Checking a solution of a differential equation I was asked to solve the differential equation (solved it):
$$\left(\sqrt{xy}-\sqrt x\right)dy+ydx=0.$$
This is a separable differential equation:
$$\frac{\left(\sqrt y-1\right)dy}{-y}=\frac{dx}{\sqrt{x}}$$ 
assuming $x\ne 0$ and $y\ne 0$.
And the solution is:
$$2\sqrt y-\ln|y|+2\sqrt x=C.$$
The next thing I was asked to do is to check if $x=0$ or $y=0$ is a solution.
How do I check that and what does it mean? I tried plugging it in but I'm not sure I'm doing it right (I think this should not be a solution but I'm wrong).
 A: To me, the problem with writing a differential equation as a differential (i.e. $f(x,y) dx + g(x,y) dy = 0$) is that it's slightly ambiguous if you're talking about 'solutions'. In particular, a 'solution' is generally a function that depends on a variable. Where am I going with this?
To check if $y=0$ and $x=0$ are solutions, this is exactly the ambiguity you encounter. If you say `$y=0$ is a solution', you're actually talking about the function $y(x) = 0$. In that case, it's best to rewrite the differential as an ordinary differential equation for a function $y(x)$, yielding
\begin{equation}
 \left( \sqrt{x\, y(x)} - \sqrt{x}\right) \frac{dy}{dx} + y(x) = 0.\tag{1}
\end{equation}
It's clear that when you substitute $y(x) = 0$ (and hence $\frac{dy}{dx} =0$), you satisfy (1). So, $y = 0$ solves the original differential equation.
Now, you can do the same with $x=0$. In this case, if you say '$x=0$ is a solution', that means that you're looking at the function $x(y) = 0$. Writing the original differential equation as an ordinary differential equation for $x(y)$ (and thus involving $\frac{dx}{dy}$) will enable you to substitute $x(y)=0$, and check whether it solves that ODE for $x(y)$.
