Question

y^2y"=y'

Answer 1

This seems to be a very difficult problem, you may do best to post this in the differential equations subsection and wait for someone who knows how to solve non-linear DE's.

Answer 2

y^2y" = y'

i wonder, assume:
\[ y = \sum_{n=0}~~ a_nx^n\]
\[y'=\sum_{n=1}a_n~nx^{n-1}\]
\[y''=\sum_{n=2}a_n~n(n-1)x^{n-2}\]

Answer 3

i wonder how y^2 would work out, it looks to be some monsterous form of

\[\sum_{k=0}^ia_k\left(\sum_{n=0}^ia_nx_n\right)\]

Answer 4

*x^n that is, not x_n

Answer 5

y=0,1,-1

Answer 6

hmmm, i wonder of this is correct for y^2 :)
\[\left(\sum_{n=0}^{\infty}a_nx^n\right)^2=\frac{\prod_{n=0}^{\infty}(a_n)^2}{(1-x)^2}\]

Answer 7

oh I didn't think to use a power series, good idea :)

Answer 8

its a nice idea, but im not sure how practical it is ;)

Answer 9

http://www.wolframalpha.com/input/?i=y%5E2+y%27%27-y%27%3D0

it definantly doesnt have a nice solution ;)