Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 15

The general solution is $$ y=e^{-\frac{1}{3}x^3+c}.$$

We have $$ y'+x^2y=0\Longrightarrow \frac{dy}{y}=-x^2dx\\ \Longrightarrow\int \frac{dy}{y}=\int-x^2dx+c\\ \Longrightarrow \ln y=-\frac{1}{3}x^3+c.$$ Hence, the general solution is $$ y=e^{-\frac{1}{3}x^3+c}.$$