Finding the constants for a PDE Here we have the straight forward PDE
$U_{xx}=\frac{1}{4}U_{tt}$
Solving it by separation of variables, we get the two ODEs:
\begin{equation}
\begin{array}
f\frac{F_{xx}}{F}=k\\
\frac{G_{tt}}{G}=k
\end{array}
\end{equation}
which are easily solved individually by finding their characteristic equation without the first order derivative term, giving the solutions for F: $\lambda=\frac{\pm\sqrt{-4\cdot(-k)\cdot 4}}{2}=\pm\frac{1}{2}\sqrt{k}$; for G: $\lambda=\frac{\pm\sqrt{-4\cdot(-k)}}{2}=\pm\sqrt{k}$.
Now we have the two solutions:
\begin{equation}
\begin{array}
rF(x)=c_1e^{1/2\sqrt{k}x}+c_2e^{-1/2\sqrt{k}x}\\
G(t)=c_3e^{\sqrt{k}t}+c_4e^{-\sqrt{k}t}
\end{array}
\end{equation}
With initial conditions, $U(0,t)=U(x,0)=0$ and $U(\frac{\pi}{2},t)=\sin2t$ we should have for F(x):
$0=c_1e+c_2e \longrightarrow c_1=c_2=A$:
$F(x)=A(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})$.
The same can be found for G, yielding:
$G(t)=B(e^{\sqrt{k}t}+e^{-\sqrt{k}t})$
But at this stage, we still have 3 unknown coefficients including k, because the next initial condition would act on the product of these two equations, since u(x,t)=f(x)y(t):
\begin{equation}
u(x,t)= A(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})B(e^{\sqrt{k}t}+e^{-\sqrt{k}t})
\end{equation}
So with the third condition, and renaming $AB=C$ we get:
\begin{equation}
\sin2t= C(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})
\end{equation}
which gives that the coefficient A:
\begin{equation}
C= \frac{\sin2t}{(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})}
\end{equation}
which would give:
\begin{equation}
u(x,t)= \frac{\sin2t(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})}{(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})}
\end{equation}
then, cancelling the common terms we get
\begin{equation}
u(x,t)= \frac{\sin2t(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})}{(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})}
\end{equation}
But this is not a normal answer for a PDE problem as such, usually the k-constant is solve for too?
When I put in an arbitrary value for k, I get the following plot:

Though this looks fine, it is strange that the k-coefficient is left unknown.
Thanks
 A: Supposing an additional condition, for instance $U_t(0,x)=0$ we can solve the problem by using the Laplace Transform. After transforming, the problem to solve for $x$ now is
$$
s^2U(x,s)=4U_{xx}(x,s),\ \ U(0,s)=0, U\left(\frac{\pi}{2},s\right)= \frac{2}{s^2+4}
$$
thus we obtain the transformed solution
$$
U(x,s) = \frac{2 e^{\frac{\pi  s}{4}-\frac{s x}{2}} \left(e^{s x}-1\right)}{\left(e^{\frac{\pi  s}{2}}-1\right) \left(s^2+4\right)}
$$
now to invert $U(x,s)$ we use the residues at the solutions for $\left(e^{\frac{\pi  s}{2}}-1\right) \left(s^2+4\right)=0$ which are $s=\{\pm 2i\}\cup\{ \pm 4i k\}, k\in \mathbb{Z}$
and the residues at $\{\pm 2i\}$ are
$$
\lim_{s\to\pm 2i}\frac{2 e^{\frac{\pi  s}{4}-\frac{s x}{2}} \left(e^{s x}-1\right)}{\left(e^{\frac{\pi  s}{2}}-1\right) \left(s\pm 2i\right)}
$$
giving $\frac{2 \sin (x)}{s^2+4}$ and the residues at $\pm 4ik,\ \ k\in \mathbb{Z}$
$$
\lim_{s\to\pm 4ik}\frac{2 e^{\frac{\pi  s}{4}-\frac{s x}{2}} \left(e^{s x}-1\right)}{\left(s^2+4\right)}
$$
giving
$$
\frac{2 \sin (2 k x) (s \sin (\pi  k)+4 k \cos (\pi  k))}{\left(4 k^2-1\right) \left(16 k^2+s^2\right)},\ \ k\in \mathbb{Z}_{\ge 0}
$$
and after inversion we have
$$
U(x,t) = \sin (2 t) \sin (x)+\sum_{k=0}^n\frac{2 \sin (2 k x) (\sin (\pi  k) \cos (4 k t)+\cos (\pi  k) \sin (4 k t))}{4 k^2-1}
$$
Follows a plot with $n=10$

