`(dy)/(dx)=2y((e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)))`

Options

A. `y=c(e^(2x)-e^(-2x))`
B. `y=c(e^(2x)+e^(-2x))`
C. `y=c(e^(4x)-1)`
D. `x=c(e^(y)+e^(-2y))`

Answer

To solve the differential equation

\[ \frac{dy}{dx} = 2y \left( \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \right), \]

we will follow these steps:

### Step 1: Separate the variables
We can rearrange the equation to separate \(y\) and \(x\):

\[ \frac{dy}{y} = 2 \left( \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \right) dx. \]

### Step 2: Integrate both sides
Now we will integrate both sides. The left side integrates to:

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

and for the right side, we need to integrate:

\[ \int 2 \left( \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \right) dx. \]

Let \(t = e^{2x}\), then \(dt = 2e^{2x} dx\) or \(dx = \frac{dt}{2t}\). Substitute \(t\) into the integral:

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

This simplifies to:

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

Now we can integrate:

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

### Step 3: Combine the results
Combining the results from both integrals, we have:

\[ \ln |y| = t - 2 \tan^{-1}(t) + C, \]

where \(C = C_2 - C_1\).

### Step 4: Solve for \(y\)
Exponentiating both sides gives:

\[ y = e^{t - 2 \tan^{-1}(t) + C} = e^C e^{t - 2 \tan^{-1}(t)}. \]

Let \(K = e^C\), then:

\[ y = K e^{t - 2 \tan^{-1}(t)}. \]

### Step 5: Substitute back for \(t\)
Recall \(t = e^{2x}\):

\[ y = K e^{e^{2x} - 2 \tan^{-1}(e^{2x})}. \]

### Final Solution
Thus, the solution to the differential equation is:

\[ y = K e^{e^{2x} - 2 \tan^{-1}(e^{2x})}. \]