# 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

$(Ⅰ)$ 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

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

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

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

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

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

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

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

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

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

$$C = 0.$$

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

### 第 $(Ⅱ)$ 问

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

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

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

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

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

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