= Solution
Write $A(t)=\int_0^t a(s)\,ds$. Multiplying the <ordinary differential equation> by the <integrating factor> $e^{A(t)}$ and using the <product rule> gives
$$
\frac{d}{dt}\bigl(e^{A(t)}y(t)\bigr)=e^{A(t)}y(t)^2.
$$
Integration from $0$ to $t$ therefore yields the <Volterra integral equation>
$$
\boxed{y(t)=be^{-A(t)}+\int_0^t e^{-(A(t)-A(s))}y(s)^2\,ds.}
$$
Conversely, a <continuous function> satisfying this <integral equation> becomes continuously differentiable after multiplication by $e^{A(t)}$, by the <fundamental theorem of calculus>, and satisfies the original <initial value problem>.
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