A catenary cable is one that is hung between two points not in the same vertical line. As depicted in

Fig. $3\mathrm{.}2$a$.$ It is subject to no loads other than its own weight. Thus, its weight $(\frac{N}{m})$ acts as a uniform

load per unit length along the cable. A free$-$body diagram of a section $AB$ is depicted on Fig. $3\mathrm{.}2$b$,$

where $T_A$ and $T_B$ are the tension forces at the end. Based on the horizontal and vertical force

balances, the following differential equation model of the cable can be derived:

$\frac{d^{2}y}{d{x}^2}=\frac{w}{T_A}\sqrt[\phantom{2}]{1+(\frac{dy}{dx}{)}^2}$

Calculus can be employed to solve this the equation for the height $y$ of the cable as a function

of distance $x,$

$y=\frac{T_A}{w}cosh(\frac{w}{T_A}x)+y_0-\frac{T_A}{w}$

Where the hyperbolic cosine can be computed by:

$coshx=\frac{1}{2}(e^x+e^{-x})$

Use either Newton$-$Raphson method, Secant method, and Fixed$-$Point method to manually

calculate the value of $T_A$ given values for the parameters $w=10$ and $y_0=5,$ such that the cable has a

height of $y=15$ at $x=50.$ Please do explain step by step and the logic or process behind it$.$