`(dy)/(dx) =y ((e^(3x)-e^(-3x))/(e^(3x) +e^(-3x)))`

Answer

To solve the differential equation

\[ \frac{dy}{dx} = y \left( \frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}} \right), \]

we will follow these steps:

### Step 1: Separate the variables

We can rearrange the equation to separate the variables \(y\) and \(x\):

\[ \frac{dy}{y} = \left( \frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}} \right) dx. \]

### Step 2: Integrate both sides

Now we will integrate both sides. The left side becomes:

\[ \int \frac{dy}{y} = \ln |y| + C_1. \]

For the right side, we need to simplify the expression:

\[ \int \left( \frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}} \right) dx. \]

Let \( t = e^{3x} \), then \( dt = 3e^{3x} dx \) or \( dx = \frac{dt}{3t} \). Substituting this into the integral gives:

\[ \int \frac{t - \frac{1}{t}}{t + \frac{1}{t}} \cdot \frac{dt}{3t} = \frac{1}{3} \int \frac{t^2 - 1}{t^2 + 1} \cdot \frac{dt}{t^2}. \]

This simplifies to:

\[ \frac{1}{3} \int \left( 1 - \frac{2}{t^2 + 1} \right) dt = \frac{1}{3} \left( t - 2 \tan^{-1}(t) \right) + C_2. \]

Substituting back \( t = e^{3x} \):

\[ \frac{1}{3} \left( e^{3x} - 2 \tan^{-1}(e^{3x}) \right) + C_2. \]

### Step 3: Combine results

Setting the integrals equal gives:

\[ \ln |y| = \frac{1}{3} \left( e^{3x} - 2 \tan^{-1}(e^{3x}) \right) + C, \]

where \( C = C_2 - C_1 \).

### Step 4: Exponentiate to solve for \(y\)

Exponentiating both sides results in:

\[ y = e^{\frac{1}{3} \left( e^{3x} - 2 \tan^{-1}(e^{3x}) \right) + C} = k e^{\frac{1}{3} \left( e^{3x} - 2 \tan^{-1}(e^{3x}) \right)}, \]

where \( k = e^C \).

### Final Solution

Thus, the solution to the differential equation is:

\[ y = k e^{\frac{1}{3} \left( e^{3x} - 2 \tan^{-1}(e^{3x}) \right)}. \]