Discussion
The solution of the homogeneous integral equation
\[ x(t) = \lambda \int_{0}^{1} t s\, x(s)\, ds, \]
for \(\lambda \pm 0\), is :
Answer: C
The solution is \(x(t) = t\), up to a constant multiple. (The stem prints '\(\lambda \pm 0\)' where \(\lambda \neq 0\) is meant.)
The kernel \(ts\) is separable. Pulling \(t\) out of the integral:
\[ x(t) = \lambda t \int_{0}^{1} s\, x(s)\, ds = c\,\lambda t \]
so every solution is a multiple of \(t\). Substituting \(x = t\) back gives \(\int_{0}^{1} s^{2}\, ds = 1/3\), so a nonzero solution exists exactly for the eigenvalue \(\lambda = 3\), and the eigenfunction is \(t\).
The forms \(t^{2}\), \(1 + t\) and \(1 + t^{2}\) cannot be solutions, because the right-hand side of the equation is always a multiple of \(t\).
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