The differential equations are :


$$ ( n\_{j,k,0} )'(x) = - \frac {jn\_{j,k,0}(x)} {a-x}, $$


$$ ( n\_{j,k,b} )'(x) = \frac { (j-b+1)n\_{j,k,b-1}(x) - (j-b)n\_{j,k,b}(x) } {a-x}, $$


for $ 0\lt b\lt c $.


$$ (f\_{j,k})'(x) = \frac{ (j-c+1)n\_{j,k,c-1}(x) }{a-x}.$$
Here, the second equation holds with $0\lt b\lt c$.



Solve the following system of differential equations
If b is an integer, you can simply solve the equations successively, starting with b=0. If b is supposed to be real, you need to explain what $n\_{j,k,b-1}$ is supposed to mean for b between 0 and 1. Also, what is the point of the index k? Nothing in the equations depends on k!


