QJEEM21Mad6ac: JEE Main 2021 Mathematics - Class 12 JEEM21Mad6ac Solution: Let y = y(x) be the solution of the differential equation xdy…

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

dividing both sides by x², we get

${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$

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

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

Integrating both side, we get

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

${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$

Given, y(1) = 0 $\Rightarrow$ at x = 1, y = 0

$\therefore$ $\Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$

$\Rightarrow$ C = 0

$\therefore$ ${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$

$\Rightarrow$ y = x sin(ln(x))

$\therefore$ Area $= \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$

Let, lnx = t

$\Rightarrow$ x = e^t

$\Rightarrow$ dx = e^t dt

New lower limit, t = ln(1) = 0

and upper limit t = ln $({e^{\pi {} }})$ = ${\pi {} }$

$\therefore$ Area = $\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$

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

$= \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$

$= {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$

$= {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$

$= \alpha {e^{2\pi {} }} + \beta$

$\therefore$ $\alpha = {1 \over 5},\beta = {1 \over 5}$

So, $10(\alpha + \beta ) = 4$