... polynomial areα1=1 +1 − 3 /4 2= 3 4 , α2=1 −1 − 3 /4 2=1 4 .Thus our two series solutions will be of the formw1= z 3 /4 ∞n=0anzn, w2= z1 /4 ∞n=0bnzn.Substituting ... + 3) For our first solution we have the difference equationa0= 1, a1= 0, a2= 0, a 3 = 0, an +4 = −an(n + 4) (n + 3) . For our second solution,b0= 0, b1= 1, b2= 0, b 3 = 0, bn +4 = ... difference equation for bnthat is of order one less than theequation for an.117812 3 4 56-1-0.50.511.512 3 4 56-1-0.50.511.5Figure 23. 3: The graph of approximations and numerical...