MathematicsMedium38×since 2002Q3628Let the solution curve of the differential equation x dy=(x2+y2+y)dx,x>0x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0x dy=(x2+y2+y)dx,x>0, intersect the line x=1x=1x=1 at y=0y=0y=0 and the line x=2x=2x=2 at y=αy=\alphay=α. Then the value of α\alphaα is :A12\frac{1}{2}21B32\frac{3}{2}23C−-−32\frac{3}{2}23D52\frac{5}{2}25Check answerSkip