Example C The second-order equation Yk+2 – Yk+1 – k²yk = 0 (3.159) - can be written in the following factored form: (E + k)(E – k)yk 0. (3.160) Note that (E – k)(E + k)yk = Yk+2 + Yk+1 – k²yk, (3.161) which shows that the order of the factors is important. To solve equation (3.160), let 21(k) = (E – k)yk; (3.162) therefore, equation (3.160) becomes (E + k)z1(k) = 0, (3.163) which has the solution 21 (k) = A(-1)*(k – 1)!, (3.164) where A is an arbitrary function. We now have (E – k)yk = A(-1)* (k – 1)!, (3.165) or Yk+1 – kyk = A(-1)*(k – 1)!. (3.166) The solution to this latter equation is k-1 A(-1)*(i – 1)! | (k – 1)! E + B (3.167) Yk i! i=1 Therefore, the general solution to equation (3.159) is k-1 Yk = +B| (k – 1)!. (3.168) i=1