Discussion
The solution of this integral equation is
Answer: D
The solution is \(u(x) = \frac{x}{1 - \lambda}\).
The integral is a constant times \(x\): write \(u(x) = x(1 + \lambda C)\) with \(C = \int_{0}^{1}e^{t}u(t)\,dt\). Substituting back:
\[ \begin{aligned} C &= (1 + \lambda C)\int_{0}^{1}te^{t}\,dt = 1 + \lambda C \\ C &= \frac{1}{1 - \lambda} \\ u(x) &= x\left(1 + \frac{\lambda}{1 - \lambda}\right) = \frac{x}{1 - \lambda} \end{aligned} \]
The same follows from the resolvent of Q.50: \(u = x + \lambda\int_{0}^{1}\frac{xe^{t}}{1-\lambda}t\,dt\), which is \(x + \frac{\lambda x}{1-\lambda}\). \(\frac{x+1}{1-\lambda}\) adds a constant the equation cannot produce, since every term is a multiple of \(x\); \(x^{2}\) cannot appear for the same reason; and \(\frac{x}{1+\lambda^{2}}\) does not satisfy the equation.
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