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Question

Let C1 be the curve obtained by solution of differential equation 2xydydx=y2x2, x>0. Let the curve C2 be the solution of 2xyx2y2=dydx. If both the curves pass through (1,1), then the area enclosed by the curves C1 and C2 is equal to :

A
π21
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B
π4+1
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C
π1
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D
π+1
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Solution

The correct option is A π21
2xydydx=y2x2
dydx=y2x22xy
Put y=vx
v+xdvdx=v2x2x22vx2=v212v
xdvdx=v212v22v=(v2+1)2v
2vv2+1dv=dxx
Integrating both sides,
ln(v2+1)=lnx+lnc
v2+1=cx
y2x2+1=cx
x2+y2=cx
It passes through (1,1)
x2+y22x=0

Similarly, for second differential equation dxdy=x2y22xy,
equation of curve is x2+y22y=0


Required area
=2×(14×π×1212×1×1)
=(π21) sq. units

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