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Q.

The curve satisfying the differential equation (x2–y2) dx + 2xy dy = 0 and passes through point (1, 1) is a circle of area ....... .

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answer is 3.14.

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Detailed Solution

x2y2dx+2xydy=0dydx=y2x22xy 

put y=vx dydx=v+xdvdxv+xdvdx=v2x2x22vx2v+xdvdx=v212v xdvdx=v212v2vdvv2+1=dxx

integrate, lnv2+1=ln|x|+lncy2x2+1=cx

passing through (1,1) then

c=2x2+y22x=0

it is a circle having radius 1

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