Question
Let y=y(x) be the solution curve of the differential equation (1+x^ 2 ) d y+ (y- ^ -1 x ) d x=0, y(0)=1. Then the value of y(1) is :
JEE Main
Mathematics
Differential Equations
2026
JEE Main 2026 (21 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. 4 e ^ / 4 + 2 -1
- B. 2 e ^ / 4 + 4 -1
- C. 2 e ^ / 4 - 4 -1
- D. 4 e^ / 4 - 2 -1
Answer
B. 2 e ^ / 4 + 4 -1
Step-by-step solution
Rewrite as dy dx + y 1+x^2 = ^ -1 x 1+x^2 . IF = e^ ^ -1 x . y e^ ^ -1 x = ^ -1 x 1+x^2 e^ ^ -1 x dx. Let t = ^ -1 x: = te^t dt = e^t(t-1) + C. y = ^ -1 x - 1 + Ce^ - ^ -1 x . Using y(0) = 1: C = 2. y(1) = 4 - 1 + 2 e^ /4 .
Related: Mathematics — Differential Equations · All PYQ Banks