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Differential Equations appeared 46 times across 3 years — 5.3% of Mathematics. This question is from Solution of Differential Equations.

Year 2026 2025 2024 Total
Questions 13 17 16 46

Let f be a twice differentiable non-negative function such that (f(x))² = 25 + ∫₀x ( (f(t))² + (f'(t))² ) dt. Then the mean of f( ₑ(1)), f( ₑ(2)), ..., f( ₑ(625)) is equal to _____.

Numerical Answer Type:
Enter a numerical value Answer: 1565 to 1565 +4 marks

Solution & Explanation

Related Formula
(d)/(dx)∫₀^x g(t)dt = g(x) (Leibniz Rule)
Core Logic

Differentiate both sides with respect to x using Leibniz rule:

2f(x)f'(x) = f²(x) + (f'(x))²

Rearranging yields a perfect square:

(f'(x))² - 2f(x)f'(x) + f²(x) = 0 (f'(x) - f(x))² = 0

Thus, f'(x) = f(x).

Step 1: Solve the Differential Equation

Integrate (f'(x))/(f(x)) = 1:

ln f(x) = x + C f(x) = ex+C = k e^x

To find k, plug x = 0 into the original integral equation:

(f(0))² = 25 + ∫₀⁰ dt = 25

Since f is non-negative, f(0) = 5. Thus, k = 5 and f(x) = 5e^x.

Step 2: Calculate Sequence Values

Evaluate f(ln n) for n=1, 2, …, 625:

f(ln n) = 5eln n = 5n
Step 3: Calculate the Mean

Mean of f(ln 1), f(ln 2), , f(ln 625):

Mean = Σn=1⁶²⁵ 5n625 = (5 (1 + 2 + + 625))/(625) = (5)/(625) × (625 × 626)/(2) = (5 × 626)/(2) = 5 × 313 = 1565
Pattern Recognition

An integral equation with x as an upper limit immediately demands Leibniz differentiation. If it results in a perfect square differential relation, an exponential solution map natively emerges.

Chapter Mix

Class 12 Maths: Differential Equations

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 10

Q11 jee_main_2024_31_jan_morning Linear Differential Equations
Let y = y(x) be the solution of the differential equation (dy)/(dx) = (( x) + y)/( x( x - x x)), x in (0, (π)/(2)) satisfying the condition y((π)/(4)) = 2. Then, y((π)/(3)) is
  • A. √(3)(2 + ₑ√(3))
  • B. √(3)2(2 + ₑ 3)
  • C. √(3)(1 + 2 ₑ 3)
  • D. √(3)(2 + ₑ 3)

Solution

Core Logic
(dy)/(dx) = (( x)/( x) + y)/( x ((1)/( x) - ( ² x)/( x))) = ( x + y x)/( x (1 - ² x)) (dy)/(dx) = ( x + y x)/( x ² x) = ² x + (2y)/( 2x) (dy)/(dx) - 2 (2x)y = ² x
Step 1: Integrating Factor

This is an LDE of form (dy)/(dx) + Py = Q.

I.F. = e∫ -2 (2x) dx

Let 2x = t 2dx = dt.

I.F. = e-∫ t dt = e-ln| (t/2)| = e-ln| x| = (1)/(| x|)
Step 2: Solution of LDE
y(I.F.) = ∫ Q(I.F.) dx + C y(1)/( x) = ∫ ² x (1)/( x) dx + C

Let x = t ² x dx = dt.

y(1)/( x) = ∫ (dt)/(t) + C = ln| x| + C y = x(ln| x| + C)
Step 3: Boundary Value

Given y(π/4) = 2:

2 = 1(ln 1 + C) C = 2

Thus, y = x (ln| x| + 2). At x = π/3:

y(π/3) = √(3)(ln√(3) + 2)
Chapter Mix

Class 12 Maths: Differential Equations

More Differential Equations Questions — jee_main_2026_23_january_morning

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)