Let y = y ( x ) be the solution of the differential equation x d y - y d x = x 2 - y 2 d x ,   x &#8805…

Given $xdy-ydx=\sqrt{x^2-y^2}dx$

$\Rightarrow \frac{xdy-ydx}{x^2}=\frac{1}{x}\sqrt{1-\frac{y^2}{x^2}}dx$

$\Rightarrow \int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}=\int \frac{dx}{x}$

$\Rightarrow {\sin }^{-1}\left(\frac{y}{x}\right)=ln\left|x\right|+c$

at $x=1,y=0\Rightarrow c=0$

$y=x\sin \left(lnx\right)$

$A={\int }_1^{e^{\pi }}x\sin \left(lnx\right)dx$

$x=e^t,dx=e^{t}dt\Rightarrow {\int }_0^{\pi }{e}^{2t}\sin \left(t\right)dt=A$

$\alpha e^{2\pi }+\beta ={\left(\frac{e^{2t}}{5}(2\sin t-\cos t)\right)}_0^{\pi }=\frac{1+e^{2\pi }}{5}$

$\alpha =\frac{1}{5}, \beta =\frac{1}{5}$

So, $10(\alpha +\beta )=4$