Solution
Let
y(x)=∑n=0∞anxn
dxdy=∑n=1∞nanxn−1=∑n=0∞(n+1)an+1xn
dx2d2y=∑n=0∞n(n+1)an+1xn−1=∑n=0∞(n+1)(n+2)an+2xn
Substitution into equation:
∑n=0∞(n+1)(n+2)an+2xn+∑n=0∞(n+1)an+1xn+1+∑n=0∞anxn+2+2∑n=0∞anxn=0
∑n=0∞(n+1)(n+2)an+2xn+∑n=1∞nanxn+∑n=2∞an−2xn+2∑n=0∞anxn=0
2a2+6a3x+2a0+3a1x+∑n=2∞(n+1)(n+2)an+2xn+∑n=2∞nanxn+∑n=2∞an−2xn+2∑n=2∞anxn=0
Coefficient near xn are equal to zero.
n=0: 2a2 + 2a0=0 => a2 = -a0
n=1: 6a3+3a1=0 => a3=-a1 /2
n>1: (n+1)(n+2)an+2+nan+an-2+2an=0 => (n+1)(n+2)an+2+(n+2)an+an-2=0
an+2 =-an/(n+1)-an-2/[(n+1) (n+2)]
Therefore for arbitrary a0 and a1 and last recurrent formula solution is
y(x)=a0+a1x −a0x2−2a1x3+∑n=2∞an+2xn+2
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