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





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.





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





Integrating both sides


Substituting the value of $t$

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


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


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

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

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


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