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B
logxky=x−yxy
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C
logxky=x−yy
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D
logxky=x+yy
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Solution
The correct option is Alogxky=x+yxy x2(1−y)dydx+y2(1+x)=0 or 1−yy2dy+1+xx2dx=0. or (1y2−1y)dy+(1x2+1x)dx=0. Integrating, we get logx−logy−(1x+1y)=c=logk (say) or logxky=x+yxy