设f(x,y)在区域D:0≤x≤1,0≤y≤1上连续,且xy[f(x,y)dσ]2=f(x,y)-1,则f(x,y)=________.

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问题 设f(x,y)在区域D:0≤x≤1,0≤y≤1上连续,且xy[f(x,y)dσ]2=f(x,y)-1,则f(x,y)=________.

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答案4xy+1

解析f(x,y)dσ=A,则A2xy=f(x,y)-1.该式两端在D上积分,有A2xydσ=f(x,y)dσ-dσ,即A2=A-1,解之得A=2,所以f(x,y)=4xy+1.
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