# Differential equation with no solution that satisfies a certain condition

It is my first week dealing with Differential Equations, and I am stuck at the following question. I am not sure how to approach this, and any help would be greatly appreciated.

Let $$L$$ be a positive number.

Please show that the equation $$(\sqrt{x}+\sqrt{y})\sqrt{y}dx=xdy$$ has no solution that satisfies $$\lim_{x\to \infty} \frac{y(x)}{x}=L$$.

$$(\sqrt{x}+\sqrt{y})\sqrt{y}dx=xdy$$

means

$$\sqrt{x}\sqrt{y}+y=xy'.$$

If we divide by $$x$$ we get

$$\sqrt{ \frac{y(x)}{x}}+\frac{y(x)}{x}=y'(x).$$

Now let $$x \to \infty$$ to derive

$$y'(x) \to \sqrt{L}+L.$$

Furthermore, from $$\lim_{x\to \infty} \frac{y(x)}{x}=L$$, we see that there is $$x_0>0$$ such that $$\frac{y(x)}{x} > L/2$$ for $$x>x_0.$$ Hence $$y(x) > \frac{L}{2}x$$ if $$x>x_0.$$

It results that $$y(x) \to \infty$$ as $$x \to \infty.$$

Then, by L'Hospital

$$L=\lim_{x\to \infty} \frac{y(x)}{x}= \lim_{x\to \infty}y'(x).$$

But this gives $$L=L + \sqrt{L}$$, hence $$L=0,$$ a contradiction.

$$(\sqrt{x}+\sqrt{y})\sqrt{y}dx=xdy$$

$$(\sqrt{\dfrac{x}{x}}+\sqrt{\dfrac{y}{x}})\sqrt{\dfrac{y}{x}}dx=dy$$

$$(1+\sqrt{\dfrac{y}{x}})\sqrt{\dfrac{y}{x}}dx=dy$$

Substituting $$y=xt$$ and $$\dfrac{dy}{dx}=t+x\dfrac{dt}{dx}$$

$$(1+\sqrt{t})\sqrt{t}=t+x\dfrac{dt}{dx}$$

$$\sqrt{t}+t=t+x\dfrac{dt}{dx}$$

$$\sqrt{t}=x\dfrac{dt}{dx}$$

$$\dfrac{dx}{x}=\dfrac{dt}{\sqrt{t}}$$

Integrating both sides

$$2\sqrt{t}=lnx+C$$

Substituting the value of $$t$$

$$2\sqrt{\dfrac{y}{x}}=\ln x+C$$

or

$$\dfrac{y}{x}=\left(\dfrac{\ln x +C}{2}\right)^2$$

Now,

$$\displaystyle \lim_{x \to \infty}\dfrac{y}{x}=\lim_{x \to \infty}\left(\dfrac{\ln x +C}{2}\right)^2$$

Since the value of $$\dfrac{y}{x}$$ is a square therfore, all the solution satisfy

$$\displaystyle \lim_{x\to \infty} \frac{y(x)}{x}=L$$