Let the solution curve of the differential equation xdy/dx - y = √(y² + 16x²), y(1) = 3 be y = y(x). Then y(2) is equal to:

Differential Equations: JEE Main 2022 Mathematics question

JEE Main 2022 · 29 Jun, Morning shift · Q3619

JEE Main 202229 Jun, Morning shiftMathematicsHardDifferential Equations163× since 2002

Let the solution curve of the differential equation

x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}

y(1) = 3

y = y(x)

. Then y(2) is equal to:

JEE Main 2022 · 29 Jun, Morning shift · Q3619

JEE Main 202229 Jun, Morning shiftMathematicsHardDifferential Equations163× since 2002

Let the solution curve of the differential equation

x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}

y(1) = 3

y = y(x)

. Then y(2) is equal to: