I have a countable infinite set of functions $\{\Gamma_n(\alpha),n\in\mathbb{N}\}$ over the interval $0<\alpha<1$. They should satisfy a linear system of differential equation (first order,homogenous), as following:
\begin{eqnarray*} \frac{\partial\Gamma_{n}\left(\alpha\right)}{\partial\alpha} & = & \sum_{k=1}^{\infty}c_{n,k}\alpha^{k-1}\Gamma_{n+k+2}\left(\alpha\right)\\ \end{eqnarray*}
with $c_{n,k}$ real coefficients and boundary conditions
\begin{eqnarray*} \Gamma_{0}(0) & = & A\\ \Gamma_{1}(0) & = & 0\\ \Gamma_{2}(0) & = & B\\ \Gamma_{n>2}(0) & = & 0\\ \Gamma_{n}(1) & = & \gamma_n\in \mathbb{R}\\ \end{eqnarray*} with $A,B,C_n$ real coefficients, too.
I've constructed the associated matrix, an upper triangular one: $$\left[\begin{array}{cccccccc} 0 & 0 & 0 & c_{0,1} & c_{0,2}\alpha & c_{0,3}\alpha^2 & \cdots & \cdots\\ \vdots & \ddots & \ddots & \ddots & c_{1,1} & c_{1,2}\alpha & \cdots & \cdots\\ \vdots & & \ddots & \ddots & \ddots & c_{2,1} & c_{2,2}\alpha & \cdots\\ \vdots & & & \ddots & \ddots & \ddots & \ddots & \ddots\\ \vdots & & & & \ddots & \ddots & \ddots & \ddots \end{array}\right]$$
But, the procedure of converting the system to an $n-$th order linear differential equation does not work here, since it is a infinite matrix. The only hint I founded is that an infinite number of $\Gamma_n$'s must be non-zero. Otherwise, the system would have a trivial solution. For example, if $\Gamma_{m>5}$'s are zero, then from the fourth row on $\frac{\partial\Gamma_m>5}{\partial\alpha}=0$ should hold, and by using boundary condition we obtain $\Gamma_m>5(\alpha)=0$. Then, returning to upper rows, they become trivial, too. Do you have any suggestion on how to proceed?
EDIT: Considering the boundary conditions (and the expected solutions' behaviour, I thought it'd have been useful to look for solutions $$\Gamma_n\left(\alpha\right)=\alpha^n f_n(\alpha)$$ so boundaries are satisfied automatically with $f_n(1)=\gamma_n$. But still, no easy equation have shown up. Does anyone have any suggestions or common tricks to solve this type of systems?
EDIT: Going further with the solution research, I guessed for a power series expansion for any $\Gamma_n$'s $$\Gamma_n(\alpha)=\sum_{k=0}^\infty a^{(n)}_k\alpha^k$$ $a^{(n)}_k$ being a series of real coefficients for every $\Gamma_n$. I obtain a infinite linear system for the $a^{(n)}$, $$\begin{eqnarray*} a_{1}^{(n)} & = & C_{n,1}a_{0}^{\left(n+3\right)}\\ 2a_{2}^{(n)} & = & C_{n,1}a_{1}^{\left(n+3\right)}+C_{n,2}a_{0}^{\left(n+4\right)}\\ 3a_{3}^{\left(n\right)} & = & C_{n,1}a_{2}^{\left(n+3\right)}+C_{n,2}a_{1}^{\left(n+4\right)}+C_{n,3}a_{0}^{\left(n+5\right)}\\ \dots & = & \dots\\ a_{k}^{(n)} & =\frac{1}{k} & \sum_{j=1}^{k}C_{n,j}a_{k-j}^{\left(n+2+j\right)} \end{eqnarray*}$$ But this are not compatible with all the boundary conditions. In fact the system is trivialized, if one notice that in the first equation $a_{0}^{\left(n+3\right)}$ corresponds to $\Gamma_{n+3}$, so it is zero. Consequently, all other coefficients are zero. And that is not compatible with \begin{eqnarray*} \Gamma_{n}(1) & = & \gamma_n\in \mathbb{R}\\ \end{eqnarray*} Does anyone see a reason to exclude the existence of a solution?