题目

Problem

\begin{align*} (\cos x)\frac{\mathrm{d}y}{\mathrm{d}x} +(\sin x)y = 2\cos^3x\sin x-3, \qquad 0\leqslant x<\frac{\pi}{2} \end{align*}

(a) Find the general solution of this differential equation. Give your answer in the form $y=f(x)$ .

(7)

(b) Find the particular solution of this differential equation for which $y=3\sqrt3$ at $x=\dfrac{\pi}{3}$ . Give your answer in the form $y=f(x)$ .

(3)

解答

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先除以 $\cos x$ ，因为题目给出 $0\leqslant x<\dfrac{\pi}{2}$ ，所以 $\cos x>0$ 。得到一阶线性方程后，积分因子是 $\sec x$ 。

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Divide by $\cos x$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} +(\tan x)y = 2\cos^2x\sin x-3\sec x \end{align*}

The integrating factor is

\begin{align*} \mathrm{e}^{\int\tan x\,\mathrm{d}x} = \mathrm{e}^{\ln(\sec x)} = \sec x \end{align*}

Multiplying by $\sec x$ ,

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

Integrate:

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

Therefore

\begin{align*} y ={}& \cos x(\sin^2x-3\tan x+C)\\[3mm] ={}& \cos x\sin^2x-3\sin x+C\cos x \end{align*}

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把 $x=\dfrac{\pi}{3}$ 和 $y=3\sqrt3$ 代入通解。注意 $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$ ， $\cos\dfrac{\pi}{3}=\dfrac12$ 。

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Using

\begin{align*} y=\cos x\sin^2x-3\sin x+C\cos x, \end{align*}

and substituting $x=\dfrac{\pi}{3}$ , $y=3\sqrt3$ :

\begin{align*} 3\sqrt3 ={}& \left(\frac12\right) \left(\frac{\sqrt3}{2}\right)^2 -3\left(\frac{\sqrt3}{2}\right) +C\left(\frac12\right)\\[3mm] ={}& \frac38-\frac{3\sqrt3}{2}+\frac{C}{2} \end{align*}

\begin{align*} \frac{C}{2} ={}& 3\sqrt3+\frac{3\sqrt3}{2}-\frac38\\[3mm] ={}& \frac{9\sqrt3}{2}-\frac38 \end{align*}

Therefore

\begin{align*} C=9\sqrt3-\frac34 \end{align*}

The particular solution is

\begin{align*} y = \cos x\sin^2x-3\sin x +\left(9\sqrt3-\frac34\right)\cos x \end{align*}