Before You Begin

Question #40608

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = y(x)$ be the solution to the differential equation $x(x+y) \frac{dy}{dx} = y(x+y) + x^2 e^{-y/x}, \quad x > 0$ satisfying the boundary condition $y(1) = \ln(K)$ , where $K = \prod_{m=1}^{\infty} 2^{ \frac{E_2(2^m!) - 2 E_2(2^{m-1}!)}{2^m} }$ and $E_p(n!)$ denotes the exponent of the highest power of prime $p$ dividing $n!$ . If $W(x)$ is the Lambert W function satisfying $W(x)e^{W(x)} = x$ , what is the value of $y\left(\frac{e}{4}\right)$ ?

Options

A.
$\frac{e}{4} W(1)$
B.
$\frac{e}{4} W(1 - \ln 4)$
C.
$\frac{e}{2} W(1)$
D.
e W(1)

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Exponent of Prime in n!|Substitution Logic in Homogeneous DE|Integration Constant Trap

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of homogeneous differential equations and discrete number theory (Legendre's Formula).

The primary trap lies in the evaluation of the integration constant $C$ .

Students often approximate $E_2(2^m!) \approx 2^m$ , ignoring the discrete $-1$ jump in the exact formula $E_2(2^m!) = 2^m - 1$ .

This approximation incorrectly zeroes out the numerator, leading to $K=1$ and $C=0$ , which cascades into the trap answer $\frac{e}{4} W(1 - \ln 4)$ .

Furthermore, the DE requires a precise substitution $y=vx$ and exact differential integration of $(1+v)e^v$ , punishing any algebraic missteps before the boundary condition is even applied.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Exponent of Prime in n!
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free