MathematicsMedium38×since 2002Q3622Let the solution curve y=y(x)y = y(x)y=y(x) of the differential equation [xx2−y2+eyx]xdydx=x+[xx2−y2+eyx]y\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y[x2−y2x+exy]xdxdy=x+[x2−y2x+exy]y pass through the points (1, 0) and (2α\alphaα, α\alphaα), α\alphaα > 0. Then α\alphaα is equal toA12exp(π6+e−1){1 \over 2}\exp \left( {{\pi \over 6} + \sqrt e - 1} \right)21exp(6π+e−1)B12exp(π6+e−1){1 \over 2}\exp \left( {{\pi \over 6} + e - 1} \right)21exp(6π+e−1)Cexp(π6+e+1)\exp \left( {{\pi \over 6} + \sqrt e + 1} \right)exp(6π+e+1)D2exp(π3+e−1)2\exp \left( {{\pi \over 3} + \sqrt e - 1} \right)2exp(3π+e−1)Check answerSkip