设z=xf(y/x)+yg(x/y),其中f,g二阶可导,证明:

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问题 设z=xf(y/x)+yg(x/y),其中f,g二阶可导,证明:

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答案dz/dx=f(y/x)-y/xf’(y/x)+g’(x/y),dz/dy=f’(y/x)+g(x/y)-x/y g’(x/y),d2z/dx2=-y/x2f’(y/x)+y/x2f’(y/x)+y2/x3f"(y/x)+1/yg"(x/y)=y3/x3f"(y/x)+1/yg"(x/y),d2z/dy2=1/xf"(y/x)-x/y2g’(x/y)+x/y2g’(x/y)+x2/y3g"(x/y)=1/xf"(y/x)+x2/y3g"(x/y),故x2d2z/dx2-y2d2z/dy2=y2/xf"(y/x)+x2/yg"(x/y)-y2/xf"(y/x)-x2/yg"(x/y)=0.

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