Differential Equations and Newtons method How can I approach this question?
For problem one this is what I did:
Given the DE,
$$p'(x) = p''(x) + \left(2\pi*\frac{f}{c}\right)^2p(x) = 0,$$
and its solution, $p(x) = \sin(kx)$, I substituted the things on the right hand side of the DE to get
$$p'(x) = -\sin(kx)\,k^2 + \cos(kx) + \left(2\pi*\frac{f}{c}\right)^2 \sin(kx) = .0$$
Then, I plugged in $x=0$ to get $\cos(kx) = 0$.
My answer does not depend on $f$ and $c$ at all, which is what the question is asking for.
What is the right approach to solving for $k$? Also, for the initial condition $p(0) = 0$, shouldn't any value of $k$ work because you will always get $\sin(0) = 0$?
Problem 1: Phonetics
The shape of the vocal tract tends to promote certain sound frequencies. For example, to produce the first vowel in the word about, the vocal tract opens widely. The cross-sectional area throughout the vocal tract is approximately the same and may be modeled by a cylinder with one end open (the lips) and the other end closed (the glottis/vocal folds).
Let $p(x)$ denote the sound pressure at position x within the cylinder starting at the lips, $x=0$, and ending at the glottis, $x=L$, where $L$ is the length of the vocal tract. Then $p(x)$ satisfies the differential equation
$$ p''+(2πfc)2p=0\tag{$*$}$$
with conditions at the endpoint $p(0)=0$ and $p'(L)=0$. This is called a boundary value problem. $f$ is the frequency of the produced sound, and $c$ is the speed of sound.
Show that $p(x)=\sin(kx)$ solves the differential equation and the first boundary condition ($p(0)=0$) when $k$ is chosen correctly. What value of $k>0$ ensures that this function is a solution? Your answer will depend on $f$ and $c$.
Use the second boundary condition $p'(L)=0$ to determine the frequencies $f$ that the vocal tract can produce. Note: your answer should be expressed in terms of an integer $n$ so that there would be infinitely many frequencies produced. Your answer will also depend on $L$ and $c$.
Problem 2: Third order differential equations and Newton's method
We are trying to solve the third order differential equation
$$y'''+3y''−y=0. \tag{$**$}$$
Inspired by earlier results in the course, we guess that the solution to this differential equation might be $y=Ae^{kx}$ where A and k are constants. Show that by plugging this guess into the differential equation we get an equation for $k$:
$k_3+3k_2−1=0$.
Find the positive root of this cubic by using three iterations of Newton's method and write down a solution to $(∗∗)$. Hint: plot your cubic to come up with a starting point for Newton's method
 A: You are being asked to find a relationship between $k$ and $f$ and $c$.    $\sin{(kx)}$ was given to show you the form of the equation but now you are asked to determine exactly what $k$ should be in this case, in terms of the other quantities in the problem.
To do this, generate the needed derivates of $\sin{(kx)}$ and substitute them into the second part of the equation (I don't know where you got the $p'$part from as it was not given in the problem below).  Now determine how to set the value of $k$ so that the equation is always true (hint: $\sin$ will not always be $0$ so the other part must be set to $0$).
Once you have replaced $k$ with quantities from the original problem, you are actually on the right track regarding the boundary conditions: you will find that a trig function must be zero, so what values of the argument will satisfy that condition?
The second problem is the same techniques applied again, just to equations of a different form.
A: You Have written that the DE is in the form $p'(x)=p''(x)+(2\pi*\frac fc)^2p(x)=0$ but in "Problem 1:Phonetics" it is stated as $p''(x)+(2\pi fc)2p(x)=0$? 
which I think is an error, since after running through the problem the it must be $p''(x)+(\frac{2\pi f}{c})^2p(x)=0$ since the sine and all trig functions are Transcendental Functions they must have a dimensionless argument, which can only be achieved through the form above, and the square just makes problem nicer. 
Now you have the problem in a form of 
$$
p''(x)+k^2p(x)=0
$$
This is an extremely important ODE that will pop up everywhere in your math career. I will give a quick introduction but for a more detailed version have a look here 
You can assume a solution of $p(x)=Ae^{nx}$ like all second order and higher DE's but when you substitute into the equation you are left with $n^2+k^2=0$ see the problem? you will end up with complex roots for n which will give you 
$$
p(x)=Ae^{ikx} 
$$
which if you have done any complex analysis you will know can be written as $p(x)=A\cos(kx)+B\sin(kx)$ via Euler's identity. So with that in mind the solution to the problem in Phonetics section can be written as
$$
p(x)=A\sin\left(\frac{2\pi fx}{c}\right)+B\cos\left(\frac{2\pi fx}{c}\right)
$$
Notice how the argument is dimensionless, Anyway now the boundary conditions can be used $p(0)=0$ and $p'(L)=0$.  The first BC will eliminate the cos term since $B=0$. Now you are left with 
$$
p'(L)=\frac{A2\pi f}c \cos\left(\frac{2\pi fL}{c}\right)=0
$$
The constant A can equal zero but this would give you a trivial solution, so if you let $cos\left(\frac{2\pi fL}{c}\right)=0$ you can exploit the periodicity of the cosine function. If you plot the cosine function you know there will be a root at every $\frac{(2n+1)\pi}{2}$ where n is an integer. Therefore you can let the argument of the cosine function equal to $\frac{(2n+1)\pi}{2}$ and solve for the frequency. 
Lastly you should notice that $\frac{(2n+1)\pi}{2}$ is true for all integer values of n and you can use this to write the frequency as an infinite series solution. 
Problem 2 use Newton's method to solve the polynomial $k^3+3k-1=0$ this is a simple numerical problem, use the formula defined in the link. When choosing a starting value, plot the function in matlab or wolfram alpha and look for an small interval that contains the root.
