题目

Problem

(a) Determine the general solution of the differential equation

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} - 3y \tan x =\,& \mathrm{e}^{4x} \sec^3 x\\[2mm] \end{align*}

giving your answer in the form $y = \mathrm{f}(x)$ .

(5)

(b) Determine the particular solution for which $y = 4$ at $x = 0$

(2)

解答

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这是标准一阶线性微分方程。先找 integrating factor：

\begin{align*} \mathrm{e}^{\int -3\tan x\,\mathrm{d}x}. \end{align*}

乘上积分因子后，左边会合成一个乘积导数，然后直接积分。

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The differential equation is

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}-3y\tan x =\mathrm{e}^{4x}\sec^3 x. \end{align*}

The integrating factor is

\begin{align*} \mathrm{e}^{\int -3\tan x\,\mathrm{d}x} =&\,\mathrm{e}^{3\ln(\cos x)}\\[2mm] =&\,\cos^3 x. \end{align*}

Multiplying the differential equation by $\cos^3 x$ ,

\begin{align*} \cos^3 x\frac{\mathrm{d}y}{\mathrm{d}x} -3y\sin x\cos^2 x =&\,\mathrm{e}^{4x}. \end{align*}

The left hand side is

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(y\cos^3 x). \end{align*}

Therefore

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(y\cos^3 x) =&\,\mathrm{e}^{4x}. \end{align*}

Integrating,

\begin{align*} y\cos^3 x =&\,\int \mathrm{e}^{4x}\,\mathrm{d}x\\[2mm] =&\,\frac{1}{4}\mathrm{e}^{4x}+C. \end{align*}

Hence

\begin{align*} \boxed{y=\left(\frac{1}{4}\mathrm{e}^{4x}+C\right)\sec^3 x}. \end{align*}

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把 $x=0$ 、 $y=4$ 代入通解即可。注意 $\sec 0=1$ ，所以常数 $C$ 会很快求出。

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Using

\begin{align*} y=\left(\frac{1}{4}\mathrm{e}^{4x}+C\right)\sec^3 x, \end{align*}

and $y=4$ when $x=0$ ,

\begin{align*} 4=&\,\left(\frac{1}{4}\mathrm{e}^0+C\right)\sec^3 0\\[2mm] 4=&\,\frac{1}{4}+C. \end{align*}

\begin{align*} C=\frac{15}{4}. \end{align*}

Therefore the particular solution is

\begin{align*} \boxed{y=\frac{1}{4}\left(\mathrm{e}^{4x}+15\right)\sec^3 x}. \end{align*}