1. Rewrite the differential equation

Given $x\,dy - y\,dx = \sqrt{x^2-y^2}\,dx, \qquad x\ge 1, \qquad y(1)=0.$

Divide by $dx$: $x\frac{dy}{dx}-y=\sqrt{x^2-y^2}.$

So, $x\,y' - y = \sqrt{x^2-y^2}.$

2. Use the substitution $y=vx$

Let $y=vx \quad \Rightarrow \quad y'=v+x\frac{dv}{dx}.$

Substitute into the equation: $x\left(v+x\frac{dv}{dx}\right)-vx=\sqrt{x^2-v^2x^2}.$

The $xv$ terms cancel: $x^2\frac{dv}{dx}=x\sqrt{1-v^2}.$

Since $x\ge 1>0$, $x\frac{dv}{dx}=\sqrt{1-v^2}.$

Separate variables: $\frac{dv}{\sqrt{1-v^2}}=\frac{dx}{x}.$

Integrate: $\sin^{-1}v=\ln x + C.$

Using $y(1)=0$, we get $v(1)=0$, so $\sin^{-1}(0)=\ln 1 + C \Rightarrow 0=0+C \Rightarrow C=0.$

Hence $\sin^{-1}v=\ln x \quad \Rightarrow \quad v=\sin(\ln x).$

Therefore, $\boxed{y=x\sin(\ln x)}.$

3. Set up the required area

The area bounded by $x=1$, $x=e^{\pi}$, $y=0$, and $y=y(x)$ is $A=\int_1^{e^\pi} y\,dx = \int_1^{e^\pi} x\sin(\ln x)\,dx.$

4. Evaluate the integral

Let $t=\ln x \Rightarrow x=e^t, \quad dx=e^t dt.$

Then $x\sin(\ln x)\,dx = e^t\sin t \cdot e^t dt = e^{2t}\sin t\,dt.$

When $x=1$, $t=0$; when $x=e^\pi$, $t=\pi$.

So, $A=\int_0^\pi e^{2t}\sin t\,dt.$

Use the standard result $\int e^{at}\sin bt\,dt = \frac{e^{at}}{a^2+b^2}(a\sin bt - b\cos bt).$

Here $a=2$, $b=1$, so $\int e^{2t}\sin t\,dt = \frac{e^{2t}}{5}(2\sin t - \cos t).$

Thus $A=\left[\frac{e^{2t}}{5}(2\sin t - \cos t)\right]_0^\pi.$

Now evaluate:

• At $t=\pi$: $\sin\pi=0$, $\cos\pi=-1$ $\frac{e^{2\pi}}{5}(0-(-1))=\frac{e^{2\pi}}{5}.$
• At $t=0$: $\sin0=0$, $\cos0=1$ $\frac{1}{5}(0-1)=-\frac{1}{5}.$

Hence $A=\frac{e^{2\pi}}{5}-\left(-\frac{1}{5}\right)=\frac{e^{2\pi}}{5}+\frac{1}{5}.$

So in the form $A=\alpha e^{2\pi}+\beta,$ we have $\alpha=\frac15, \qquad \beta=\frac15.$

Therefore, $10(\alpha+\beta)=10\left(\frac15+\frac15\right)=10\cdot \frac25=4.$

5. Comparison with stored answer

Derived answer: $\boxed{4}$.

Stored correct answer: $4$.

They agree.