TITLE: What is the degree of the differential equation $\left|\frac{dy}{dx}\right| + \left|y\right| = 0$?
QUESTION [5 upvotes]: Consider the differential equation $$\left|\frac{dy}{dx}\right| + \left|y\right| = 0$$ 
where $\left|\cdot\right|$ means the absolute value function. I have to find the degree of the above differential equation. Can I say the degree of this differential equation is not defined as it is not a polynomial in $y'$? 
If we further solve it, we get $$\left|\frac{dy}{dx}\right| = -\left|y\right|$$ Then taking square we get  $ \left(\frac{dy}{dx}\right)^{2}- y^{2} =0$ which has degree $2$. 
Now can I say that the two differential equation are not same? So the degree of the first one is not defined. Am I right?

REPLY [3 votes]: Okay. Let's do it nice and easy:
$$
\left|\frac{dy}{dx}\right| + \left|y\right| = 0
$$
Consider instead the simplified equation:
$$
\left|A\right| + \left|B\right| = 0
$$
Do you agree that this is exactly the same as:
$$
\begin{cases} A = 0 \\ B = 0 \end{cases}
$$
In your case, substitute:
$$
A = \frac{dy}{dx} \quad \Longrightarrow \quad \frac{dy}{dx} = 0 \quad \Longrightarrow \quad y = \mbox{constant} \\
B = y \quad \Longrightarrow \quad y = 0
$$
Therefore $y(x)=0$ is the only solution, as has been argued already in both the other answers and some of the comments as well.
But yes, the expression contains a derivative. So, strictly speaking, it's a differential equation.
And since the whole is equivalent with the equation $\,y=0$ , I'd say the degree of that ODE is zero.