Question
Let y=y(x) be the solution of the differential equation x^ 4 ~d y+ (4 x^ 3 y+2 x ) d x=0, x>0, y ( 2 )=0. Then ^ 4 y ( 3 ) is equal to :
JEE Main
Mathematics
Differential Equations
2026
JEE Main 2026 (23 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. 72
- B. 92
- C. 64
- D. 81
Answer
D. 81
Step-by-step solution
x^4\,dy + 4x^3 y\,dx + 2 x\,dx = 0 d(x^4 y) + 2 x\,dx = 0. Integrating: x^4 y - 2 x = C. y( /2) = 0: ( /2)^4 0 - 2 ( /2) = 0 C = 0. x^4 y = 2 x. At x = /3: ( /3)^4 y( /3) = 2 ( /3) = 1. ^4 y( /3) = 81 ( /3)^4 y( /3) 1 ( /3)^4 ^4 81 ... Directly: y( /3) = 81 ^4 , so ^4 y( /3) = 81.
Related: Mathematics — Differential Equations · All PYQ Banks