Answer
$y=\dfrac{1}{3}-\dfrac{4}{3}e^{-x^{3}}$
Work Step by Step
The standard form of the given equation is:
$y'+3x^2 y=x^2$ ....(1)
The integrating factor is: $e^{\int (3x^2) dx}=e^{x^3}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [e^{x^3}y]' =\int x^2 e^{x^3}dx $
or, $e^{x^3}y=(1/3)e^{x^3}+c$
Applying the initial conditions, we get $c=\dfrac{-4}{3}$
Thus, we have
$e^{x^3}y=(1/3)e^{x^3}-\dfrac{4}{3}$
Hence, $y=\dfrac{1}{3}-\dfrac{4}{3}e^{-x^{3}}$