Answer
$y=-\frac{1}{3}+Ce^{x^3}$
Work Step by Step
The integrating factor for the given linear differential equation is $I=e^{\int -3x^2 dx}=e^{-x^3}$.
Solve for $y$:
$y'-3x^2y=x^2$ (Multiply by the integrating factor)
$y'e^{-x^3}+(-3x^2)e^{-x^3}y=x^2e^{-3x^2}$
$\frac{d(ye^{-x^3})}{dx}=x^2e^{-x^3}$ (Separate the variable)
$d(ye^{-x^3})=x^2e^{-x^3}dx$ (Integrate)
$\int d(ye^{-x^3})=\int x^2e^{-x^3}dx$
$ye^{-x^3}=-\frac{e^{-x^3}}{3}+C$ (Multiply by $e^{x^3}$)
$y=-\frac{1}{3}+Ce^{x^3}$
Thus, the solution is $y=-\frac{1}{3}+Ce^{x^3}$.
