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Part BCSIR NET June 2025log-x-turns-the-interval-into-0-to-2pi-so-the-eigenvalues-are-n-squared-over-4

Log x turns the interval into 0 to 2pi so the eigenvalues are n squared over 4

Let (λn)n∈N(\lambda_{n})_{n}\in\mathbb{N} be the sequence of eigenvalues of the Sturm-Liouville problem (d/dx)(x dy/dx)+(λ/x)y=0,1<x<) + (\lambda/x)y = 0, 1 < x < e2πe^{2\pi}, y(1) = 0, y(e2π)y(e^{2\pi}) = 0. Then ∑n\sum_{n}₌1∞1/λn_{1}^\infty 1/\lambda_{n} is equal to

  1. A.π2/12\pi^{2}/12
  2. B.2π2/32\pi^{2}/3✓ Correct
  3. C.π2/4\pi^{2}/4
  4. D.π2/16\pi^{2}/16

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Every boundary value problem has a Green's function

More on this topic

The chapter behind this: Sturm–Liouville problems and Green's functions — free to read

From Ordinary Differential Equations › Sturm–Liouville problems and Green's functions

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