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The problem requires solving the partial differential equation (PDE) using either the Laplace transform or separation of variables. The PDE is given as:

\[
\frac{\partial u}{\partial t} = 3 \frac{\partial u}{\partial x}
\]

with the initial condition:

\[
u(x, 0) = 4e^{-2x}
\]

Let's solve this using both methods for clarity.

### Method 1: Separation of Variables
1. **Assume a solution** of the form \( u(x, t) = X(x)T(t) \).
2. Substitute into the PDE:
   \[
   \frac{\partial}{\partial t} (X(x)T(t)) = 3 \frac{\partial}{\partial x} (X(x)T(t))
   \]
   This becomes:
   \[
   X(x) \frac{dT}{dt} = 3 T(t) \frac{dX}{dx}
   \]
3. **Separate the variables**:
   \[
   \frac{1}{T(t)} \frac{dT}{dt} = 3 \frac{1}{X(x)} \frac{dX}{dx} = -\lambda
   \]
   where \(\lambda\) is a separation constant.

4. **Solve the ODEs**:
   - For \(T(t)\):
     \[
     \frac{dT}{dt} = -\lambda T \implies T(t) = A e^{-\lambda t}
     \]
   - For \(X(x)\):
     \[
     \frac{dX}{dx} = -\frac{\lambda}{3} X \implies X(x) = B e^{-\frac{\lambda}{3} x}
     \]

5. The general solution is:
   \[
   u(x, t) = C e^{-\lambda t} e^{-\frac{\lambda}{3} x}
   \]

6. **Apply the initial condition** \( u(x, 0) = 4 e^{-2x} \):
   \[
   u(x, 0) = C e^{-\frac{\lambda}{3} x} = 4 e^{-2x}
   \]
   This implies \(\lambda = 6\) and \(C = 4\).

7. **Final solution** using separation of variables:
   \[
   u(x, t) = 4 e^{-6t} e^{-2x}
   \]

### Method 2: Laplace Transform
1. **Take the Laplace transform** with respect to \(t\):
   \[
   \mathcal{L}\left(\frac{\partial u}{\partial t}\right) = 3 \mathcal{L}\left(\frac{\partial u}{\partial x}\right)
   \]
   Let \( \mathcal{L}\{u(x, t)\} = \tilde{u}(x, s) \).

2. Use the property of the Laplace transform:
   \[
   s \tilde{u}(x, s) - u(x, 0) = 3 \frac{\partial \tilde{u}}{\partial x}
   \]

3. Substitute the initial condition \( u(x, 0) = 4 e^{-2x} \):
   \[
   s \tilde{u}(x, s) - 4 e^{-2x} = 3 \frac{\partial \tilde{u}}{\partial x}
   \]

4. **Solve the resulting first-order ODE** in \(x\):
   This is a linear first-order ODE for \(\tilde{u}(x, s)\):
   \[
   \frac{\partial \tilde{u}}{\partial x} - \frac{s}{3} \tilde{u} = -\frac{4}{3} e^{-2x}
   \]

5. Solve this using an integrating factor:
   \[
   \tilde{u}(x, s) = A(s) e^{\frac{s}{3} x} + \int -\frac{4}{3} e^{-2x} e^{\frac{s}{3} x} \, dx
   \]

6. Find \(A(s)\) by applying boundary conditions or simplifying further, then take the inverse Laplace transform to get \(u(x, t)\).

Both methods should yield a solution that is consistent with the initial condition, resulting in:

\[
u(x, t) = 4 e^{-6t} e^{-2x}
\]

Would you like a detailed explanation of any specific step? Here are some related questions:

1. How does separation of variables work in solving PDEs?
2. What is the significance of the Laplace transform in solving PDEs?
3. How do you apply the Laplace transform to time-dependent PDEs?
4. How do initial conditions influence the solution of a PDE?
5. Why do we assume \(u(x, t) = X(x)T(t)\) in separation of variables?

**Tip:** When solving PDEs using separation of variables, always check if the boundary and initial conditions are compatible with the assumed solution form.

By using Laplace transform or separation of variables, solve the PDE: ∂u/∂t = 3 ∂u/∂x with u(x, 0) = 4e^(-2x)

Learn how to solve the partial differential equation (PDE) ∂u/∂t = 3∂u/∂x with the initial condition u(x, 0) = 4e^(-2x) using both the separation of variables method and the Laplace transform. This problem covers essential mathematical techniques for solving PDEs.

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