The given partial differential equation is:

$u_t + 3t^2 u_x = 0, \quad u(x,0) = 2 + t^3$

This is a first-order linear PDE. To solve it, we can use the method of characteristics.

Step 1: Characteristics equation

The PDE can be written as:

$\frac{\partial u}{\partial t} + 3t^2 \frac{\partial u}{\partial x} = 0$

The corresponding characteristic equations are:

$\frac{dx}{dt} = 3t^2, \quad \frac{du}{dt} = 0$

Step 2: Solving the characteristic equations

For $\frac{dx}{dt} = 3t^2$:

We can integrate this to find $x$ as a function of $t$:

$\frac{dx}{dt} = 3t^2 \quad \Rightarrow \quad x = t^3 + C_1$

where $C_1$ is a constant determined by the initial condition.

For $\frac{du}{dt} = 0$:

This equation implies that $u$ is constant along the characteristic curves, meaning:

$u = C_2$

Step 3: Applying the initial condition

The initial condition is given as $u(x, 0) = 2 + t^3$, which implies:

$u(x, 0) = 2$

From the characteristic equation $x = t^3 + C_1$, when $t = 0$, we get $x = C_1$. Therefore, $C_1 = x$ at $t = 0$, and thus along each characteristic curve, the solution remains constant.

Thus, the solution is:

$u(x, t) = 2 + t^3$

Let me know if you need further details or clarifications.

Related Questions:

1. How do we apply the method of characteristics to more complex PDEs?
2. How can initial conditions influence the solution of a PDE?
3. What are characteristic curves, and why are they important in solving PDEs?
4. Can this method be extended to nonlinear PDEs?
5. What types of boundary conditions can be applied to first-order PDEs?

Tip:

For first-order PDEs, the method of characteristics is a powerful technique that reduces the partial differential equation to a set of ordinary differential equations.