kaoyan1basic 高等数学 第85题
📝 题目
### 第85题 设 $u=u\left(\sqrt{x^{2}+y^{2}}\right)\left(r=\sqrt{x^{2}+y^{2}}>0\right)$ 有二阶连续的偏导数，且满足 $$ $\displaystyle \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}-\frac{1}{x} \frac{\partial u}{\partial x}+u=x^{2}+y^{2}$ $$ 则 $u\left(\sqrt{x^{2}+y^{2}}\right)=$ $\_\_\_\_$ ．
💡 答案解析
**答案**：$\displaystyle u = \frac{1}{4}r^2 + C_1 \ln r + C_2$，其中$r = \sqrt{x^2+y^2}$ **解析**： 步骤1：令$r = \sqrt{x^2+y^2}$，则$u = u(r)$，计算偏导：$\displaystyle \frac{\partial u}{\partial x} = u' \cdot \frac{x}{r}$，$\displaystyle \frac{\partial^2 u}{\partial x^2} = u'' \cdot \frac{x^2}{r^2} + u' \cdot \frac{r^2 - x^2}{r^3}$，同理$\displaystyle \frac{\partial^2 u}{\partial y^2} = u'' \cdot \frac{y^2}{r^2} + u' \cdot \frac{r^2 - y^2}{r^3}$。 步骤2：代入方程$\displaystyle \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} - \frac{1}{x}\frac{\partial u}{\partial x} + u = x^2+y^2$，得$\displaystyle u'' + \frac{u'}{r} - \frac{u'}{r} + u = r^2$，即$u'' + u = r^2$。 步骤3：解二阶常系数线性微分方程$u'' + u = r^2$，齐次解$u_h = C_1 \cos r + C_2 \sin r$，特解设$u_p = Ar^2 + B$，代入得$2A + Ar^2 + B = r^2$，比较得$A=1, B=-2$，故$u = C_1 \cos r + C_2 \sin r + r^2 - 2$。 **难度**：★★★★☆