How do you solve this differential equation with power series? #y''+x^2y'+xy=0#

1 Answer
Oct 27, 2017

See below.

Explanation:

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.