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Differential Equations · Question 1

JEE Advanced 2026 Paper 2MCQMedium📈 Topic in 33% papers+4−1
1

Let y:(- , ) (0, ) be the solution of the differential equation dydx= e5xy3+y3ex+exy4, satisfying y(0)= 1 2. Then the value of y( e 2) is

  1. 5+ 352
  2. 7+ 532
  3. 7+ 532
  4. 5+ 352

Answer

(2) 7+ 532

Solution

Variables separable: dydx= y3(e5x+1)ex(1+y4).

(y-3+y )dy= (e4x+e-x )dx - 12y2+ y22= e4x4-e-x+C.

At x=0, y2= 12: -1+ 14= 14-1+C C=0.

At x= e 2: RHS = 164- 12= 72. Put u=y2: u- 1u=7 u2-7u-1=0 u= 7+ 532 (u>0).

y( e 2)= 7+ 532.

Answer: (B)

JEE Advanced 2026 Paper 2

Keep practising — Differential Equations

2

Let y=f(x) be the real valued function defined on the interval (0, ), satisfying y(1)=0 and the differential equation x dydx=y-x3. Then which of the following statements is (are) TRUE ?

  1. The function f has a local minimum at x= 1 3
  2. The function f has a local maximum at x= 1 3
  3. The function f is increasing in the interval (1, 2)
  4. If g(x)=4x3-5x2+ 32x for x>0, then the number of elements in the set \x (0, ): f(x)=g(x)\ is 2
JEE Advanced 2026 Paper 2
3

For all x > 0, let y1(x), y2(x), and y3(x) be the functions satisfying dy1dx - ( x)2y1 = 0, y1(1) = 5, dy2dx - ( x)2y2 = 0, y2(1) = 13, dy3dx - ( 2 - x3x3 )y3 = 0, y3(1) = 35e, respectively. Then x 0+ y1(x)y2(x)y3(x) + 2xe3x x is equal to .

JEE Advanced 2025 Paper 1
4

Let y(x) be the solution of the differential equation x2 dydx+xy=x2+y2, x> 1e, satisfying y(1)=0. Then the value of 2 (y(e))2y(e2) is .

JEE Advanced 2025 Paper 2
5

Let f(x) be a continuously differentiable function on the interval (0, ) such that f(1)=2 and t x t10f(x)-x10f(t)t9-x9=1 for each x>0. Then, for all x>0, f(x) is equal to

  1. 3111x- 911x10
  2. 911x+ 1311x10
  3. -911x+ 3111x10
  4. 1311x+ 911x10
JEE Advanced 2024 Paper 1

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