Changes

<math>(2a_2)+(6a_3-a_1)x+(12a_4-2a_2+a_0)x^2 + \dots </math>

<math>(2a_2)+(6a_3-a_1)x+(12a_4-2a_2+a_0)x^2 + \dots </math>

so that for <math>n\geq 2</math>, the coefficient of <math>x^n</math> on the right hand side is  <math>(n+1)(n+2)a_{n+2}-na_n+a_{n-2}</math>.  So we need only set this equal to <math>1/(n!)</math> to find an infinite series representation of our solution.  If we're given the values of <math>y(0), y'(0)</math>, as we frequently are in applications (remember what we said about having two "degrees of freedom" in a DEQ involving second-order derivatives?) we can solve this completely.  Let's say <math>y(0)=0, y'(0)=1</math>.  We have

so that for <math>n\geq 2</math>, the coefficient of <math>x^n</math> on the right hand side is  <math>(n+1)(n+2)a_{n+2}-na_n+a_{n-2}</math>.  So we need only set this equal to <math>1/(n!)</math> to find an infinite series representation of our solution.  If we're given the values of <math>y(0), y'(0) \ </math>, as we frequently are in applications (remember what we said about having two "degrees of freedom" in a DEQ involving second-order derivatives?) we can solve this completely.  Let's say <math>y(0)=0, y'(0)=1</math>.  We have

<math>a_0=0 \ </math>

<math>a_0=0 \ </math>