Chapter 6 - Differential Equations - 6.2 Exercises - Page 412: 8

$y=Ce^{\frac{1}{2} x^2} -1$

Begin by separating the variable to each side in terms of x's and y's. $\frac{dy}{dx}=x(1+y)$ $ \frac{1}{1+y} dy= x dx$ $ \int \frac{1}{1+y} dy = \int xdx$ Integrate $ \ln{|1+y|} = \frac{1}{2}x^2 +C$ $1+y= e^{\frac{1}{2}x^2 +C}$ Let $e^C = C$ $ y= Ce^{\frac{1}{2}x^2} -1$