Let the solution curve of the differential equation x {\d} y=(\sqrt{x^2+y^2}+y) {\d} x, x > 0, intersect the line x=1 at y=0 and the line x=2 at y=\alpha. Then the value of \alpha is : [28-Jul-2022-Shift-1]

JEE Mains 28-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 1 of 90

Marks: +1, -0

Let the solution curve of the differential equation xdy=(√x2+y2+y)dx,x>0, intersect the line x=1 at y=0 and the line x=2 at y=α. Then the value of α is :

[28-Jul-2022-Shift-1]

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