Discussion
The initial value problem corresponding to the integral equation :
\[ x(t) = x - \cos x + \int_{0}^{x} (x - t)\, x(t)\, dt \]
is :
Answer: C
The initial value problem is \(x'' - x = \cos t\), \(x(0) = -1\), \(x'(0) = 1\).
The stem uses \(x\) both for the unknown function and the variable; the options show it means \(x(t) = t - \cos t + \int_{0}^{t}(t - s)x(s)\,ds\). Put \(t = 0\): \(x(0) = 0 - 1 + 0 = -1\). Differentiate using Leibniz's rule (the integrand vanishes at \(s = t\)):
\[ \begin{aligned} x'(t) &= 1 + \sin t + \int_{0}^{t} x(s)\,ds \\ x''(t) &= \cos t + x(t) \end{aligned} \]
So \(x'(0) = 1\) and \(x'' - x = \cos t\). Options A and B have the wrong forcing term and \(x(0) = 1\), which contradicts \(x(0) = -1\), and option D adds an \(x'\) term that the single integral cannot produce.
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