y''(1)=1^2(1)+3=4 \end{align*}

Also,

\begin{align*} y'''(1) =&\, 2(1)(1)+1^2(3)+4\\[3mm] =&\, 9 \end{align*}

and, using part (a),

\begin{align*} y''''(1) =&\, 9+1^2(4)+4(1)(3)+2(1)\\[3mm] =&\, 27 \end{align*}

Therefore

\begin{align*} y =&\, 1+3(x-1)+\frac{4}{2!}(x-1)^2 +\frac{9}{3!}(x-1)^3\\[3mm] &\,\hspace{2pt}+\frac{27}{4!}(x-1)^4\\[3mm] =&\, 1+3(x-1)+2(x-1)^2 +\frac32(x-1)^3 +\frac98(x-1)^4 \end{align*}

FP2 2026 Jan Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程