题目

Problem

(a) Determine the general solution of the differential equation

\begin{align*} (x + 1) \frac{\mathrm{d}y}{\mathrm{d}x} - xy = \mathrm{e}^{3x} \qquad x > -1 \end{align*}

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

(7)

(b) Determine the particular solution of the differential equation for which $y = 5$ when $x = 0$ .

(2)

解答

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先把微分方程写成一阶线性标准形式

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

这里 $x>-1$ ，所以 $\ln(x+1)$ 可以直接使用，不需要写成 $\ln|x+1|$ 。

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The differential equation is

\begin{align*} (x+1)\frac{\mathrm{d}y}{\mathrm{d}x}-xy=\mathrm{e}^{3x}. \end{align*}

Divide by $x+1$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} -\frac{x}{x+1}y =&\,\frac{\mathrm{e}^{3x}}{x+1}. \end{align*}

The integrating factor is

\begin{align*} I(x) =&\,\mathrm{e}^{\int -\frac{x}{x+1}\,\mathrm{d}x}. \end{align*}

Rewrite the integrand:

\begin{align*} -\frac{x}{x+1} =&\,-\frac{x+1-1}{x+1}\\[2mm] =&\,-1+\frac{1}{x+1}. \end{align*}

\begin{align*} \int -\frac{x}{x+1}\,\mathrm{d}x =&\,\int\left(-1+\frac{1}{x+1}\right)\,\mathrm{d}x\\[2mm] =&\,-x+\ln(x+1), \end{align*}

since $x>-1$ . Hence

\begin{align*} I(x) =&\,\mathrm{e}^{-x+\ln(x+1)}\\[2mm] =&\,(x+1)\mathrm{e}^{-x}. \end{align*}

Multiply the linear differential equation by $(x+1)\mathrm{e}^{-x}$ :

\begin{align*} (x+1)\mathrm{e}^{-x} \frac{\mathrm{d}y}{\mathrm{d}x} -x\mathrm{e}^{-x}y =&\,\mathrm{e}^{2x}. \end{align*}

The left hand side is

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[y(x+1)\mathrm{e}^{-x}\right]. \end{align*}

Therefore

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[y(x+1)\mathrm{e}^{-x}\right] =&\,\mathrm{e}^{2x}. \end{align*}

Integrate:

\begin{align*} y(x+1)\mathrm{e}^{-x} =&\,\int \mathrm{e}^{2x}\,\mathrm{d}x\\[2mm] =&\,\frac12\mathrm{e}^{2x}+C. \end{align*}

\begin{align*} y =&\,\frac{\frac12\mathrm{e}^{2x}+C} {(x+1)\mathrm{e}^{-x}}\\[4mm] =&\,\frac{\mathrm{e}^{3x}}{2(x+1)} +\frac{C\mathrm{e}^{x}}{x+1}. \end{align*}

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把 $x=0$ 、 $y=5$ 代入 (a) 的 general solution，就能求出常数 $C$ 。

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From part (a),

\begin{align*} y =&\,\frac{\mathrm{e}^{3x}}{2(x+1)} +\frac{C\mathrm{e}^{x}}{x+1}. \end{align*}

Use $x=0$ and $y=5$ :

\begin{align*} 5 =&\,\frac{\mathrm{e}^{0}}{2(1)} +\frac{C\mathrm{e}^{0}}{1}\\[2mm] =&\,\frac12+C. \end{align*}

Thus

\begin{align*} C=\frac92. \end{align*}

Therefore the particular solution is

\begin{align*} y =&\,\frac{\mathrm{e}^{3x}}{2(x+1)} +\frac{9\mathrm{e}^{x}}{2(x+1)}. \end{align*}