Equation for stationary string 
I have some doubts on the following derivation of the EOM of a stationary string.
Let $F_x, F_y$ be horizontal and vertical tension of the string
$\mu$ be the mass per unit length of the string [kg/m]
$$\frac{F_{2y}}{F_{2x}} = \tan\theta_+ = y'(x + dx)$$
$$\frac{F_{1y}}{F_{1y}} = \tan\theta_- = y'(x)$$
$$dm = \mu dx$$
No net force,
$$F_{1x} = F_{2x} = F$$
$$F_{2y} - F_{1y} = gdm = \mu gdx$$
$$F(y'(x + dx) - y'(x)) = \mu gdx$$
$$F\frac{d^2y}{dx^2} = \mu g$$
Solving gives
$$y = \frac{\mu g}{2F}x^2 + Ax + B$$
I am new to calculus, so I have some questions, please explain in a way that a beginner can understand, thanks!:

*

*why do we use $dm = \mu dx$ instead of $dm = \mu \sqrt{1 + (\frac{dy}{dx})^2} dx$ here? because it seems to me that $\mu \cdot dx$ doesn't describe the mass at some point on a curve, but $\mu \cdot ds$ does.

*Isn't a string held at both end called catenary which is not a parabola? then why in this case it is a parabola? What are the difference between a catenary and this graph?

I might be making some very basic conceptual mistakes because I am mostly self learning now. So please help me clear my doubt.
 A: $$\newcommand{\md}{\mathrm{d}}$$Yes, you are right. In the most general case, we must use $\md m = \mu\, \md s$ for the mass of the infinitesimally small element of the rope. In essence, what that equation is saying is that the mass of the small section of rope is the equal to the mass per unit length times the length of the small section.
I don't know where you got the derivation from, but they probably used $\md m = \mu\, \md x$ for the mass because they assumed that the rope sags very little. (They may have even said that. I would advise that you read their explanation closely to see what assumptions they made.)
If the rope is almost taut (i.e. little to no sagging), then $\sqrt{1 + \left( \dfrac{\md y}{\md x}\right)^2}\, \md x \approx 1 \,\md x$, which implies that $\md s \approx \md x$.
A catenary is the name given to equations of the form
$$y = a\cosh \left(\dfrac{x}{a}\right)$$
where $\cosh x$ is the hyperbolic cosine function, which is defined as $$\cosh x = \dfrac{e^x + e^{-x}}{2}$$
It is different from the parabola, as we can see from the graph below:

In the graph, $y = \cosh x$ is shown in red and $y = x^2$ is in blue.
Here's another graph showing how a parabola can be an approximation to a catenary, if the "sag" is relatively small:

Here, $y = \cosh x$ is in red once again, but the equation of the blue curve is $y = \dfrac{1}{2}x^2 + 1$. As we can see, if $x$ is close to $0$, then the parabola is a reasonably good approximation of the catenary.
