I am not sure, if there is an explicit form, but if there is, how do I get it?
This is the formula: $$c_n=\frac{1-n \cdot c_{n-1}}{\lambda}$$
where $\lambda \in \mathbb{R}$ and $n \in \mathbb{N}$
I already tried some forms for c via trail and error, but I couldn't find a solution ...
On
We have $$\begin{align} c_{10} & = \tfrac 1\lambda - \tfrac{10}\lambda c_9 \\ & = \tfrac 1\lambda - \tfrac{10}{\lambda^2} (1-9c_8) \\ & = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^2} c_8 \\ & = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^3} -\tfrac{10\cdot 9\cdot 8}{\lambda^4}c_7 \\ & = \ldots = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^3} -\tfrac{10\cdot 9\cdot 8}{\lambda^4}+\tfrac{10\cdot 9\cdot 8\cdot 7}{\lambda^5} - \ldots + (-1)^{10} \frac{10!}{\lambda^{11}}\cdot c_0 \end{align}$$
So if I have not done any mistakes, the solution is
$$c_n = \sum_{k=0}^{n-1} \left((-1)^k \frac{n!}{(n-k)!\cdot \lambda^{k+1}} \right) + (-1)^n \frac{n!}{\lambda^{n+1}} c_0$$
The expression for $c_n$ is $$c_n=\frac{(-1)^n n!}{\lambda^{n+1}}\left\{S_{n}(-\lambda)+\lambda c_0-1\right\}$$ where \begin{equation}S_n(x)=\sum_{k=0}^n\frac{x^k}{k!} \end{equation} for all $x \in \mathbb{R}$. I prove it inductively.
For $n=1$ the proposed expression begets $$-\frac{1}{\lambda^2}(1-\lambda+\lambda c_0-1)=\frac{1-c_0}{\lambda}=c_1$$ So the statement is true for $n=1$. Let it be true for $n=k$. Now we have from the recursion relation \begin{equation} \begin{split} c_{k+1}=& \frac{1}{\lambda}-\frac{k+1}{\lambda}c_k \\ \ =& \frac{1}{\lambda}-\frac{(-1)^k (k+1)!}{\lambda^{k+2}}S_{k}(-\lambda)-\frac{(-1)^k (k+1)!}{\lambda^{k+2}}(\lambda c_0-1)\\ \ =& \frac{(k+1)!}{\lambda^{k+2}}\left\{\frac{\lambda^{k+1}}{(k+1)!}+(-1)^{k+1}S_k(-\lambda)+(-1)^{k+1}(\lambda c_0-1)\right\}\\ \ =&\frac{(-1)^{k+1}(k+1)!}{\lambda^{k+2}}\left(S_{k+1}(-\lambda)+\lambda c_0-1\right) \end{split} \end{equation} Hence the hypothesis is proved.