JEE Main202621 January 2026Morning ShiftMathematicsDifferential EquationsActual
Let y=y(x) be the solution curve of the differential equation (1+x² ) d y+ (y- ⁻¹ x ) d x=0, y(0)=1 . Then the value of y(1) is :
Options
- A4 e ^ / 4 + 2 -1
- B2 e ^ / 4 + 4 -1
- C2 e ^ / 4 - 4 -1
- D4 e^ / 4 - 2 -1
Correct answer
B. 2 e ^ / 4 + 4 -1
Step-by-step solution
Rewrite as dy dx + y 1+x^2 = ⁻¹x 1+x^2 . IF = e^ ⁻¹x . y e^ ⁻¹x = ⁻¹x 1+x^2 e^ ⁻¹x dx . Let t = ⁻¹x : = te^t dt = e^t(t-1) + C . y = ⁻¹x - 1 + Ce^ - ⁻¹x . Using y(0) = 1 : C = 2 . y(1) = 4 - 1 + 2 e^ /4 .