Making
#y = sum_(k=0)^n a_k x^k# and substituting into the differential equation
#sum_(k=2)^n k(k-1)a_k x^(k-2)+x^2 sum_(k=1)^n k a_k x^(k-1) + x sum_(k=0)^n a_k x^k = 0# or
#sum_(k=2)^n k(k-1)a_k x^(k-2)+sum_(k=1)^n k a_k x^(k+1) + sum_(k=0)^n a_k x^(k+1) = 0# or
#sum_(k=-1)^n (k+3)(k+2)a_(k+3) x^(k+1)+sum_(k=1)^n k a_k x^(k+1) + sum_(k=0)^n a_k x^(k+1) = 0# or
#sum_(k=-1)^n((k+3)(k+2)a_(k+3)+(k+1)a_k)x^(k+1) = 0#
with #a_(-1) = 0# and then
#(k+3)(k+2)a_(k+3)+(k+1)a_k=0# for #k = 0,1,2,3, cdots#
This infinite recurrence equation needs the a priori definition of #a_0, a_1# which can be associated to initial conditions.
