Discussion
The boundary value problem
\[ y'' + \lambda y = x, \; y(0) = 0, \; y'(1) = 0 \]
is equivalent to the Fredholm integral equation.
where \(k(x, t) = \begin{cases} t; & 0 \lt t \lt x \\ x; & x \lt t \lt 1 \end{cases}\)
Answer: A
The equivalent equation is \(y(x) = \frac{1}{6}(x^{3} - 3x) + \lambda\int_{0}^{1} k(x, t)\, y(t)\, dt\).
Write the problem as \(y'' = x - \lambda y\) with \(y(0) = 0\), \(y'(1) = 0\). For these boundary conditions the Green's function of \(y'' = g\) is \(-\min(x, t) = -k(x, t)\), so
\[ \begin{aligned} y(x) = &-\int_{0}^{1} k(x, t)\,t\,dt \\ &+ \lambda\int_{0}^{1} k(x, t)\,y(t)\,dt \end{aligned} \]
The first integral is
\[ \begin{aligned} \int_{0}^{x} t^{2}\,dt + x\int_{x}^{1} t\,dt &= \frac{x^{3}}{3} + \frac{x(1 - x^{2})}{2} \\ &= \frac{x}{2} - \frac{x^{3}}{6} \end{aligned} \]
so the free term is \(\frac{x^{3}}{6} - \frac{x}{2} = \frac{1}{6}(x^{3} - 3x)\). Check: it vanishes at 0 and its derivative \(\frac{x^{2} - 1}{2}\) vanishes at 1. The other free terms fail these boundary checks, and their kernels carry extra factors.
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