show that the ODE $(\sqrt x + \sqrt y)\sqrt y dx = xdy$ has no solution $y(x)$ such that $\lim_{x\to \infty}\frac{y(x)}{x}=L$ when $ L>0 \in \Bbb R$ I need to show that the ODE $(\sqrt x + \sqrt y)\sqrt y dx = xdy$ has no solution $y(x)$ such that $\lim_{x\to \infty}\frac{y(x)}{x}=L$ when $ L>0 \in \Bbb R$
I solved it by finding the general solution $y(x) = \frac{1}{4} (4 c_1^2 x + 4 c_1 x \log(x) + x \log^2(x))$ and then its easy to see that $\lim_{x\to \infty}\frac{y(x)}{x}=\infty$ and we finished.
I'm asking if there is a more elegant proof to this problem instead of the straight forward solution
 A: To make the ODE well-defined, all $y(x)$ in my answer are non-negative and $L\geq0$, too.
Write the ODE as:
$$y’=\frac{(\sqrt x+\sqrt y)\sqrt y}{x}=\left(1+\sqrt{\frac yx}\right)\cdot \sqrt{\frac yx}.$$
If $\lim_{x\to\infty}\frac yx=L$, then $\lim_{x\to\infty}y’(x)$ exists and
$$\lim_{x\to\infty}y’(x)=(1+\sqrt L)\sqrt L.$$
By L'Hôpital's rule,
$$L=\lim_{x\to\infty}\frac{y(x)}x=\lim_{x\to\infty}y’(x)=(1+\sqrt L)\sqrt L,$$
hence $L=0$.
Now, notice that $y’\geq \frac yx$. So $y\equiv 0$, or, $y$ never hits $0$ and
$$\left(\ln\frac yx\right)’=(\ln y-\ln x)’=\frac{y’}y-\frac1 x\geq 0.$$
For the latter case, $\ln \frac yx$ is non-decreasing and converges to $-\infty$ (recall that $\frac yx\to0$,) which is absurd. Therefore, the only solution satisfying $\lim_{x\to\infty}\frac{y(x)}x=L\in\mathbb R_{\geq0}$ is $y\equiv0$, and $L$ also must be $0$.
Edit. Since OP changed one of the conditions in the question: he changed “$L\in\mathbb R$” to “$L>0$”, after I posted this answer (Although I don’t think it is a good idea to change the question after answers posted.) Now things become simpler. We get a contradiction immediately from the second paragraph, when we deduced $L= (1+\sqrt L)\sqrt L=\sqrt L+L$, which contradicts to the assumption $L>0$. I’ll remain the existence of the third paragraph, where we proved that if $L=0$ then $y\equiv0$.
A: $$(\sqrt x + \sqrt y)\sqrt y dx - xdy=0$$
$$\frac{dy}{dx}=\frac{\sqrt{xy}+y}{x}$$
It id homogeneous ODE, take $y=vx$, then
$$v'x+v=\sqrt{v}+v \implies \int \frac{dv}{\sqrt v}=\int \frac{dx}{x}+C \implies 2\sqrt{y/x}=\ln x+C$$
Finally we get $$y(x)=\frac{x}{4}(
\ln x+C)^2 \implies \lim_{x\rightarrow \infty} \frac{y(x)}{x} \rightarrow \infty $$
So the limit does not exist.
