 Solution

Factor the right side as $(x-1{)}^2(x-3)$. On $-1<x<1$, division by $x^2-1$ gives a first-order linear differential equation with integrating factorMultiplication makes the equation $(\mu y{)}^{\prime }=3-x$, so $\mu y=3x-x^2/2+C$. At zero, $\mu (0)y(0)=1$, hence $C=1$. Simplifying givesIts derivative isThe derivation solves the initial-value problem on $(-1,1)$. The simplified expression also extends smoothly through $x=1$ and satisfies the original undivided equation there, giving a solution on all $x>-1$; its pole prevents continuation through $x=-1$.