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解:这个函数是复合函数$u=u(r)$$r=\sqrt{x^2+y^2}$而成。根据复合函数求导法则:
$\dfrac{\partial u}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{\partial r}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{\sqrt{x^2+y^2}}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{r}$
$\dfrac{\partial u}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{\partial r}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{\sqrt{x^2+y^2}}=\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{x}{r}$$\dfrac{1}{x}\cdot\dfrac{\partial u}{\partial x}=\dfrac{1}{r}\cdot\dfrac{\textrm{d}u}{\textrm{d}r}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)=\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{r}\right)=\dfrac{x}{r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}\right)=$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)=\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{x}{r}\right)=\dfrac{x}{r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}\right)=\dfrac{x}{r}\cdot\dfrac{\partial}{\partial r}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)\dfrac{\partial r}{\partial x}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r-x\cdot(\partial r/\partial x)}{r^2}=\dfrac{x^2}{r^2}\cdot\dfrac{\textrm{d}^2u}{\textrm{d}r^2}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r^2-x^2}{r^3}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right)=\dfrac{\partial}{\partial y}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{y}{r}\right)=\dfrac{y}{r}\cdot\dfrac{\partial}{\partial y}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{\partial}{\partial y}\left(\dfrac{y}{r}\right)=\dfrac{y}{r}\cdot\dfrac{\partial}{\partial r}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)\dfrac{\partial r}{\partial y}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r-y\cdot(\partial r/\partial y)}{r^2}=\dfrac{y^2}{r^2}\cdot\dfrac{\textrm{d}^2u}{\textrm{d}r^2}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r^2-x^2}{r^3}$
代入不等式:$\dfrac{x^2+y^2}{r^2}\cdot\dfrac{\textrm{d}u^2}{\textrm{d}r^2}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{2r^2-x^2-y^2}{r^3}-\dfrac{1}{r}\cdot\dfrac{\textrm{d}u}{\textrm{d}r}+u=x^2+y^2$
代入$x^2+y^2=r^2$$\dfrac{\textrm{d}^2u}{\textrm{d}r^2}+u=r^2$,为二阶线性常系数微分方程。
通解为$u=C_1\cos r+C_2\sin r+r^2-2$
$u(\sqrt{x^2+y^2})=C_1\cos\sqrt{x^2+y^2}+C_2\sin\sqrt{x^2+y^2}+x^2+y^2-2$
\subsubsection{特殊值反代}