Solution: The given PDE is the heat equation: \[ u_t = u_{xx}, \quad 0 < x < \pi, \quad t > 0, \] with boundary conditions: \[ u(0, t) = 0, \quad u(\pi, t) = 0, \quad t > 0, \] and initial condition: \[ u(x, 0) = \sin(3x), \quad 0 \leq x \leq \pi. \]

Step 1: Assume a separable solution. We assume a solution of the form: \[ u(x, t) = X(x) T(t). \] Substituting into the heat equation, we get: \[ X(x) T'(t) = X''(x) T(t). \] Dividing both sides by \( X(x) T(t) \), we obtain: \[ \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda, \] where \( \lambda > 0 \) is a separation constant. If $\lambda \leq 0$, then one can show that $X(x) = 0$, which will imply the trivial solution. Thus, we will consider $\lambda > 0$.

Step 2: Solve the spatial ODE. The spatial part of the equation is: \[ X''(x) + \lambda X(x) = 0. \] The boundary conditions \( u(0, t) = 0 \) and \( u(\pi, t) = 0 \) imply: \[ X(0) = 0, \quad X(\pi) = 0. \] The general solution to this ODE is: \[ X(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x). \] Applying the boundary conditions, we find: \[ X(0) = A = 0, \quad X(\pi) = B \sin(\sqrt{\lambda} \pi) = 0. \] For non-trivial solutions, \( \sin(\sqrt{\lambda} \pi) = 0 \), which implies: \[ \sqrt{\lambda} = n, \quad n = 1, 2, 3, \dots. \] Thus, the eigenvalues are \( \lambda_n = n^2 \), and the corresponding eigenfunctions are: \[ X_n(x) = \sin(nx). \]

Step 3: Solve the time dependent ODE. The temporal part of the equation is: \[ T'(t) + \lambda T(t) = 0. \] For \( \lambda_n = n^2 \), the solution is: \[ T_n(t) = C_n e^{-n^2 t}. \]

Step 4: Construct the general solution. The general solution to the heat equation is: \[ u(x, t) = \sum_{n=1}^{\infty} C_n \sin(nx) e^{-n^2 t}. \]

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Step 5: Apply the initial condition. The initial condition \( u(x, 0) = \sin(3x) \) implies: \[ \sin(3x) = \sum_{n=1}^{\infty} C_n \sin(nx). \] Thus, we have \[ C_n = \begin{cases} 1, & n = 3, \\ 0, & n \neq 3. \end{cases} \] Therefore, the solution to the initial-boundary value problem is: \[ u(x, t) = \sin(3x) e^{-9t}. \]