设二元函数z=f(x,y)满足:=excosy+y,又fx(x,0)=x/(1+4x2),f(0,y)=y,则f(x,y)=________.

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问题 设二元函数z=f(x,y)满足:=excosy+y,又fx(x,0)=x/(1+4x2),f(0,y)=y,则f(x,y)=________.

选项

答案exsiny+xy2/2+(1/8)㏑(1+4x2)+y-siny

解析
由fx(x,0)=x/(1+4x2)得φ(x)=x/(1+4x2),即fx=exsiny+y2/2+x/(1+4x2),
从而f(x,y)=exsiny+xy2/2+1/8㏑(1+4x2)+h(y),
再由f(0,y)=y得h(y)=y-siny,故f(x,y)=exsiny+xy2/2+(1/8)㏑(1+4x2)+y-siny.
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