What is the degree of the differential equation $\left|\frac{dy}{dx}\right| + \left|y\right| = 0$? 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?
 A: 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.
A: Considering that the initial equation is only true for $y'=0=y$, there is no sense in computing degrees or orders.
A: Your typographical picture is not a differential equation but a clumsy way of defining the function $x\mapsto y(x):\equiv0$. Neither the bounty nor the upvotes to the question, nor squaring, can improve this situation.
A differential equation is of the form $y'=f(x,y)$, or $F(x,y,y')=0$, and defines for each point $(x,y)$ of some open set $\Omega\subset{\mathbb R}^2$ a slope $y'\in{\mathbb R}$ (maybe several admissible slopes). We then ask for the functions $x\mapsto y(x)$ that satisfy $y'(x)=f\bigl(x,y(x)\bigr)$, resp., $F\bigl(x,y(x),y'(x)\bigr)=0$  for all $x$ in the domain of $y(\cdot)$.
