Part BCSIR NET December 2024one-homogeneous-solution-is-constant

One homogeneous solution is constant

Let $\lambda \in \mathbb{R}$ , and $K : [0, 1] \times [0, 1] \to \mathbb{R}$ be a function such that every solution of the boundary value problem $(d^{2}u/$ dx $^{2})(x) + \lambda{}u(x) = 0; ($ du/dx)(0) = u(0), (du/dx)(1) = 0 satisfies the integral equation $u(x) + \lambda\int_{0}^{1}K(x, t)u(t)$ dt = 0. Then

A.K(x, t) = (1 + x)(1 − t) for 0 ≤ x ≤ t ≤ 1, and (1 + t)(1 − x) for 0 ≤ t < x ≤ 1
B.K(x, t) = −1 − x for 0 ≤ x ≤ t ≤ 1, and −1 − t for 0 ≤ t < x ≤ 1✓ Correct
C. $K(x, t) = 1 - x^{2}$ for 0 ≤ x ≤ t ≤ 1, and $1 - t^{2}$ for 0 ≤ t < x ≤ 1
D.K(x, t) = (1 + x)(t − 1) for 0 ≤ x ≤ t ≤ 1, and (1 + t)(x − 1) for 0 ≤ t < x ≤ 1

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

One homogeneous solution is constant

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