Solution: We will use separation of variables. Let
\begin{equation}\label{eq:17May2024-2}
u(x,t) = X(x) T(t), \quad 0 \lt x \lt a \text{ and } t > 0.
\end{equation}
Substituting $u(x,t)$ into the given heat equation (Equation \ref{eq:17May2024-1}), we have
\begin{align*}
XT' = k X''T & \implies \frac{X''}{X} = \frac{T'}{kT} = -\lambda \\
& \implies X'' + \lambda X = 0 \text{ and } T' + k \lambda T = 0.
\end{align*}
Now we have wo ordinary differential equations. Let us determine the boundary conditions.
\begin{align*}
u(0,t) = 0 \implies X(0)T(t) = 0 \implies X(0) = 0 \\
u(a,t) = 0 \implies X(a)T(t) = 0 \implies X(a) = 0.
\end{align*}
Thus, we have an ODE
\begin{equation}\label{eq:17May2024-3}
\begin{gathered}
X'' (x) + \lambda X(x) = 0 \quad 0 \lt x \lt a \\
X(0) = 0 = X(a).
\end{gathered}
\end{equation}
Note that if $\lambda \leq 0$, then $u\equiv 0$, which is not possible. Thus, $\lambda >0$. So, we assume that $\lambda =s^2$ for some $s\in \mathbb{R} \setminus \{ 0 \} $.
Solving Equation \eqref{eq:17May2024-3}, we get
\[
X(x) = C_1 \cos sx + C_2 \sin sx .
\]
Applying the given boundary conditions, we obtain
\begin{align*}
X(0) = 0 & \implies C_1 = 0 \text{ and } X(a) = 0 \implies C_2 \sin sa = 0 .
\end{align*}
If $C_2 = 0$, then $X\equiv 0$ which implies $u\equiv 0$. Thus, $\sin sa=0$. So,
\[
sa = n\pi \implies s = \frac{n\pi }{a} \implies \lambda = s^2 = \frac{n^2 \pi ^2}{a^2}.
\]
For each $\lambda _n$ the corresponding eigenfunction will be
\begin{equation}\label{eq:17May2024-4}
X_n(x) = \sin \left( \frac{n\pi x}{a} \right), \quad n\geq 1.
\end{equation}
Similarly the other differential equation
\[
T_n'(t) + \lambda _n k T_n(t) = 0
\]
will have solution given by
\[
T_n(t) = e^{-\lambda _n kt} = e^{\frac{-n^2 \pi ^2}{a^2}kt}.
\]
Thus,
\begin{equation}\label{eq:17May2024-5}
u_n(x,t) = X_n(x) T_n(t) = e^{\frac{-n^2 \pi ^2}{a^2}kt} \sin \left( \frac{n\pi x}{a} \right),\quad n\geq 1.
\end{equation}
So, using the superposition of these solutions, we get
\begin{equation}\label{eq:17May2024-6}
u(x,t) = \sum_{n=1}^{\infty} u_n(x,t) = \sum_{n=1}^{\infty} b_n \sin \left( \frac{n\pi x}{a} \right)e^{\frac{-n^2 \pi ^2}{a^2}kt},
\end{equation}
where $b_n$ is determined by the initial condition of Equation \eqref{eq:17May2024-1}. Let $u(x,0) = f(x)$. Then substituting $t = 0$ in Equation \eqref{eq:17May2024-6}, we obtain
\begin{align*}
f(x) = u(x,0) = \sum_{n=1}^{\infty} b_n \sin \left( \frac{n\pi x}{a} \right).
\end{align*}
Hence, we need to find the Fourier sine series of $u(x, 0)$ on the interval $[0,a]$, and the coefficients are
\begin{align*}
b_n & = \frac{2}{a} \int _0^a f(x) \sin \left( \frac{n\pi x}{a} \right), \mathrm{d} x, \ n\geq 1 \\
& = \frac{2}{a} \left[ \int _0^{\frac{a}{2}} \sin \left( \frac{n\pi x}{a} \right) \mathrm{d} x + \int_{\frac{a}{2}}^a 2 \sin \left( \frac{n\pi x}{a} \right) \mathrm{d}x \right] \\
& = \frac{2}{a} \left[ -\cos \left( \frac{n\pi x}{a} \right) \right]_0 ^ {\frac{a}{2}} + \frac{4}{a} \left[ -\cos \left( \frac{n\pi x}{a} \right) \right] _{\frac{a}{2}} ^a \\
& = \frac{2}{n\pi } \left[ 1 - \cos \left(\frac{n\pi }{2}\right) \right] + \frac{4}{n\pi }\left[ \cos \left( \frac{n\pi }{2} \right) - \cos (n \pi ) \right] \\
& = \frac{2}{n\pi } + \frac{2}{n\pi } \cos \left( \frac{n\pi }{2} \right) - \frac{4}{n\pi } \cos (n\pi ) \\
& = \frac{2}{n\pi } + \frac{2}{n\pi } \cos \left( \frac{n\pi }{2} \right) - \frac{4}{n\pi } (-1)^n.
\end{align*}
Thus, the solution of Equation \eqref{eq:17May2024-1} is given by
\begin{gather*}
u(x,t) = \sum_{n=1}^{\infty} b_n \sin \left( \frac{n\pi x}{a} \right)e^{\frac{-n^2 \pi ^2}{a^2}kt},\ 0 \lt x \lt a \text{ and } t>0,
\end{gather*}
where
\[
b_n = \frac{2}{n\pi } + \frac{2}{n\pi } \cos \left( \frac{n\pi }{2} \right) - \frac{4}{n\pi } (-1)^n.
\]