= Solution
An <integrating factor> is
$$
\mu(x)=\exp\left(\int3x^2\,dx\right)=e^{x^3}.
$$
Multiplication by this <integrating factor> turns the <first-order linear differential equation> into $(e^{x^3}y)'=x^2e^{x^3}$. Integrating from zero to $x$ gives
$$
e^{x^3}y(x)-a=\frac13(e^{x^3}-1).
$$
Thus the <initial value problem> has the solution
$$
\boxed{y(x)=\frac13+\left(a-\frac13\right)e^{-x^3},\qquad \lim_{x\to+\infty}y(x)=\frac13.}
$$
Indeed, direct <differentiation> gives $y'=-3x^2(a-1/3)e^{-x^3}$, which verifies both the <differential equation> and the <initial condition>.
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