= Solution
One version of <dominated differentiation under the integral sign> is the following. Let the parameter range be open. Suppose $f(\cdot,y)$ is integrable, $\partial_yf(x,y)$ exists for almost every $x$, and near each parameter value its absolute value is bounded by an integrable <function> of $x$ independent of $y$. Then
$$
F'(y)=\int_I\partial_yf(x,y)\,dx.
$$
This follows by applying the <dominated convergence theorem> to difference quotients, bounded by that same integrable <function> through the <mean value theorem>.
On any <compact> parameter <interval> $0<y\leq B$, the original integrand is bounded by $B/(1+x^4)$, using $|\sin(x^2y)|\leq Bx^2$. Its first and second parameter derivatives are
$$
\frac{\cos(x^2y)}{1+x^4},\qquad
-\frac{x^2\sin(x^2y)}{1+x^4}.
$$
Their absolute values are bounded respectively by $(1+x^4)^{-1}$ and $x^2/(1+x^4)$, both integrable on $(0,\infty)$. Applying the theorem twice proves that $F$ is twice differentiable and gives
$$
F''(y)-F(y)
=-\int_0^\infty\frac{\sin(x^2y)}{x^2}\,dx.
$$
This last integral is absolutely convergent: its integrand is bounded near zero and bounded by $x^{-2}$ at infinity. Substituting $t=x\sqrt y$ yields
$$
\boxed{F''(y)-F(y)=c\sqrt y},\qquad
c=-\int_0^\infty\frac{\sin(t^2)}{t^2}\,dt.
$$
The expression for $c$ is finite and independent of $y$, which is all that is required.
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