Answer
$c_0\Sigma^\infty_{n=0} \dfrac{x^{3n}}{3^n n!}=c_{0} e^{x^3/3}$
Work Step by Step
Here, $y=\Sigma^\infty_{n=0}C_{n} x^{n}$
This gives: $y'=\Sigma^\infty_{n=1}n C_n x^{n-1}$
Now, $y' =x^2y$
$\Sigma^\infty_{n=1}nC_{n} x^{n-1}= x^2[ \Sigma^\infty_{n=0}C_{n}x^{n}]$
$\implies \Sigma^\infty_{n=2}C_{n+1}(n+1)x^{n}-\Sigma^\infty_{n=0}C_{n}x^{n+2}=0$
$\implies C_1+2C_2x+\Sigma^{2}_{n=0}((n+1)-C_{n-2}) x^{n}=0$
n=2 : $C_{3} =\dfrac{C_{0}}{3}$;
n=3: $C_{4} = \dfrac{C_{1}}{4}=0$;
n=8 : $C_{9} = (\dfrac{1}{9}) \times (\dfrac{1}{6}) \times \dfrac{C_0}{3}=0$
The pattern gives: $C_{3n} = \dfrac{1}{3^n n!}C_0$
Since, $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Thus, we get $y=C_0\Sigma^\infty_{n=0} \dfrac{x^{3n}}{3^n n!}$
or, $y=C_{0} e^{x^3/3}$
