Answer
$C_0\Sigma^\infty_{n=0} \dfrac{x^{3n}}{3^n n!}=C_{0} e^{x^3/3}$
Work Step by Step
Given: $y' =x^2y$
Since, $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
Here, we have
$y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$
Then, the given equation $y' =x^2y$ becomes :
$\Sigma^\infty_{n=1}nC_{n} x^{n-1}= x^2[ \Sigma^\infty_{n=0}C_{n}x^{n}]$
or, $\Sigma^\infty_{n=2}c_{n+1}(n+1)x^{n}-\Sigma^\infty_{n=0}c_{n}x^{n+2}=0$
or, $c_1+2c_2x+\Sigma^{2}_{n=0}((n+1)-c_{n-2}) x^{n}=0$
For n=2 , we have $C_{3} =\dfrac{C_{0}}{3}$;
For n=3 , we have $C_{4} = \dfrac{C_{1}}{4}=0$;
For n=8 , we have $C_{9} = (\dfrac{1}{9}) (\dfrac{1}{6}) \times \dfrac{1}{3} C_{0}=0$
Thus, we get $C_{3n} = \dfrac{C_0}{3^n n!}$
We know that $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, $y=C_0\Sigma^\infty_{n=0} \dfrac{x^{3n}}{3^n n!}=C_{0} e^{x^3/3}$
