Solve the initial value problem: $$ u_{tt} - u_{xx} = 0 \\ u(x,0) = 0 \\ u_t(x,0) = \begin{cases} \cos \pi (x-1) & \text{if } 1 < x < 2 \\ 0 & \text{otherwise} \end{cases} $$ Find and draw the solution at $t=3$ for
(a). $x\in (-\infty, \infty)$.
(b). $x\in (0, \infty)$ with $u_x(0,t) = 0$.
(c). $x\in (-\infty, 3)$ with $u(3,t) = 0$.
(d). $x\in (0,3)$ with $u_x(0,t)=u_x(3,t)=0$.
For (a) since we are in the entire real line, we can just use d'Alembert's solution and obtain $u = \sin(\pi(x-3t))-\sin(\pi(x+3t))$, but how do we do it for the other cases that we have semi infinite domain?
For (d) we can use separation of variables, correct?
You should only ask questions about a single problem. This is not the place for a homework dump.
With that being said, I'm going to solve the first problem. First, define the primitive of $u_t(x,0)=g(x)$ as
$$ G(s) = \int g(s)\ ds = \begin{cases} \dfrac{1}{\pi}\sin\big(\pi(s-1)\big), & 1 < s < 2 \\ 0, & \text{otherwise} \end{cases} $$
Then the solution given by d'Alembert's formula is
$$ u(x,t) = \frac{G(x+t) - G(x-t)}{2} $$
where you have to check case by case for both $x+t$ and $x-t$. Below is a graph ($x$ vs. $t$) of the 4 characteristic lines going through $x=1$ and $x=2$.
The numbered regions are as follows:
\begin{array}{rrr} \text{I}: && x - t < 1, && x + t < 1 \\ \text{II}: && x - t < 1, && 1 < x + t < 2 \\ \text{III}: && x - 1 < 1, && x + t > 2 \\ \text{IV}: && 1 < x - t < 2, && 1 < x + t < 2 \\ \text{V}: && 1 < x - t < 2, && x + t > 2 \\ \text{VI}: && x - t > 2, && x + t > 2 \\ \end{array} Then, you can simplify the solution
$$ u(x,t) = \begin{cases} 0, && (x,t) \in \text{I, III, VI} \\ \dfrac{1}{2\pi}\sin\big(\pi(x+t-1)\big), && (x,t) \in \text{II} \\ \dfrac{1}{2\pi}\Big[\sin\big(\pi(x+t-1)\big) - \sin\big(\pi(x-t-1)\big) \Big], && (x,t) \in \text{IV} \\ -\dfrac{1}{2\pi}\sin\big(\pi(x-t-1)\big), && (x,t) \in \text{V} \end{cases} $$