Answer
$y=C_{0} e^{x^3/3}$
Work Step by Step
We have $y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$
and $\Sigma^\infty_{n=2}c_{n+1}(n+1)x^{n}-\Sigma^\infty_{n=0}c_{n} \times x^{n+2}=0$
$\implies c_1+2 \times c_2x+\Sigma^{2}_{n=0}((n+1)-c_{n-2}) x^{n}=0$
$C_{3} =\dfrac{C_{0}}{3}(For n=2 )\\C_{4} = \dfrac{C_{1}}{4}=0 (n=3)$;
and $C_{9} = (\dfrac{1}{9}) (\dfrac{1}{6}) \times \dfrac{1}{3} C_{0}=0; (n=8)$
This implies that $C_{3n} = \dfrac{C_0}{3^n n!}$
Use formula: $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence,we get $y=C_0\Sigma^\infty_{n=0} \dfrac{x^{3n}}{3^n n!}$
or, $y=C_{0} e^{x^3/3}$
