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Question

Find the solution of (x2yx2)dydx+(y2+xy2)=0

A
logxky=x+yxy
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B
logxky=xyxy
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C
logxky=xyy
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D
logxky=x+yy
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Solution

The correct option is A logxky=x+yxy
x2(1y)dydx+y2(1+x)=0
or 1yy2dy+1+xx2dx=0.
or (1y21y)dy+(1x2+1x)dx=0.
Integrating, we get
logxlogy(1x+1y)=c=logk (say)
or logxky=x+yxy

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