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 __________.

$xdy - ydx = \sqrt {{x^2} - {y^2}} dx$<br><br>dividing both sides by x², we get<br><br>${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$<br><br>$$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$$<br><br>$$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$$<br><br>Integrating both side, we get<br><br>$$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $$<br><br>${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$<br><br>Given, y(1) = 0 $\Rightarrow$ at x = 1, y = 0<br><br>$\therefore$ $\Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$<br><br>$\Rightarrow$ C = 0<br><br>$\therefore$ ${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$<br><br>$\Rightarrow$ y = x sin(ln(x))<br><br>$\therefore$ Area $= \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$<br><br>Let, lnx = t<br><br>$\Rightarrow$ x = e^t<br><br>$\Rightarrow$ dx = e^t dt<br><br>New lower limit, t = ln(1) = 0<br><br>and upper limit t = ln$({e^{\pi {} }})$ = ${\pi {} }$<br><br>$\therefore$ Area = $\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$<br><br>$= \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$<br><br>$$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$$<br><br>$= {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$<br><br>$= {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$<br><br>$= \alpha {e^{2\pi {} }} + \beta$<br><br>$\therefore$ $\alpha = {1 \over 5},\beta = {1 \over 5}$<br><br>So, $10(\alpha + \beta ) = 4$