We are given the partial differential equation (PDE):

$U_t(t, x) - 2tx U_x(t, x) = u(t, x) + t \quad \text{for} \quad (t, x) \in \mathbb{R}^2,$ with the initial condition:

$u(0, x) = -2x \quad \text{for} \quad x \in \mathbb{R}.$

To solve this problem, we will use the method of characteristics. The given equation is a first-order linear PDE, and we can solve it by reducing it to a system of ordinary differential equations (ODEs) along the characteristic curves.

Step 1: Set up the characteristic equations

The PDE can be written in characteristic form as:

$\frac{dt}{1} = \frac{dx}{-2tx} = \frac{du}{u(t,x) + t}.$

This suggests that we will follow the characteristic curves defined by:

1. $\frac{dt}{1} = \frac{dx}{-2tx}$,
2. $\frac{dt}{1} = \frac{du}{u(t,x) + t}$.

Step 2: Solve for $x(t)$ along the characteristic curves

Start by solving the first equation:

$\frac{dx}{dt} = -2tx.$

This is a separable differential equation. Rearranging the terms:

$\frac{dx}{x} = -2t \, dt.$

Integrating both sides:

$\ln |x| = -t^2 + C_1 \quad \Rightarrow \quad x = C_2 e^{-t^2},$

where $C_2 = e^{C_1}$ is a constant determined by initial conditions.

Step 3: Solve for $u(t)$ along the characteristic curves

Now consider the second equation:

$\frac{du}{dt} = u(t,x) + t.$

To solve this, first notice that this is a linear equation. We can use the integrating factor method. The equation can be rewritten as:

$\frac{du}{dt} - u = t.$

The integrating factor is $e^{-t}$, so we multiply both sides of the equation by $e^{-t}$:

$e^{-t} \frac{du}{dt} - e^{-t} u = t e^{-t}.$

This simplifies to:

$\frac{d}{dt} \left( e^{-t} u \right) = t e^{-t}.$

Integrating both sides with respect to $t$:

$e^{-t} u = \int t e^{-t} \, dt.$

Using integration by parts for the integral $\int t e^{-t} \, dt$, we get:

$\int t e^{-t} \, dt = -(t + 1)e^{-t} + C_3.$

Thus:

$e^{-t} u = -(t + 1)e^{-t} + C_3,$

which simplifies to:

$u(t) = -(t + 1) + C_3 e^{t}.$

Step 4: Use the initial condition

The initial condition is given by $u(0, x) = -2x$. At $t = 0$, $x = C_2 e^{-t^2} = C_2$, so we have:

$u(0, C_2) = -2C_2.$

From the solution for $u(t)$, at $t = 0$:

$u(0) = -(0 + 1) + C_3 e^{0} = -1 + C_3.$

Thus, we have:

$-1 + C_3 = -2C_2.$

Solving for $C_3$:

$C_3 = 2C_2 - 1.$

Step 5: Final Solution

Now we can write the general solution:

$u(t, x) = -(t + 1) + (2C_2 - 1)e^{t}.$

From the equation $x = C_2 e^{-t^2}$, we have $C_2 = x e^{t^2}$. Therefore, the final solution is:

$u(t, x) = -(t + 1) + \left(2x e^{t^2} - 1\right) e^{t}.$

Conclusion

Thus, the solution to the PDE is:

$u(t, x) = -(t + 1) + \left(2x e^{t^2} - 1\right) e^{t}.$

Would you like more details or clarification on any part of the solution? Here are some related questions to deepen your understanding:

1. What is the method of characteristics and how does it apply to solving PDEs?
2. Can you explain how the integrating factor method works in solving linear ODEs?
3. How does the initial condition affect the form of the solution in the context of PDEs?
4. What happens if we consider different boundary conditions in this type of PDE?
5. Could you apply a similar method to solve higher-order PDEs?

Tip: Always check the solution against the initial condition to verify its correctness!