Solve
y''= x^3/y
I tried by using let u=y'
then I get
u'=x^3/y .. now something tells me this is not going to be an easy ride home.
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
$\displaystyle \chi$ $\displaystyle \sigma$