Discussion
The Volterra integral equation
\[ x(t) = t + \int_{0}^{t} (s - t)\, x(s)\, ds \]
is equivalent to the initial value problem :
Answer: A
The equivalent problem is x'' + x = 0, x(0) = 0, x'(0) = 1.
Differentiate \(x(t) = t + \int_{0}^{t}(s - t)x(s)\, ds\) using Leibniz's rule:
\[ \begin{aligned} x'(t) &= 1 + (t - t)x(t) - \int_{0}^{t} x(s)\, ds \\ &= 1 - \int_{0}^{t} x(s)\, ds \\ x''(t) &= -x(t) \end{aligned} \]
At \(t = 0\) the integrals vanish, so \(x(0) = 0\) and \(x'(0) = 1\). The solution is \(x = \sin t\).
The options with \(x'' - x\) have the wrong sign, those with \(x'\) in place of x differentiate wrongly, and \(x(0) = 1,\ x'(0) = 0\) swaps the initial values.
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