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 $y = \mathrm{f}(x)$

(8)

解答

展开

先把方程除以 $x^2$ ，写成一阶线性方程：

\begin{align*} y'+P(x)y=Q(x). \end{align*}

这里 $0<x<\pi$ ，所以 $x>0$ 且 $\sin x>0$ ，后面使用 $\ln x$ 和 $\ln(\sin x)$ 都是自然的。

展开

The differential equation is

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

Divide by $x^2$ :

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

The integrating factor is

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

Multiplying through by $x^2\sin x$ gives

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

Integrate both sides:

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

Therefore

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

展开

观察原微分方程的左边（LHS）形式： $x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + (x\cot x+2)xy$ 。如果将括号展开，可得 $x^2 y' + x^2(\cot x)y + 2xy$ 。

注意到 $x^2 y' + 2xy$ 正好是 $x^2 y$ 的导数，因此可以引入新变量 $z = x^2 y$ ，将方程改写为关于 $z$ 的一阶线性微分方程，其形式比直接处理 $y$ 要精简得多。

展开

Expand and group the left-hand side of the differential equation:

\begin{align*} x^2\frac{\mathrm{d}y}{\mathrm{d}x} + \left(x^2\cot x + 2x\right)y =&\,\, 4\sin x\\[3mm] \left(x^2\frac{\mathrm{d}y}{\mathrm{d}x} + 2xy\right) + x^2(\cot x)y =&\,\, 4\sin x \end{align*}

Let $z = x^2 y$ . By the Product Rule:

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

Substitute $z = x^2 y$ and $\dfrac{\mathrm{d}z}{\mathrm{d}x}$ into the equation:

\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} + (\cot x)z =&\,\, 4\sin x \end{align*}

This is a first-order linear differential equation in $z$ . The integrating factor is:

\begin{align*} \mathrm{I.F.} =&\,\, \mathrm{e}^{\int \cot x\,\mathrm{d}x}\\[3mm] =&\,\, \mathrm{e}^{\ln(\sin x)}\\[3mm] =&\,\, \sin x \end{align*}

(which is strictly positive on the interval $0 < x < \pi$ ).

Multiply the differential equation through by $\sin x$ :

\begin{align*} \sin x \frac{\mathrm{d}z}{\mathrm{d}x} + (\cos x)z =&\,\, 4\sin^2 x\\[3mm] \frac{\mathrm{d}}{\mathrm{d}x}(z\sin x) =&\,\, 4\sin^2 x \end{align*}

Integrate both sides with respect to $x$ :

\begin{align*} z\sin x =&\,\, \int 4\sin^2 x\,\mathrm{d}x\\[3mm] =&\,\, \int 2(1-\cos 2x)\,\mathrm{d}x\\[3mm] =&\,\, 2x - \sin 2x + C \end{align*}

Substitute $z = x^2 y$ back into the equation:

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

Therefore, the general solution is:

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