Taking the limit of an implicitly defined function The original excercise was to solve the
$$\frac{y'(x)}{x}+2y(x)y'(x) \log(x)+\frac{y^2(x)}{x}-\frac{y(x)}{x^2}+1=0$$
Differential equation with $y(0)=0$.
The general solution is
$$\frac{y(x)}{x}+y^2(x) \log(x) + x =C$$
This is not defined in the $x=0$ point, but $C$ might be defined to met the $\lim \limits_{x \to 0} y(x)=0$ condition. So i took the limit of both sides as $(x,y) \to (0,0)$. $x \to 0$, $y^2 \log(x) \to 0$, but the limit of $\frac{y}{x}$ as $(x,y) \to (0,0)$ is path-dependent, so there is not any $C$ that could satisfy it. But I realized that $y$ is a function of $x$, so $y$ "can't go in every way" to the $0$; so the $C$ might exist.
Is there a way to prove if there is a $C$ which can satisfy the limit, without transforming the equation into an explicit form?
 A: As Gribouillis  commented, solving the quadratic in $y$ gives
$$y_1=-\frac{1+\sqrt{4 c x^2 \log (x)-4 x^3 \log (x)+1}}{2 x \log (x)}$$
$$y_2=-\frac{1-\sqrt{4 c x^2 \log (x)-4 x^3 \log (x)+1}}{2 x \log (x)}$$ Developing as Taylor series around $x=0$ gives
$$y_1=-\frac{1}{x \log (x)}-c x+x^2+O\left(x^3\right)$$
$$y_2=c x-x^2+O\left(x^3\right)$$
A: Following the suggestion in the comment of Gribouillis, multiply the integrated equation through with $4x^2\ln(x)$ to get
$$2(x\ln(x)y(x))+(2x\ln(x)y(x))^2 = 4(C-x)x^2\ln(x)$$
or
$$
(2x\ln(x)y(x)+1)^2=1+4(C-x)x^2\ln(x)
$$
leading to the solution with $y(x)\approx 0$ for $x\approx 0$ 
$$
2x\ln(x)y(x)=\sqrt{1+4(C-x)x^2\ln(x)}-1=\frac{4(C-x)x^2\ln(x)}{1+\sqrt{1+4(C-x)x^2\ln(x)-4x^3\ln x}}
\\
\implies y(x)=\frac{2x(C-x)}{1+\sqrt{1+4(C-x)x^2\ln(x)}}
$$

Ignoring how to find explicit solutions per quadratic equations, you can try to insert $y(x)=Cx+Dx^2+o(x^2)$ into the original ODE,
$$
\frac{xy'-y}{x^2}+2yy'\ln x+\frac{y^2}x+1=0
\\
0=[D+o(1)]+[2C^2x\ln x+o(x)]+[C^2x+o(x)]+1=D+1+o(1)
$$
which implies $D=-1$ and provides no restriction on $C$. From the explicit solution we see that the next term is $\sim x^3\ln x$ so that higher order terms can not be computed using a simple power series approach.
A: I think you can obtain developments without solving the equation by examining the order of magnitude of the different terms. Suppose we have
$$\frac{y}{x} + y^{1+\alpha}\log(x) + x = c$$
for some $\alpha>0$, so that we cannot solve explicitly the equation. We can write this as
$$y = cx - x^2 - y^{1+\alpha} x\log(x)$$
Suppose that we know that $y$ is bounded when $x\to 0$. We see from this equation that $y\to 0$ because all the terms on the rhs tend to $0$. As a consequence we get $y^{1+\alpha} = o(y)$, hence
$$y = O(x) + o(y)\quad \Longrightarrow y = O(x)$$
As a consequence, $y^{1+\alpha} x \log(x) = O(x^{2+\alpha}\log(x)) = o(x^2)$. Hence
$$y = c x - x^2 + o(x^2)$$
One can get more terms by writing $y = c x - x^2 + x^2 z$. It leads to
$$y = c x - x^2 - c^{1+\alpha} x^{2+ \alpha} \log(x) + o(x^{2+\alpha}\log(x))$$
