= Solution
{c}
A useful <differentiation under the integral sign> theorem is this: if $f(\cdot,y)$ is integrable, $f$ is differentiable in $y$ almost everywhere, and on every <compact> parameter interval $|\partial_yf(x,y)|$ is bounded by an <integrable function> of $x$, then $F'(y)=\int_I\partial_yf(x,y)dx$, under the usual pointwise <continuity> hypotheses. It follows by applying <dominated convergence> to difference quotients. The theorem can be iterated when a corresponding bound holds for the second derivative.
On a <compact> interval $0<a\leq y\leq b$, the original integrand is bounded by $b/(1+x^4)$, using $|\sin(x^2y)|\leq bx^2$. Its first and second parameter derivatives are
$$
\frac{\cos(x^2y)}{1+x^4},\qquad-\frac{x^2\sin(x^2y)}{1+x^4},
$$
with integrable bounds $(1+x^4)^{-1}$ and $x^2/(1+x^4)$. Thus $F$ is twice differentiable. Subtracting the integrals gives
$$
F''(y)-F(y)=-\int_0^\infty\frac{\sin(x^2y)}{x^2}dx
=-\sqrt y\int_0^\infty\frac{\sin t^2}{t^2}dt.
$$
The final integral is finite: its integrand tends to one at zero and is absolutely bounded by $t^{-2}$ at infinity. Consequently
$$
\boxed{F''(y)-F(y)=c\sqrt y,\quad c=-\int_0^\infty\frac{\sin t^2}{t^2}dt,}
$$
which establishes the required parameter-independent constant without needing an explicit Fresnel integral evaluation.
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