题目

Problem

Obtain the general solution of the equation

\begin{align*} x^2\frac{\mathrm{d}y}{\mathrm{d}x} +(x\cot x+2)xy=4\sin x \qquad 0<x<\pi \end{align*}

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

(8)

解答

解法一

思路

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这是一个一阶线性微分方程。先除以 $x^2$ ，化成 $\dfrac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)$ 。积分因子来自 $\int(\cot x+\frac2x)\,\mathrm{d}x$ 。由于 $0<x<\pi$ ， $\sin x>0$ ，所以 $\ln(\sin x)$ 不需要额外讨论绝对值符号。

### 答题过程

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Divide by $x^2$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} +\left(\cot x+\frac2x\right)y =&\,\frac{4\sin x}{x^2} \end{align*}

The integrating factor is

\begin{align*} \mathrm{IF} =&\,\mathrm{e}^{\int\left(\cot x+\frac2x\right)\,\mathrm{d}x}\\[4mm] =&\,\mathrm{e}^{\ln(\sin x)+2\ln x}\\[4mm] =&\,x^2\sin x \end{align*}

Multiplying the differential equation by $x^2\sin x$ gives

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left(x^2\sin x\,y\right) =&\,x^2\sin x\cdot \frac{4\sin x}{x^2}\\[4mm] =&\,4\sin^2x \end{align*}

Integrate:

\begin{align*} x^2\sin x\,y =&\,\int 4\sin^2x\,\mathrm{d}x\\[4mm] =&\,\int 2(1-\cos2x)\,\mathrm{d}x\\[4mm] =&\,2x-\sin2x+C \end{align*}

Therefore

\begin{align*} y=\frac{2x-\sin2x+C}{x^2\sin x} \end{align*}

IAL 2020 Oct Q6

题目

解答

解法一

思路