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The given problem is a first-order partial differential equation:

\[
2u_t + 3u_x = 0
\]

with the auxiliary condition:

\[
u(x, 0) = \sin(\sin(x)).
\]

This is a linear first-order PDE, and we can solve it using the method of characteristics. The general form of a linear first-order PDE is:

\[
a u_t + b u_x = 0,
\]

where \(a = 2\) and \(b = 3\) in this case.

### Step 1: Solve the characteristic equations

The characteristic equations are given by:

\[
\frac{dt}{a} = \frac{dx}{b} = \frac{du}{0}.
\]

Substituting \(a = 2\) and \(b = 3\), we get:

\[
\frac{dt}{2} = \frac{dx}{3} = \frac{du}{0}.
\]

From \(\frac{du}{0}\), we see that \(u\) remains constant along the characteristic curves.

Next, solve for the characteristic curves:

\[
\frac{dt}{2} = \frac{dx}{3} \implies 3 dt = 2 dx \implies x - \frac{3}{2}t = \text{constant}.
\]

Thus, the characteristic curves are given by:

\[
x - \frac{3}{2} t = \xi,
\]

where \(\xi\) is a constant along each characteristic.

### Step 2: General solution

Since \(u\) is constant along each characteristic, the solution must be a function of \(x - \frac{3}{2}t\), i.e.,

\[
u(x,t) = f\left(x - \frac{3}{2}t\right),
\]

where \(f\) is determined by the initial condition.

### Step 3: Apply the initial condition

The initial condition is \(u(x,0) = \sin(\sin(x))\). Using this in the general solution:

\[
u(x,0) = f(x) = \sin(\sin(x)).
\]

Thus, the function \(f(x)\) is:

\[
f(x) = \sin(\sin(x)).
\]

Therefore, the solution to the PDE is:

\[
u(x,t) = \sin\left(\sin\left(x - \frac{3}{2}t\right)\right).
\]

### Final Answer:

The solution for the PDE \(2u_t + 3u_x = 0\) with the auxiliary condition \(u(x,0) = \sin(\sin(x))\) is:

\[
u(x,t) = \sin\left(\sin\left(x - \frac{3}{2}t\right)\right).
\]

Would you like further explanation or have any questions about the method used?

Here are 5 related questions to extend this:

1. How do characteristic curves help in solving first-order PDEs?
2. What is the physical interpretation of characteristic lines?
3. How would the solution change if the coefficients of \(u_t\) and \(u_x\) were different?
4. Can this method be applied to nonlinear first-order PDEs?
5. What happens if a boundary condition is added to this problem?

**Tip:** For solving first-order linear PDEs, the method of characteristics is a powerful tool that reduces the partial differential equation to ordinary differential equations along characteristic curves.

Solve the first-order partial differential equation 2u_t + 3u_x = 0 with the auxiliary condition u(x, 0) = sin(sin(x)).

Learn how to solve the first-order partial differential equation 2u_t + 3u_x = 0 using the method of characteristics, with an initial condition u(x, 0) = sin(sin(x)). This detailed guide covers the characteristic curves and general solution step-by-step, suitable for undergraduate students.

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