MathematicsHard38×since 2002Q3639Let y=y(x)y=y(x)y=y(x) be a solution curve of the differential equation. (1−x2y2)dx=ydx+xdy\left(1-x^{2} y^{2}\right) d x=y d x+x d y(1−x2y2)dx=ydx+xdy. If the line x=1x=1x=1 intersects the curve y=y(x)y=y(x)y=y(x) at y=2y=2y=2 and the line x=2x=2x=2 intersects the curve y=y(x)y=y(x)y=y(x) at y=αy=\alphay=α, then a value of α\alphaα is :A1+3e22(3e2−1)\frac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}2(3e2−1)1+3e2B3e22(3e2−1)\frac{3 e^{2}}{2\left(3 e^{2}-1\right)}2(3e2−1)3e2C1−3e22(3e2+1)\frac{1-3 e^{2}}{2\left(3 e^{2}+1\right)}2(3e2+1)1−3e2D3e22(3e2+1)\frac{3 e^{2}}{2\left(3 e^{2}+1\right)}2(3e2+1)3e2Check answerSkip