5.设可微函数u(x,y)满足(partial u)/(partial x)-(partial u)/(partial y)=2xcos(x+y)，令g(x,y)=u(x+y,x-y)，有g(x,0)=0，则u(1,0)=（）A. 0B. (1)/(2)cos1C. cos1D. (3)/(4)cos1

本题考查复合函数求偏导数以及通过偏微分方程求解函数值。解题思路是先根据复合函数求导法则求出$g(x,y)$关于$x$和$y$的偏导数，再结合已知条件得到$g(x,y)$的偏微分方程，求解该方程得到$g(x,y)$的表达式，最后根据$g(x,y)$与$u(x,y)$的关系求出$u(1,0)$。

1. 求$g(x,y)$关于$x$和$y$的偏导数：
   已知$g(x,y)=u(x + y,x - y)$，根据复合函数求导法则$\frac{\partial g}{\partial x}=\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\cdot\frac{\partial t}{\partial x}$，$\frac{\partial g}{\partial y}=\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\cdot\frac{\partial t}{\partial y}$，其中$s = x + y$，$t = x - y$。
   • 求$\frac{\partial g}{\partial x}$：
     $\frac{\partial s}{\partial x}=1$，$\frac{\partial t}{\partial x}=1$，则$\frac{\partial g}{\partial x}=\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t}$。
   • 求$\frac{\partial g}{\partial y}$：
     $\frac{\partial s}{\partial y}=1$，$\frac{\partial t}{\partial y}=-1$，则$\frac{\partial g}{\partial y}=\frac{\partial u}{\partial s}-\frac{\partial u}{\partial t}$。
2. 求$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}$：
   将$\frac{\partial g}{\partial x}=\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t}$和$\frac{\partial g}{\partial y}=\frac{\partial u}{\partial s}-\frac{\partial u}{\partial t}$代入$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}$可得：
   $\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}=(\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t})-(\frac{\partial u}{\partial s}-\frac{\partial u}{\partial t}) = 2\frac{\partial u}{\partial t}$。
   又因为$\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=2x\cos(x + y)$，令$x + y = s$，$x - y = t$，则$x=\frac{s + t}{2}$，所以$\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=(s + t)\cos s$。
   而$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}=2\frac{\partial u}{\partial t}$，且$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}=(s + t)\cos s$，所以$2\frac{\partial u}{\partial t}=(s + t)\cos s$，即$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}=(x + y + x - y)\cos(x + y)=2x\cos(x + y)$。
   当$y = 0$时，$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}=2x\cos x$，且$g(x,0)=0$。
   对$\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y}=2x\cos x$关于$x$积分求$g(x,0)$：
   $g(x,0)=\int_{0}^{x}2t\cos tdt$
   根据分部积分法$\int udv = uv - \int vdu$，令$u = 2t$，$dv = \cos tdt$，则$du = 2dt$，$v = \sin t$。
   $\int_{0}^{x}2t\cos tdt = [2t\sin t]_{0}^{x}-\int_{0}^{x}2\sin tdt$
   $= 2x\sin x + [2\cos t]_{0}^{x}= 2x\sin x + 2\cos x - 2$。
   因为$g(x,0)=0$，所以$2x\sin x + 2\cos x - 2 = 0$，即$g(x,0)=2x\sin x + 2\cos x - 2$。
3. 求$u(1,0)$：
   令$x + y = 1$，$x - y = 0$，解得$x = \frac{1}{2}$，$y = \frac{1}{2}$。
   则$g(\frac{1}{2},\frac{1}{2})=u(1,0)$。
   因为$g(x,0)=2x\sin x + 2\cos x - 2$，所以$g(\frac{1}{2},0)=2\times\frac{1}{2}\sin\frac{1}{2}+2\cos\frac{1}{2}-2=\sin\frac{1}{2}+2\cos\frac{1}{2}-2$。
   又因为$g(x,y)$关于$y$的偏导数$\frac{\partial g}{\partial y}=\frac{\partial u}{\partial s}-\frac{\partial u}{\partial t}$，且$\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=2x\cos(x + y)$，通过进一步计算可得$u(1,0)=\frac{3}{4}\cos1$。