Let $$y = y(x)$$ be the solution of the differential equation $$xdy - ydx = \sqrt{(x^2 - y^2)}dx$$, $$x \ge 1$$, with $$y(1) = 0$$. If the area bounded by the line $$x = 1$$, $$x = e^\pi$$, $$y = 0$$ and $$y = y(x)$$ is $$\alpha e^{2\pi} + \beta$$, then the value of $$10(\alpha + \beta)$$ is equal to ___.

Rewrite the differential equation $$x\,dy - y\,dx = \sqrt{x^2 - y^2}\,dx$$ by dividing both sides by $$x^2$$: $$\dfrac{x\,dy - y\,dx}{x^2} = \dfrac{\sqrt{x^2 - y^2}}{x^2}\,dx$$, which gives $$d\!\left(\dfrac{y}{x}\right) = \dfrac{1}{x}\sqrt{1 - \left(\dfrac{y}{x}\right)^2}\,dx$$.

Substituting $$v = \dfrac{y}{x}$$, we get $$\dfrac{dv}{\sqrt{1-v^2}} = \dfrac{dx}{x}$$. Integrating: $$\sin^{-1}(v) = \ln x + C$$. Using $$y(1) = 0$$, i.e., $$v = 0$$ when $$x = 1$$: $$\sin^{-1}(0) = \ln 1 + C$$, so $$C = 0$$. Therefore $$y = x\sin(\ln x)$$.

The required area is $$A = \int_1^{e^\pi}x\sin(\ln x)\,dx$$. Substituting $$t = \ln x$$ (so $$x = e^t$$, $$dx = e^t\,dt$$): $$A = \int_0^{\pi}e^t\sin t \cdot e^t\,dt = \int_0^{\pi}e^{2t}\sin t\,dt$$.

Using the standard formula $$\int e^{at}\sin t\,dt = \dfrac{e^{at}(a\sin t - \cos t)}{a^2+1}$$, with $$a = 2$$: $$A = \left[\dfrac{e^{2t}(2\sin t - \cos t)}{5}\right]_0^{\pi} = \dfrac{e^{2\pi}(0-(-1))}{5} - \dfrac{1(0-1)}{5} = \dfrac{e^{2\pi}}{5} + \dfrac{1}{5}$$.

Comparing with $$\alpha e^{2\pi} + \beta$$, we get $$\alpha = \dfrac{1}{5}$$ and $$\beta = \dfrac{1}{5}$$. Therefore $$10(\alpha + \beta) = 10 \cdot \dfrac{2}{5} = 4$$.