Lets write the second order DE as...

\(\displaystyle y^{''} y = x^{3}\) (1)

... and suppose that the solution is analytic in \(\displaystyle x=0\) so that is...

\(\displaystyle y(x)= a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} + \dots \) (2a)

\(\displaystyle y^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots + n a_{n} x^{n-1} + \dots\) (2b)

\(\displaystyle y^{''} (x) = 2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + \dots + n(n-1) a_{n} x^{n-2} + \dots\) (2c)

The \(\displaystyle a_{n}\) are found inserting the (2a) and (2c) in (1), once we know the 'initial conditions' \(\displaystyle y(0)=a_{0}\) and \(\displaystyle y^{'} (0) = a_{1}\) ...

\(\displaystyle 2 a_{0} a_{2} =0 \rightarrow a_{2} = 0\)

\(\displaystyle 6 a_{0} a_{3} + 2 a_{1} a_{2} = 0 \rightarrow a_{3}=0\)

\(\displaystyle 12 a_{0} a_{4} + 6 a_{1} a_{3} + 2a_{2}^{2}= 0 \rightarrow a_{4}=0\)

\(\displaystyle 20 a_{0} a_{5} + 12 a_{1} a_{4} + 8a_{2} a_{3}= 1 \rightarrow a_{5}= \frac{a_{0}}{20}\)

\(\displaystyle 30 a_{0} a_{6} + 20 a_{1} a_{5} + 12a_{2} a_{4} + 6 a_{3}^{2} = 0 \rightarrow a_{6}= - \frac{a_{1}}{30}\)

... and so one...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)