# 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).

• The second line of your argument does not work to check if $x=0$ or $y=0$ is a solution since you are assuming $x\not=0$ and $y\not=0$ to make your differential equation separable. Is there another strategy that you can use to solve the differential equation so that you can consider the cases when $x=0$ or $y=0$? – Mee Seong Im Aug 17 '17 at 14:23
• @MeeSeongIm I'm trying to understand the concept. when I plug the values to the original question I can see they both produce equality. but for $y=0$ I can't plug in the solution equation since $y\gt 0$ since we have $ln|y|$ but according to my book this is a solution – segevp Aug 17 '17 at 14:44
• the solution containes the product logarithm – Dr. Sonnhard Graubner Aug 17 '17 at 14:51
• @Cyclone according to my book it is a solution. what about $x=0$? – segevp Aug 22 '17 at 9:11

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 $$\left( \sqrt{x\, y(x)} - \sqrt{x}\right) \frac{dy}{dx} + y(x) = 0.\tag{1}$$ 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 $x(y)$.