Because y(0)=0 and y'(0)=1 then c(0)=0 and c(1)=1
Diff and sub in ODE : ∑ [n=2,∞] n(n−1)c(n)x^(n−2) + x²∑ [n=0,∞] c(n)x^n = 0
Constant term : (2*1)c(2) = 0 → c(2)= 0
x term : (3*2)c(3) = 0 → c(3)=0
xⁿ term for n≥2 : (n+2)(n+1)c(n+2) + c(n−2) = 0 → c(n+2) = −c(n−2)/{(n+1)(n+2)}
Or more conveniently c(n+4) = −c(n)/{(n+3)(n+4)} for n≥0 … (i)
From recurrence and because c(0)=c(2)=c(3)=0 then all c(n)=0 unless n=4k+1
Setting n=1,5,9... in (i) gives the series
y = x − 1/(4*5)x^5 + 1(4*5*8*9)x^9 − 1/(4*5*8*9*12*13)x^(13) …...
edit : The recurrence is not needed for c(0) and c(1). c(0) comes from setting x=0 in series and c(1)=1 comes from differentiating series and setting x=0.
Hence the first two terms c(0)+c(1)x become 0+1*x = x.