Discussion
The initial value problem
\[ x''(t) - x(t) = 0,\ x(0) = 1,\ x'(0) = 1 \]
is equivalent to the integral equation :
Answer: D
The equivalent integral equation is \(x(t) = 1 + t + \int_{0}^{t} (t - s)\, x(s)\, ds\).
Integrate \(x'' = x\) twice from 0, using the initial conditions:
\[ \begin{aligned} x'(t) &= 1 + \int_{0}^{t} x(s)\, ds \\ x(t) &= 1 + t + \int_{0}^{t}\!\int_{0}^{\tau} x(s)\, ds\, d\tau \\ &= 1 + t + \int_{0}^{t} (t - s)\, x(s)\, ds \end{aligned} \]
The last step is the standard formula that turns a repeated integral into a single one with kernel \(t - s\).
Check: \(x = e^{t}\) solves the initial value problem, and \(1 + t + \int_{0}^{t}(t - s)e^{s} ds = 1 + t + (e^{t} - 1 - t) = e^{t}\). The other options use \((t + s)\) or squared kernels, or drop the \(t\) that comes from \(x'(0) = 1\).
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