Let $C_1$ be the curve obtained by the solution of differential equation $2xy\frac{dy}{dx}=y^2-x^2,x>0.$ Let the curve $C_2$ be the solution of $\frac{2xy}{x^2-y^2}=\frac{dy}{dx}.$ If both the curves pass through (1, 1), then the area enclosed by the curves $C_{1}and{C}_2$ is equal to:

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ put $y=vx$

$v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2v{x}^2}=\frac{v^2-1}{2v}x\frac{dv}{dx}=\frac{v^2-1-2v^2}{2v}=\frac{-(v^2+1)}{2v}$

$\Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}\mathcal{l}n(v^2+1)=-\mathcal{l}nx+\mathcal{l}nc\Rightarrow v^2+1=\frac{c}{x}$

$\Rightarrow \frac{y^2}{x^2}+1=\frac{c}{x}\Rightarrow x^2+y^2=cx$ If pass through $\left(1,1\right)$

$\therefore x^2+y^2-2x=0$

Similarly for second differential equation $\frac{dx}{dy}=\frac{x^2-y^2}{2xy}$

Equation of curve is $x^2+y^2-2y=0$

Now the required area is $=(\frac{1}{4}\times \pi \times 1^2-\frac{1}{2}\times 1\times 1)\times 2=(\frac{\pi }{2}-1)$ sq. units