设函数f(x,y)可微,,求f(x,y).

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问题 设函数f(x,y)可微,,求f(x,y).

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答案[*] 由[*],得C=0,即f(0,y)=siny. 又由[*],得lnf(x,y)=-x+lng(y), 即f(x,y)=φ(y)e-x,由f(0,y)=siny,得φ(y)=siny,所以f(x,y)=e-xsiny.

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