Answer
$y=\dfrac{1}{3}-\dfrac{4}{3}e^{-x^{3}}$
Work Step by Step
Here, we have $y'+3(x^2y)=x^2$
The integrating factor is: $I=e^{\int (3x^2) dx}=e^{x^3}$
Now, $e^{x^3}[y'+3x^2 y]=(e^{x^3})(x^2)$
$\int [e^{x^3}y]' =\int (x^2) (e^{x^3}) dx $
$\implies e^{x^3}y=(\dfrac{1}{3})e^{x^3}+c$
Now, apply the initial conditions.
Thus, we have $c=\dfrac{-4}{3}$
$(e^{x^3})(y)=(\dfrac{1}{3})e^{x^3}-(\dfrac{4}{3})$
So, $y=\dfrac{1}{3}-\dfrac{4}{3}e^{-x^{3}}$
