已知可微函数f(u,v)满足=2(u-v)e-(u+v),且f(u,0)=u2e-u. 记g(x,y)=f(x,y-x),求

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问题 已知可微函数f(u,v)满足=2(u-v)e-(u+v),且f(u,0)=u2e-u
记g(x,y)=f(x,y-x),求

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答案由g(x,y)=f(x,y-x)得[*]=f’1(x,y-x)-f’2(x,y-x),又[*]=2(u-v)e-(u+v),则2(x-y+x)e-(x+y-x)=2(2x-y)e-y,即[*]=2(2x-y)e-y

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