Let C_1 be the curve obtained by the solution of differential equation 2xy dy dx=y^2-x^2,x>0. Let the curve C_2 be the solution of 2xy x^2-y^2= dy…

$$\frac{dy}{dx}=\frac{{y}^{2}-{x}^{2}}{2xy},x\in (0,\infty )$$dxdy​=2xyy2−x2​,x∈(0,∞)

put $y=vx$y=vx

$$x\frac{dv}{dx}+v=\frac{{v}^{2}-1}{2v}$$xdxdv​+v=2vv2−1​

$$\frac{2v}{{v}^{2}+1}dv=-\frac{dx}{x}$$v2+12v​dv=−xdx​

Integrate,

$$ln({v}^{2}+1)=-lnx+C$$ln(v2+1)=−lnx+C

$$ln(\frac{{y}^{2}}{{x}^{2}}+1)=-lnx+C$$ln(x2y2​+1)=−lnx+C

put $x=1,y=1,C=ln2$x=1,y=1,C=ln2

$$ln(\frac{{y}^{2}}{{x}^{2}}+1)=-lnx+ln2$$ln(x2y2​+1)=−lnx+ln2

$\Rightarrow {x}^{2}+{y}^{2}-2x=0$⇒x2+y2−2x=0 (Curve ${C}_{1}$C1​)

Similarly,

$$\frac{dy}{dx}=\frac{2xy}{{x}^{2}-{y}^{2}}$$dxdy​=x2−y22xy​

Put $y=vx$y=vx

$${x}^{2}+{y}^{2}-2y=0$$x2+y2−2y=0

Required area $\displaystyle =2{\int }_{0}^{1}(\sqrt{2x-{x}^{2}}-x)dx=\frac{\pi }{2}-1$=2∫01​(2x−x2​−x)dx=2π​−1 sq. units