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Step 1: Write down the first differential equation
The first curve, denoted by $C_{1}$, is obtained by solving the differential equation
$$2xy \frac{dy}{dx} \;=\; y^2 \;-\; x^2,\quad x > 0.$$
We rearrange it to the form
$$\frac{dy}{dx} \;=\; \frac{y^2 - x^2}{2xy}.$$
Step 2: Use the substitution $y = vx$
Let us substitute $y = v x$. Then $y' = \frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation above gives
$$v + x \frac{dv}{dx} \;=\; \frac{v^2 x^2 - x^2}{2v \, x^2} \;=\; \frac{v^2 - 1}{2v}.$$
Step 3: Simplify to find $v$ in terms of $x$
Rearranging,
\[
x \,\frac{dv}{dx} \;=\; \frac{v^2 - 1}{2v} \;-\; v
\;=\; \frac{v^2 - 1 - 2v^2}{2v}
\;=\; -\,\frac{v^2 + 1}{2v}.
\]
Thus,
\[
\frac{2v}{v^2 + 1} \,dv
\;=\; -\,\frac{dx}{x}.
\]
Step 4: Integrate both sides
Integrating, we have
\[
\int \frac{2v}{v^2 + 1}\, dv
\;=\;
\int -\,\frac{dx}{x}.
\]
On the left side,
\[
\int \frac{2v}{v^2 + 1}\, dv
\;=\; \ln\bigl(v^2 + 1\bigr),
\]
while on the right side,
\[
\int -\,\frac{dx}{x}
\;=\; -\,\ln x.
\]
Thus,
\[
\ln\bigl(v^2 + 1\bigr) = -\,\ln x + \ln C,
\]
for some constant $C$. This can be rewritten as
\[
v^2 + 1
\;=\; \frac{C}{x}.
\]
Step 5: Express in terms of $x$ and $y$ and apply the initial condition
Recall that $v = \frac{y}{x}$, so $v^2 = \frac{y^2}{x^2}$. Hence,
\[
\frac{y^2}{x^2} + 1
= \frac{C}{x}
\;\;\Longrightarrow\;\;
x^2 + y^2 = Cx.
\]
Since $C_{1}$ passes through the point $(1, 1)$, substitute $x=1$ and $y=1$ to find $C$:
\[
1^2 + 1^2 = C \cdot 1
\;\;\Longrightarrow\;\;
2 = C.
\]
So the equation of the first curve is
\[
x^2 + y^2 = 2x
\;\;\Longrightarrow\;\;
x^2 + y^2 - 2x = 0.
\]
Step 6: Solve the second differential equation
The second curve, denoted by $C_{2}$, is given by the differential equation
$$\frac{dy}{dx} \;=\; \frac{2xy}{x^2 - y^2}.$$
One can similarly solve (or accept the provided solution) to find that its general form is
\[
x^2 + y^2 = 2y.
\]
Passing through $(1, 1)$ confirms that no further constant shift is needed, and so the equation of $C_{2}$ becomes
\[
x^2 + y^2 - 2y = 0.
\]
Step 7: Identify the region to find the enclosed area
We have two curves:
$C_{1}$: $x^2 + y^2 - 2x = 0.$
$C_{2}$: $x^2 + y^2 - 2y = 0.$
The enclosed region is the area between these curves from $x=0$ to $x=1$, where each curve can be expressed appropriately in terms of $y$ or $x$.
Step 8: Set up the integral for the area
According to a standard procedure for area between curves, one needs to integrate the difference of the functions (top curve minus bottom curve) over the appropriate interval. From the provided work, the required area is:
The integral shown is
\[
\text{Area}
\;=\;
2 \int_{0}^{1} \bigl(\sqrt{2x - x^2} \;-\; x\bigr)\,dx.
\]
Step 9: Evaluate the integral
Upon evaluating this definite integral, it is found (as per the provided calculations) to be
\[
\frac{\pi}{2} - 1.
\]
Hence, the area enclosed by the two curves $C_{1}$ and $C_{2}$ is
\[
\boxed{\frac{\pi}{2} \;-\; 1}.
\]
