A function $f$ is called *holonomic* if it satisfies some linear differential equation with polynomial coefficients $$p_n(x) f^{(n)}(x)+\dots+p_1(x)f'(x)+p_0(x)f(x)=0.$$ Now if $f,g$ are holonomic then so are their sum and product. To obtain a differential equation for $h=fg$, first observe that
$$ h^{(k)} = \sum_{i=0}^k {k\choose i} f^{(i)} g^{(k-i)}.$$ Since each $f^{(i)}$ is a linear combintation of $f,f',\dots,f^{(n-1)}$ with rational coefficients (where $n$ is the order of the ODE satisfied by $f$), and analogously each $g^{(i)}$ is a linear combination of $g,g',\dots,g^{(m-1)}$, then $h,h',h'',\dots$ span a finite-dimensional vector space of dimension at most $d=mn$, and hence there exists a nontrivial relation of the form
$$ r_d(x)h^{(d)}+\dots+r_1(x)h'+r_0(x)h=0$$with $r_i(x)$ rational functions.

Now suppose that the ODE which $f$ satisfies had singular points $u_1,\dots,u_n$, while the equation of $g$ had singular points $w_1,\dots,w_m$. The ODE for $h$ might have additional singular points besides $u_1,\dots,u_n,w_1,\dots,w_m$. For example, if
$$(x-a)f''(x)+cf(x)=0\\\\ (x-b)g''(x)+dg(x)=0$$ then $h=fg$ satisfies an ODE of order 4 with leading coefficient $$(x-a)^3 (x-b)^3 \biggl((c-d)x-(bc-ad)\biggr)h^{(4)}+\dots$$ (I calculated this using C.Mallinger's *GeneratingFunctions* Mathematica package).

Is it possible to construct a holonomic ODE for $h$ (in the above example and in general) without introducing additional singular points?