题目

Problem

7. (a) Show that the transformation $y = xv$ transforms the equation

\begin{align*} &\,3 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{6}{x} \frac{\mathrm{d}y}{\mathrm{d}x}\\[2mm] &\,\hspace{2pt}+ \frac{6y}{x^2} + 3y\\[2mm] =&\, x^2 \qquad x \neq 0 \qquad \text{(I)} \end{align*}

into the equation

\begin{align*} 3 \frac{\mathrm{d}^2v}{\mathrm{d}x^2} + 3v =&\, x \qquad \text{(II)} \end{align*}

(6)

(b) Hence obtain the general solution of the differential equation (I), giving your answer in the form $y = f(x)$

(6)

题目中文翻译

(a) 证明代换 $y=xv$ 将方程

\begin{align*} &\,3\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -\frac6x\frac{\mathrm{d}y}{\mathrm{d}x}\\[2mm] &\,\hspace{2pt}+\frac{6y}{x^2}+3y\\[2mm] =&\,x^2,\qquad x\neq0\qquad\text{(I)} \end{align*}

化为方程

\begin{align*} 3\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+3v=&\,x\qquad\text{(II)}. \end{align*}

(6)

(b) 由此求微分方程 (I) 的通解，并将答案写成 $y=f(x)$ 的形式。

(6)

解答

(a)

解法一

思路

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题目已经给出代换 $y=xv$ 。先求 $y'$ 和 $y''$ ，再代入原方程。

关键是代入后含 $v$ 和 $v'$ 的项会抵消，留下只含 $v''$ 和 $v$ 的方程。

答题过程

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Given

\begin{align*} y=xv, \end{align*}

differentiate with respect to $x$ :

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

Differentiate again:

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{\mathrm{d}v}{\mathrm{d}x} +\Big(\frac{\mathrm{d}v}{\mathrm{d}x} +x\frac{\mathrm{d}^2v}{\mathrm{d}x^2}\Big)\\[2mm] =&\,2\frac{\mathrm{d}v}{\mathrm{d}x} +x\frac{\mathrm{d}^2v}{\mathrm{d}x^2}. \end{align*}

Substitute these into equation (I):

\begin{align*} &\,3\Big(2\frac{\mathrm{d}v}{\mathrm{d}x} +x\frac{\mathrm{d}^2v}{\mathrm{d}x^2}\Big)\\[2mm] &\,\hspace{2pt}-\frac{6}{x} \Big(v+x\frac{\mathrm{d}v}{\mathrm{d}x}\Big)\\[2mm] &\,\hspace{4pt}+\frac{6(xv)}{x^2}+3xv=x^2. \end{align*}

Expanding,

\begin{align*} &\,6\frac{\mathrm{d}v}{\mathrm{d}x} +3x\frac{\mathrm{d}^2v}{\mathrm{d}x^2}\\[2mm] &\,\hspace{2pt}-\frac{6v}{x}-6\frac{\mathrm{d}v}{\mathrm{d}x}\\[2mm] &\,\hspace{4pt}+\frac{6v}{x}+3xv=x^2. \end{align*}

The $v'$ terms and the $\dfrac{v}{x}$ terms cancel:

\begin{align*} 3x\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+3xv=&\,x^2. \end{align*}

Since $x\neq 0$ , divide by $x$ :

\begin{align*} \boxed{ 3\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+3v=x }. \end{align*}

This is equation (II).

(b)

解法一

思路

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先解 (II)，再用 $y=xv$ 回到原来的 $y$ 。

方程

\begin{align*} 3v''+3v=x \end{align*}

可以先除以 $3$ 。齐次解来自 $v''+v=0$ ，特解可以试一次式 $v=kx+l$ 。

答题过程

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Equation (II) is

\begin{align*} 3\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+3v=x. \end{align*}

Divide by $3$ :

\begin{align*} \frac{\mathrm{d}^2v}{\mathrm{d}x^2}+v=\frac{x}{3}. \end{align*}

The complementary function satisfies

\begin{align*} \frac{\mathrm{d}^2v}{\mathrm{d}x^2}+v=0, \end{align*}

with auxiliary equation

\begin{align*} \lambda^2+1=&\,0,\\[2mm] \lambda^2=&\,-1,\\[2mm] \lambda=&\,\pm\mathrm{i}. \end{align*}

Thus

\begin{align*} v_c=A\cos x+B\sin x. \end{align*}

For a particular integral, try

\begin{align*} v_p=kx+l. \end{align*}

Differentiating gives $v_p'=k$ and $v_p''=0$ . Substitution into equation (II) gives

\begin{align*} 3(0)+3(kx+l)=&\,x, \end{align*}

\begin{align*} kx+l=\frac{x}{3}. \end{align*}

Thus

\begin{align*} k=\frac{1}{3},\qquad l=0. \end{align*}

Therefore

\begin{align*} v=A\cos x+B\sin x+\frac{x}{3}. \end{align*}

Since $y=xv$ ,

\begin{align*} \boxed{ y=x\Big(A\cos x+B\sin x+\frac{x}{3}\Big) }. \end{align*}

FP2 2022 June Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程