 Solution

Expanding the derivative and dividing by $1-x$ givesThe only finite singular point is $\boxed{x=1}$. There $(x-1)p(x)=1$ and $(x-1{)}^{2}q(x)=c^2(x-1{)}^2-\lambda (x-1)$ are analytic, so it is a regular singular point. At infinity the coefficient of $y$ tends to $c^2\ne 0$, giving an irregular singular point after reciprocal substitution; the finite classification is independent of $\lambda$.