2019年考研数二第17题解析:一阶线性微分方程、旋转体的体积

题目

设函数 $y(x)$ 是微分方程 $y^{‘} – xy = \frac{1}{2 \sqrt{x}} e^{\frac{x^{2}}{2}}$ 满足条件 $y(1) = \sqrt{e}$ 的特解.

$(Ⅰ)$ 求 $y(x)$;

$(Ⅱ)$ 设平面区域 $D = { (x, y) | 1 \leqslant x \leqslant 2, 0 \leqslant y \leqslant y(x) }$, 求 $D$ 绕 $x$ 轴旋转所得旋转体的体积.

解析

第 $(Ⅰ)$ 问

由于:

$$
y^{‘} – xy = \frac{1}{2 \sqrt{x}} e^{\frac{x^{2}}{2}}.
$$

于是,根据一阶线性微分方程的求解公式,可得:

$$
y = \Bigg[ \int \frac{1}{2 \sqrt{x}} e^{\frac{x^{2}}{2}} \cdot e^{ \int -x \mathrm{d} x} \mathrm{d} x + C \Bigg] e^{- \int -x \mathrm{d} x} \Rightarrow
$$

$$
y = \Bigg[ \int \frac{1}{2 \sqrt{x}} e^{\frac{x^{2}}{2}} \cdot e^{\frac{- x^{2}}{2}} \mathrm{d} x + C \Bigg] e^{\frac{x^{2}}{2}} \Rightarrow
$$

$$
y = \Bigg[ \int \frac{1}{2 \sqrt{x}} (e^{\frac{x^{2}}{2}} \cdot e^{\frac{- x^{2}}{2}}) \mathrm{d} x + C \Bigg] e^{\frac{x^{2}}{2}} \Rightarrow
$$

$$
y = \Bigg[ \int \frac{1}{2 \sqrt{x}} \mathrm{d} x + C \Bigg] e^{\frac{x^{2}}{2}} \Rightarrow
$$

$$
{\color{Red}
y = (\sqrt{x} + C) e^{\frac{x^{2}}{2}}}.
$$

又:

$$
y(1) = \sqrt{e}.
$$

于是:

$$
(1 + C) e^{\frac{1}{2}} = \sqrt{e} \Rightarrow
$$

$$
(1 + C) \sqrt{e} = \sqrt{e} \Rightarrow
$$

$$
C = 0.
$$

即:

$$
{\color{Red}
y(x) = \sqrt{x} \cdot e^{\frac{x^{2}}{2}}}.
$$

第 $(Ⅱ)$ 问

由题可得:

$$
V = \pi \int_{1}^{2} y^{2}(x) \mathrm{d} x \Rightarrow
$$

$$
V = \pi \int_{1}^{2} (\sqrt{x} \cdot e^{\frac{x^{2}}{2}})^{2} \mathrm{d} x \Rightarrow
$$

$$
V = \pi \int_{1}^{2} (x \cdot e^{x^{2}}) \mathrm{d} x \Rightarrow
$$

$$
{\color{White}
(e^{x^{2}})^{‘} = e^{x^{2}} \cdot 2x} \Rightarrow
$$

$$
V = \pi \cdot \frac{1}{2} \cdot e^{x^{2}} |_{1}^{2} \Rightarrow
$$

$$
{\color{Red}
V = \frac{\pi}{2} (e^{4} – e)}.
$$