Discussion
The initial value problem
\(y'' + y = 0\), \(y(0) = y'(0) = 0\)
is equivalent to the integral equation :
Answer: C
The equivalent integral equation is \(y(t) = \int_{0}^{t} (s - t)\, y(s)\, ds\).
Write \(y'' = -y\). Integrating twice from 0 and using \(y(0) = y'(0) = 0\):
\[ \begin{aligned} y(t) &= \int_{0}^{t} (t - s)\, y''(s)\, ds \\ &= -\int_{0}^{t} (t - s)\, y(s)\, ds = \int_{0}^{t} (s - t)\, y(s)\, ds \end{aligned} \]
The first step is the standard formula for a repeated integral. Option A has the sign wrong, which would correspond to \(y'' = y\), and options B and D carry an extra factor \(t\) that the double integration does not produce. (Of course the only solution here is \(y \equiv 0\), but the question asks for the equivalent equation.)
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