JEE Main202228 Jun 2022Morning ShiftMathematicsDifferential EquationsActual
Let the solution curve y = y x of the differential equation, x x 2 - y 2 + e y x x d y d x = x + x x 2 - y 2 + e y x y pass through the points 1 , 0 and 2 α , α , α > 0 . Then α is equal to
Options
- A1 2 exp π 6 + e - 1
- B1 2 exp π 3 + e - 1
- Cexp π 6 + e + 1
- D2 exp π 3 + e - 1
Correct answer
A. 1 2 exp π 6 + e - 1
Step-by-step solution
Given x x x 2 − y 2 + e y x d y d x = y x x 2 − y 2 + e y x + x Taking x common & cancelling them we get, d y d x × 1 1 − y x 2 + e y x = y x 1 1 − y x 2 + e y x + 1 Let y = v x ⇒ d y d x = v + x d v d x v + x d v d x 1 1 - v 2 + e v = v 1 1 - v 2 + e v + 1 v + x d V d x = v + 1 1 1 - v 2 + e v x d v d x = 1 1 1 - v 2 + e v ⇒ 1 1 - v 2 + e v d v = d x x Integrating both side we get, ⇒ ∫ 1 1 - v 2 + e v d v = ∫ d x x ⇒    sin - 1 v + e v = ln