Let C1 be the curve obtained by solution of differential equation 2xydydx=y2−x2,x>0. Let the curve C2 be the solution of 2xyx2−y2=dydx. If both the curves pass through (1,1), then the area enclosed by the curves C1 and C2 is equal to :
A
π2−1
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B
π4+1
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C
π−1
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D
π+1
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Solution
The correct option is Aπ2−1 2xydydx=y2−x2 ⇒dydx=y2−x22xy
Put y=vx v+xdvdx=v2x2−x22vx2=v2−12v ⇒xdvdx=v2−1−2v22v=−(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+y2−2x=0
Similarly, for second differential equation dxdy=x2−y22xy,
equation of curve is x2+y2−2y=0
Required area =2×(14×π×12−12×1×1) =(π2−1) sq. units