 Solution

Set $u=1/y$, so $u^{\prime }=-y^{\prime }/y^2$. The equation becomes $x(x^2-1)u^{\prime }+u=-x^3$. For $x>1$ its integrating factor is $\sqrt{x^2-1}/x$, givingThe initial value is $u(2)=-2$, forcing $C=0$. HenceSubstitution into the original equation verifies this answer throughout $x>0$, including $x=1$ where the divided linear equation was singular. Its maximal connected interval through the initial point is $(0,\infty )$.