Jee Main
Differential Equations
JEE Main 2022 (Online) 29th June Morning Shift
MCQ+4 / -12022
Let the solution curve of the differential equation
\(x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}\), \(y(1) = 3\) be \(y = y(x)\). Then y(2) is equal to:
