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 $0\lt b\lt c$.