Solve the following differential equation: `e^xtany\ dx+(1-e^x)sec^2y\ dy=0`

Explanation

To solve the differential equation

\[ e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0, \]

we can rearrange it into a more manageable form.

### Step 1: Rearranging the Equation
We can rewrite the equation as:

\[ e^x \tan y \, dx = (e^x - 1) \sec^2 y \, dy. \]

### Step 2: Separating Variables
Next, we separate the variables \(x\) and \(y\):

\[ \frac{e^x}{e^x - 1} \, dx = \frac{\sec^2 y}{\tan y} \, dy. \]

### Step 3: Simplifying the Right Side
We know that \(\sec^2 y = 1 + \tan^2 y\), so we can express \(\frac{\sec^2 y}{\tan y}\) as:

\[ \frac{\sec^2 y}{\tan y} = \frac{1}{\tan y} + \tan y. \]

Thus, we can write:

\[ \frac{e^x}{e^x - 1} \, dx = \left(\frac{1}{\tan y} + \tan y\right) dy. \]

### Step 4: Integrating Both Sides
Now we integrate both sides.

1. **Left Side**: Let \(u = e^x - 1\), then \(du = e^x \, dx\), and we can rewrite the left side:

\[ \int \frac{e^x}{e^x - 1} \, dx = \int \frac{1}{u} \, du = \ln |u| + C = \ln |e^x - 1| + C. \]

2. **Right Side**: For the right side, we can split the integral:

\[ \int \left(\frac{1}{\tan y} + \tan y\right) dy = \int \cot y \, dy + \int \tan y \, dy = \ln |\sin y| - \ln |\cos y| + C = \ln |\tan y| + C. \]

### Step 5: Combining Results
Combining the results from both integrations, we have:

\[ \ln |e^x - 1| = \ln |\tan y| + C. \]

### Step 6: Exponentiating Both Sides
Exponentiating both sides gives us:

\[ |e^x - 1| = C |\tan y|. \]

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

\[ e^x - 1 = C \tan y, \]

where \(C\) is a constant.