What is the solution of the differential equation `3e^(x) tan y dx + (1 + e^(x)) sec^(2) y dy = 0` ?

Options

A. `(1 + e^(x)) tan y = C `
B. `(1 + e^(x))^(3) tan y = C `
C. `(1 + e^(x))^(2) tan y = C `
D. `(1 + e^(x)) sec^(2) y = C `

Answer

To solve the differential equation \(3e^{x} \tan y \, dx + (1 + e^{x}) \sec^{2} y \, dy = 0\), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate \(dx\) and \(dy\):
\[ 3e^{x} \tan y \, dx = - (1 + e^{x}) \sec^{2} y \, dy \]
This can be rewritten as:
\[ \frac{dx}{dy} = -\frac{(1 + e^{x}) \sec^{2} y}{3e^{x} \tan y} \]

### Step 2: Separating Variables
Next, we separate the variables \(x\) and \(y\):
\[ \frac{3e^{x}}{1 + e^{x}} \, dx = -\frac{\sec^{2} y}{\tan y} \, dy \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{3e^{x}}{1 + e^{x}} \, dx = -\int \frac{\sec^{2} y}{\tan y} \, dy \]

### Step 4: Solving the Left Integral
For the left side, we can use the substitution \(u = 1 + e^{x}\), which gives \(du = e^{x} \, dx\):
\[ \int \frac{3e^{x}}{1 + e^{x}} \, dx = 3 \int \frac{1}{u} \, du = 3 \ln |u| + C = 3 \ln |1 + e^{x}| + C \]

### Step 5: Solving the Right Integral
For the right side, we know that:
\[ -\int \frac{\sec^{2} y}{\tan y} \, dy = -\ln |\tan y| + C \]

### Step 6: Combining the Results
Now we combine both results:
\[ 3 \ln |1 + e^{x}| = -\ln |\tan y| + C \]

### Step 7: Exponentiating Both Sides
Exponentiating both sides gives:
\[ |1 + e^{x}|^3 = \frac{C}{|\tan y|} \]

### Step 8: Final Form
This can be rearranged to express the relationship between \(x\) and \(y\):
\[ C = |1 + e^{x}|^3 |\tan y| \]

### Conclusion
Thus, the general solution of the differential equation is:
\[ C = |1 + e^{x}|^3 |\tan y| \]