Solving a wave equation (Partial Differential equations) A string, stretched between the points $0$ and $\pi$ on the $x$ axis and initially chosen so that at rest is released from the position $y=f(x)$. Its motion is opposed by air resistance which is proportional to the velocity at each point. Let the unit of time be chosen so that the equation of motion becomes
$$y_{tt}(x,t)=y_{xx}(x,t)-2\beta y_{t}(x,t)$$
$$(0<x<\pi ,~ t>0)$$
where $\beta$ is a positive constant assuming that $0<\beta<1$ derive the following expression in the from of
$$y(x,t)=e^{-\beta t}\sum_{n=1}^{\infty}b_{n}\left(\cos \alpha_{n}t+\frac{\beta}{\alpha_{n}} \sin{a_n}t\right) \sin(nx)$$
What I tried:
First my boundary conditions are
\begin{align} y(0,t) &=0\\y(\pi,t) &=0\\y(x,0) &=f(x)\\ y_{t}(x,0) &=g(x)\end{align}
Before solving for the problem I would like to first verify whether my boundary conditions are correct. Also I tried using the separation of variables method for this problem but it doesn't seem to work so should I use the   de-Alembert solution instead? And how do I use it for this problem?
 A: The separation of variables technique works all right. What's the problem you have with it? Your boundary/initial conditions are rigth except for the last one: you mentioned that the string is at rest at the beginning, so $g(x)=0.$ 
The equation is easily solved by the method of separation of variables. It works as follows: you find first a general solution to the equation (forgetting about the initial condition) with your boundary conditions of the form $y(x,t)=X(x)T(t).$ You will find that there are infinitely many possible solutions of this type, $y_1(x,t)=X_1(x)T_1(t), \hspace{2mm} y_2(x,t)=X_2(x)T_2(t),\ldots, y_n(x,t)=X_n(x)T_n(t), \ldots$ Then, since your equation is linear and thus any linear superposition of solutions is still a solution, you write an Ansatz for the solution of the problem with initial conditions as:
$$y(x,t)=\sum_{n=1}^{\infty} a_n X_n(x)T_n(t),$$ 
where $a_{n}$ are some coefficients you have to adjuts so that $y(x,0)=f(x)$ and $\dfrac{\partial y(x,t)}{\partial t} \vert_{t=0}=0.$
For what is to follow, I will write just $X$ and $T$ instead of $X(x)$ and $T(t)$ and it will be implied that the derivatives $X'$ and $T'$ are of course with respect to variables $x$ and $t$ respectively. In your concrete problem, the equation is $y_tt=y_xx-2by_t,$ and introducing the Ansatz $y=XT$ we get:
$$XT''=TX''-2bXT'$$
by reordering this a bit we have:
$$\dfrac{T''+2bT'}{T}=\dfrac{X''}{X}$$
And here comes the crucial idea of the technique of separation of variables. Since the left hand side only depend on $t$ and the right hand side only depends on $x,$ we conclude that both expressions must be equal to a certain constant $K.$ Therefore, we have reduce our Partial Diferential Equations problem to two Ordinary Differential Equations:
$$T''+2bT'-KT=0$$
$$X''-KX$$
Taking into account your boundary conditions, it is already easy to see that $X$ must be a sine function with argument $nx,$ for $n$ an integer. That is, $K=-n^2.$ You can plug this information into the first equation for $T$ and solve it. 
I will omit these step since I believe it is pretty straigth-forward from here, but please ask if you have any difficulty with it!
It is particularly easy since they already telling you the form of the solutions!
Next you just combine the solutions linearly with some coefficients. What you want to do now is to find the coefficients $b_n$ that make that $\sum b_n \sin(nx)=f(x).$ The answer is that these are Fourier coefficients. I don't know if you know about this, but even if you don't, the recipe is pretty easy to follow and you can look up the key formulae in the wikipedia
