The theory about Liouvillian extensions tells us that a Picard-Vessiot extension $L \supset k$ is Liouvillian if and only if the identity component $G^°$ of $G = Gal(L / k)$ is solvable.
I think I found a Liouvillian extension, that has a non-solvable subgroup, which seems to be absurd. I don't see any mistake, so please help me out!
Counterexample Let $k = \mathbb{C}$ be the complex functions with zero-derivation, i.e. it is its own field of constants. Let $\mathbb{C}(z)$ be the function field in one variable, with the common derivation $z' = 1$. According to Wikipedia or this paper the inverse Galois problem for $\mathbb{C}(z)$ is solved positively, so we can choose an algebraic extension $L \supset \mathbb{C}(z)$, such that $H := Gal(L / \mathbb{C}(z))$ is not solvable (e.g. $H = A_5$). Now $L$ is algebraic over $\mathbb{C}(z)$, so there is a unique way to extend the derivation to $L$. In particular, $L / \mathbb{C}(z)$ is a Picard-Vessiot extension, so $L / \mathbb{C}$ is one as well. Choosing a primitve element $t \in L$ such that $L = \mathbb{C}(z, t)$, we see that $L / \mathbb{C}$ is in fact Liouvillian. Its differential Galois group $Gal(L / \mathbb{C})$ is connected, because $\mathbb{C}$ is algebraically closed, but it also contains the closed non-solvable subgroup $H = Gal(L / \mathbb{C}(z)) \subset Gal(L / \mathbb{C})$.
So this seems to contradict the fact that $Gal(L / \mathbb{C})$ should be solvable.
PS: If someone comes up with such an explicit extension $L \supset \mathbb{C}(z)$ I would be grateful, because this simplifies the argument quite a bit. I had not yet time to think about this.
So I discussed this with my professor, and our consensus was, that $\mathbb{C}(z,l) / \mathbb{C}$ is not a Liouvillian extension. To give a counterexample, take $l$ to be a square root of $z$, i.e. $l^2 = z$. Then $l$ generated $\mathbb{C}(z,l) = \mathbb{C}(l)$, but will never satisfy a differential equation
$$ l^{(n)} + c_1l^{(n-1)} + \dotsc + c_{0}l = 0,$$ because differentiating $l$ produces higher powers of $l^{-1}$.