Determine whether ODE admits a unique solution locally Consider the initial value problem,
$$ 5y'' + \bigg(\frac{y'}{x}\bigg)^2 + 4y^2 = 0, \;\; y(0) = 0, \; y'(0) = 0.$$
Determine whether or not this ODE has a unique solution in the neighborhood of the origin.

It seems like we can not use Picard-Lindelof here because of the nonlinear term, namely that it is not Lipschitz at the origin.  Beyond this, I have no intuition for how one might prove or disprove this.  Any ideas? Thanks!
EDIT: Looking into it some, I've seen some people suggest using asymptotic methods near the origin as a way to show that one (or more) solutions exist.  Any thoughts?
 A: $\newcommand{\NN}{{\mathbb{N}}}\newcommand{\CC}{{\mathbb{C}}}\newcommand{\RR}{{\mathbb{R}}}$
Besides the obvious solution $y=0$, we are looking for a 
series solution $y(x)\in x^2\CC[[x]]$.
Claim: There is a one parameter family of convergent series solutions of the form
$$y(x)=-\tfrac52x^2+b\,x^3+z(x)\mbox{ where }z(x)\in x^4\CC[[x]]\mbox{ and }b\mbox{ is arbitrary.}$$
To show this, we insert $y(x)=ax^2+b\,x^3+z(x)$, $z(x)\in M:=x^4\CC[[x]]$, in the equation, 
compare coefficients and 
find in a first step the conditions $10a+4a^2=0$ and $30b+12ab=0$. The first one is satisfied for
$a=0$ which probably leads to $y=0$ and by $a=-5/2$ which we use from now on. The second equation
is then satisfied by any $b$.
Inserting now $y(x)=-\frac52x^2+b\,x^3+z$ and multiplying by $x^2$, we obtain the equation
$$5x^2z''-10xz'+9b^2x^4+6bx^2z'+z'^2+x^2(-\tfrac52x^2+bx^3+z)^2=0$$
for a series $z\in M$.
This leads to consider the operator $L:M\to M$, $u\to 5x^2u''-10xu'$ which maps
$u(x)=\sum_{k=4}^\infty u_kx^k$ to $\sum_{k=4}^\infty (5k-15)ku_kx^k$. Therefore it has an inverse
$K=L^{-1}:M\to M$ which maps $u(x)$ to $\sum_{k=4}^\infty \frac{u_k}{(5k-15)k}x^k$.
The above equation can now rewritten in ''fixed point form'' for $r=Lz$.
$$r=-9b^2x^4-6bx^2(Kr)'-(Kr)'^2-x^2(-\tfrac52x^2+bx^3+Kr)^2=:F(x,r).$$
The existence of a unique formal solution $r\in M$ now follows by comparing the coefficients
of $x^k$, $k=4,5,...$ on both sides. On the left hand side, we have $r_k$. The coefficient 
$[F(x,r)]_k$ of $x^k $ on the right hand side only depends upon $r_4,...,r_{k-1}$. 
This means that $r_k$ are determined recursively.
In order to prove convergence of $r(x)$ we introduce a notation of majorant series.
For $f(x)=\sum_{n=0}^\infty f_nx^n\in\CC[[x]]$ and $g(x))=\sum_{n=0}^\infty g_nx^n\in\RR_+[[x]]$ 
we write $f\ll g$ if $|f_n|\leq g_n$ for all nonnegative integers $n$. This relation is compatible 
with addition, multiplication and differentiation. For the operator $K$ we show easily
$$f\ll g\mbox{ implies }Kf\ll g, (Kf)'\ll \frac1x g $$
for $f,g$ divisible by $x^4.$ This leads us to the majorant equation
$$g=9|b|^2x^4+6|b|x g+\left(\frac gx\right)^2+x^2(\tfrac52x^2+|b|x^3+g)^2=:H(x,g)$$
It has a formal solution in $x^4\RR_+[[x]]$ determined by $g_k=[H(x,g)]_k$, $k=4,5,...$.
The main point is that, by construction, $r\ll g$. A formal proof proceeds by induction but 
I omit it here. It remains to show the convergence of $g$ (which implies that of $r$ and then that of $z$ and $y$). We put $g=x^4 h$ and obtain the equation
$$h=9|b|^2+6|b|x h + x^2h^2+(\tfrac52x+|b|x^2+x^3h)^2$$
By the implicit function theorem, this equation has a holomorphic solution $h$ with $h(0)=9|b|^2$.
Its power series is the a convergent formal solution of the $h$-equation. Thus also $g=x^4h$ is 
convergent and the proof is complete.
