General solution to ODE $ y''-Ay^5=0 $ What is the solution of $$ y''-Ay^5=0  $$
I got the solution $ y = {(3/4A)}^{1/4} x^{-1/2}$ using trial and error but how to solve this type of problem in general?
 A: Let us try to replace the trial and error approach and assume that $y=ax^k$; so, replacing, we get $$a (k-1) k x^{k-2}-a^5 A x^{5 k}=0$$ So, you could set $k-2=5k$ (that is to say $k=-\frac 12$ and $a(k-1)k=a^5 A$ which leads to five different solutions for $a$, two of them being imaginary, and one of them being $0$. Among the real, one solution is your.
But, as said in comments, this is just a very particular solution.
The problem is very complex as showed in the link provided by Mathlover.
A: Set it to $$\frac{d^2y}{dx^2} = Ay^5$$
If we multiply this ODE by $2 \frac{dy}{dx}$ we get
$2 \frac{dy}{dx} \frac{d^2y}{dx^2} = 2A y^5 \frac{dy}{dx}$
Thus $\left( \frac{dy}{dx} \right)^2 = \frac{Ay^6}{6} + K$.
Now, in the simple case (where we say have initial conditions such as) when $y = 1$, $\frac{dy}{dx} = \frac{1}{\sqrt{3}}$ so $K = 0$. 
i.e. $\displaystyle \left( \frac{dy}{dx} \right)^2 = \frac{y^6}{3} $
so  $ \left( \frac{dy}{dx} \right) = \frac{y^3}{\sqrt{3}} $ 
Separating, $\frac{dy}{y^3} = \frac{dx}{\sqrt{3}}$ so
$ - \frac{1}{2} y^{-2} = \frac{x}{\sqrt{3}} + A$. 
As $y = 1$ when $x = 0$, $A = -\frac{1}{2}$ so
$ - \frac{1}{2} y^{-2} = \frac{x}{\sqrt{3}} -\frac{1}{2}$ 
i.e.  $ = \left( 1 - \frac{2x}{\sqrt{3}} \right)^{-1/2}$.
That would be a specific solution (which I've seen before which is why I mention it here, however life isn't that simple and you don't mention any inital conditions so it would amount to solving (which is trickier but not impossible)
\begin{equation}
\frac{dy}{dx} = \sqrt{\frac{Ay^{6}}{6} + K}
\end{equation}
A: In this equation, $x$ is missing and its solution is routine (check any DE textbooks). Let $u=y'$ and then $u'=y''$. Treating $u$ as a function of $y$ and then noting
$$ u'=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=\frac{du}{dy}y'=\frac{du}{dy}u $$
we have
$$ \frac{du}{dy}u=Ay^5 $$
which is a separable DE whose solution is standard. I think you can handle it.
