Since you only need a5 & a7, take y'' = 2xy - y we know that y(0) = 0 & y'(0) = 1
The solution is y = f(x) = f(0) + f'(0)x + f''(0)x2/2! + f'''(0)x3/3! + fiv(0)x4/4! + fv(0)x5/5! + fvi(0)x6/6! + fvii(0)x7/7!......
with a5 = fv(0)/5! and a7 = fvii(0)/7!
we know that y''(0) = f(0) = 2(0)(0) - (0) = 0 . Taking the other derivatives in similar fashion and finding a5 & a7, we find that
a5 = 5/5! = 1/4! = 1/24 and
a7 = 45/7! = 1/112