Answer
\[y^2+3y-x^3+x=C\]
Work Step by Step
\[y'=\frac{3x^2-1}{3+2y}\;\;\;...(1)\]
\[\frac{dy}{dx}=\frac{3x^2-1}{3+2y}\]
Separating variables,
\[(3+2y)dy=(3x^2-1)dx\]
Integrating,
\[\int(3+2y)dy=\int(3x^2-1)dx+C\]
Where $C$ is constant of integration
\[3y+y^2=x^3-x+C\]
\[y^2+3y-x^3+x=C\]
Hence general solution of (1) is
\[y^2+3y-x^3+x=C\;\;.\]
